Chapter 2 Analysis Methods


 Joan West
 1 years ago
 Views:
Transcription
1 Chapter Analysis Methods. Nodal Analysis Problem.. Two current sources with equal internal resistances feed a load as shown in Fig... I a ¼ 00 A; I b ¼ 00 A; R ¼ 00 X; R L ¼ 00 X: (a) Find the current through the load resistor R L. (b) Find the node voltage value. (a) Parallelconnected current sources, KCL applies, 00 Akð00 AÞ! 00 A By current division, (b) Applying Ohm s law, I RL ¼ 00 00k00 00 þ 00k00 00 ¼ 00 ¼ 50 A: 00 þ 00 V x ¼ 00 X 50 A ¼ 5000 V: Problem.. Find the values of currents and voltages in the circuit shown in Fig.., for R =X. Springer International Publishing AG 07 A.Ü. Keskin, Electrical Circuits in Biomedical Engineering, DOI 0.007/ _ 85
2 86 Analysis Methods Fig.. The circuit for Problem.. Fig.. The circuit for Problem.. Fig..3 The circuit for Problem..3 By Kirchhoff s current law, i 0 þ 5 i 0 ¼ 0! i 0 ¼5A v x ¼ i 0 R ¼5 ¼0 V: Problem..3 (a) Find the value of V 0 in the circuit shown in Fig..3. (b) If the gain constant of dependent source is k, what are the limiting values of k,if I 0 has always a positive value? Resistor value is 4 X. (a) KCL: 3 þ 0:I 0 I 0 ¼ 0! I 0 ¼ 3:75 A V 0 ¼ 4I 0 ¼ 4 3:75 ¼ 5 V (b) 3 þ ki 0 I 0 ¼ 0! I 0 ¼3=ðk Þ A; 0 k\: Problem..4 In the circuit shown in Fig..4, the coefficient of currentcontrolled current source is A/A. If the node voltage is V, find the value of the voltage source, the current through R =X, and current through R =X.
3 . Nodal Analysis 87 Fig..4 The circuit for Problem..4 Applying KCL at the node, and using given component values, V in V x R þ ki V x R ¼ 0 V in V x þ ðv in V x Þ V x ¼ 0! V in þ ðv in Þ ¼ 0 V in ¼ :67 V i ¼ V x ¼ ¼ 0:5A i ¼ V in V x ¼ V in V x ¼ :67 0:5 ¼ 0:667 A Problem..5 Determine the node voltages in the circuit shown in Fig..5. Use Cramer s rule, if necessary. (a) For I ¼ A; I ¼ A; R ¼ = X; R ¼ =8 X; R 3 ¼ =4 X (b) For I ¼ A; I ¼ 4A; R ¼ 5 X; R ¼ X; R 3 ¼ 0 X: (a) For I ¼ A; I ¼ A; R ¼ = X; R ¼ =8 X; R 3 ¼ =4 X Node equations are or V 4ðV V Þ ¼ 0 8V þ 4ðV V Þ ¼ 0 6V 4V ¼ 4V þ V ¼ : Fig..5 The circuit for Problem..5
4 88 Analysis Methods In matrix form, V V ¼ : Applying Cramer s rule to solve this matrix equation yields D ¼ 7 6 ¼ 56; D ¼ þ 8 ¼ 0; D ¼ þ 4 ¼ 6 V ¼ D D ¼ 5 4 ¼ 0:357 V; V ¼ D D ¼ 4 4 ¼ 0:86 V : (b) For I ¼ A; I ¼ 4A; R ¼ 5 X; R ¼ X; R 3 ¼ 0 X: Nodal equations in matrix form can be formed using analysis by inspection ; þ þ 7 5 V ¼ V 4! Applying Cramer s rule to solve this matrix equation yields D ¼ ¼ ¼ 7 00 ; D ¼ 6 0 þ 4 0 ¼ 0 þ 4 0 ¼ 6 0 D ¼ þ ¼ 0 0 þ 0 ¼ V ¼ D D ¼ 0 7 ¼ 60 7 ¼ 9:4 V; V ¼ D D ¼ V ¼ V 4 0 ¼ 40 7 ¼ 8:35 V: Problem..6 Determine the value of current I in the circuit shown in Fig..6. Use Cramer s rule, when necessary R ¼ R ¼ R 3 ¼ R 4 ¼ = X; R 5 ¼ =4 X; I ¼ A; I ¼ A. : Fig..6 The circuit of Problem..6
5 . Nodal Analysis 89 Nodal matrix equation of the circuit is obtained by applying analysis by inspection method, G þ G þ G 3 G 3 G 3 G 3 þ G 4 þ G 5 V V ¼ I I 6 8 V V! þ þ þ þ 4 ¼ : V V ¼ Applying Cramer s rule to solve this matrix equation yields D ¼ 6 8 ¼ 44; D ¼ 8 ¼ ; D ¼ 6 ¼ 4 V ¼ D D ¼ 44 ¼ 0:73 V; V ¼ D D ¼ 4 44 ¼ 0:38 V : I ¼ V V 0:38 0:73 ¼ ¼ 0:09 A R Problem..7 Find the values of voltages at the nodes of the circuit shown in Fig..7. G ¼ 0:5S; G ¼ 4 S; G 3 ¼ 0:4S; G 4 ¼ 5 S; G 5 ¼ S; I ¼ 5A; I ¼ 4A: G A ¼ G þ G ¼ 0:5 þ 0:5 ¼ 0:75 S; G B ¼ G 4 þ G 5 ¼ 0: þ ¼ :S G A þ G 3 G 3 V ¼ I 5 :5 0:4 ¼! G 3 G B þ G 3 V I 4 0:4 :6 V V 5 : ¼ 4 of this matrix equation for unknown voltages yields V ¼ 3:8095 V; V ¼:5476 V: Fig..7 The circuit of Problem..7
6 90 Analysis Methods Fig..8 The circuit of Problem..8 Problem..8 What is the voltage across resistor R 3 (in mv)? Use analysis by inspection and Cramer s rule if necessary (i = i = i 3 =A,R = R 3 = R 5 =Ω, R = ½ Ω, R 4 = R 6 = /6 Ω). Check the result using SPICE (Nodal.cir) (Fig..8). ½G½V ¼ ½; I G þ G þ G 6 G G 6 V þ G G þ G 3 þ G 4 G 4 54 V 5 ¼ ¼ G 6 G 4 G 4 þ G 5 þ G 6 V þ þ 6 6 V þ þ V 5 ¼ 4 0 5! þ þ 6 V D ¼ 053 þð7þþð7þ½34 þ 34 þ 5 ¼ ¼ D ¼ 0 6 ¼ ðþð6þð6þ½ð3þðþðþ ¼ 7 ð5þ ¼ V ¼ D D ¼ 4 ¼ 0:593 V ¼ 593 mv 09 V V V ¼ : SPICE file: *Operating point anaysis Nodal.cir I0 I3 I303 R0 R 0.5 R30 R R530 R Problem..9 Use node voltage method to find the values for the voltage at node C (=V c ) and the current through the resistor R 6 (=i). (R = R = R 3 =Ω, R 4 = R 5 = R 6 =4Ω, i S = A) (Sim_Lin_Eq_Solve.m, matrix_solve.xlsx) (Fig..9).
7 . Nodal Analysis 9 Fig..9 The circuit of Problem..9 3 G þ G4 G4 0 4 G4 G þ G4 þ G5 G5 5 V 3 3 A 5 4 V B 5 ¼ G5 G3 þ G5 þ G6 V C 0 þ þ 4 4 þ V A V B 5 ¼ þ 4 4 þ 5 V C :5 0: :5 :50 0:5 5 V 3 3 A 4 V B 5 ¼ 4 0 5: 0 0:5 :50 V C 0 of this matrix equation by either manually using Cramer s rule or by employing available software (see, MATLAB m file or EXCEL file) yields V A ¼ :657 V; V B ¼ 0:84 V; V C ¼ 0:04734 V; i ¼ V C ¼ 0:04734 ¼ 0:0834 A: R 6 4 Problem..0 (a) In the circuit shown in Fig..0, find the voltage gain, i.e., V /V =? (b) If R =.5 kx, R o =0kX, R L =0kX, k = 50, find the numerical value of (V /V ). Check the result using SPICE (cccs8.cir). Fig..0 The circuit of Problem..0
8 9 Analysis Methods (a) I ¼ V R ; V ¼k I ðr o ==R L Þ¼k V R ðr o ==R L Þ V ¼k R o==r L ¼ k R o==r L R V R L R o þ R L ðbþ V ¼50 0==0 5 ¼50 ¼00 V=V V :5 :5 SPICE netlist (cccr8.cir); *DC Operating point analysis *current controlled current sourcenodal analysis *fx N+ N Vy Value *Parameters: *x Name of the source *N+ : Name of positive node *N : Name of negative node. Current flows from the + node * through the source to the  node *Vr : Name of the voltage source where the controlling current flows. * The direction of positive control current is * from + node through the source to the  node of Vr=0 *Value: Current gain v0m f0vr50 Vr30 R 3 0.5k R0 0 0k RL 0 0k Problem.. In the circuit shown in Fig.., f ¼ ; R ¼ R ¼ R 3 ¼ X; i S ¼ A: V ¼?; V ¼?; i ¼? (Use node voltages method.) Check the result using SPICE (cccs7.cir). Fig.. The circuit of Problem..
9 . Nodal Analysis 93 i S V R V V R 3 ¼ 0 V V V þ ð V V Þ ¼ 0: R R 3 R Substituting given values of components into these equation yields V þ V ¼ 0 3V 4V ¼ 0 : Simplifying, V V ¼ 3V 4V ¼ 0 : Solving this simultaneous set of linear equations for unknown voltage values yields V ¼ 0:8V; V ¼ 0:6V; i ¼ V V ¼ 0:8 0:6 ¼ 0:A : R SPICE Netlist,cccs7.cir Analysis: DC Operating Point *fx N+ N Vy Val * x Name of the source, N+ : positive node *N : Name of negative node. Current flows from the + node * through the source to the  node *Vy : Name of the voltage source *The direction of positive control current is *from + node through the source to the  node of Vy=0 *Val: Current gain i0 f0vy Vy30 R3 R0 R30 Problem.. In the circuit shown in Fig.., I S =A,R = R = R 3 = R 4 = Ω, f = 4 A/A. Find the values of currents flowing through resistors R 3 and R 4.
10 94 Analysis Methods Fig.. The circuit of Problem.. KCL at nodes and : i ¼ V R 3 ¼ V ¼ V : 4V V ðv V Þ ¼ 0 ð:þ 4V þ ðv V ÞV V ¼ ð0þ: ð:þ Simplifying (.) and (.), V þ 3V ¼ V þ V ¼ 0: ð:3þ ð:4þ Solving these equations for unknowns yields V ¼V; V ¼ V; i 4 ¼ V ¼ A; i 3 ¼ V ¼ A: R 4 R 3 Problem..3 In the circuit shown in Fig..3, R = R = R 3 = R 4 =Ω, Is = A, and f = 4 A/A. V =?,V =?,V 3 =?,i =? Use node voltages method. Check your results using SPICE (cccs4). KCL at nodes and with i = V /R 3 : V R f V R 3 V V R ¼ 0 ð:5þ
11 . Nodal Analysis 95 Fig..3 The circuit of Problem..3 f V R 3 þ V V R V R 3 ¼: ð:6þ Using given values, these equations become V 4V V þ V ¼ 0 4V þ V V V ¼: ð:7þ ð:8þ Simplifying, V þ 3V ¼ 0 V þ V ¼: ð:9þ ð:0þ of this set of simultaneous equations yields V =3V,V = V. V R4 ¼ ðaþr 4 ¼ ðþðþ¼v V 3 ¼ V V R4 ¼ðÞðÞ ¼ þ ¼V Netlist for SPICE check: i ¼ V R 3 ¼ ¼A *cccs4.cir *Analysis: DC Operating Point current controlled current sourcenodal analysis i03 f vref 4 vref R0 R R34 R43
12 96 Analysis Methods Fig..4 The circuit of Problem..4 Problem..4 In the circuit shown in Fig..4, R = R = R 3 =Ω, I S =A, f = 4 A/A. V =?,V =?,i =?,i R =? Use nodal analysis method. i s f V R 3 V R V V R ¼ 0 ð:þ f V R 3 þ V V R V R 3 ¼ 0: ð:þ Substituting the values, 4V V V þ V ¼ 0 ð:3þ 4V þ V V V ¼ 0: ð:4þ Simplifying, V 3V ¼ 0 ð:5þ V þ V ¼ 0: ð:6þ In matrix form, ¼ 3 0 V V : ð:7þ of this matrix equation yields
13 . Nodal Analysis 97 V ¼ V; V ¼V; i ¼ V R 3 ¼ ¼A i R ¼ V V ¼ ð Þ ¼ 3A: R Problem..5 (a) Determine the voltagetocurrent ratio (the input resistance) in the circuit shown in Fig..5: R x ¼ V x I x ¼? (b) Determine the value of current through the voltage source, if R ¼ R ¼ X; R 3 ¼ X; R 4 ¼ X; V x ¼ V; k ¼ 0:5S: 4 ðaþ ðr þ R ÞR 3 V ¼ I x R þ R þ R 3 V x ¼ V V : ð:8þ ð:9þ Node : I x kv V R 4 ¼ 0 V ¼R 4 I x kv R 4 ð:0þ ð:þ Equation (.)! (.9) Fig..5 The circuit of Problem..5
14 98 Analysis Methods V x ¼ V þ R 4 I x þ kv R 4 ¼ V ð þ kr 4 ÞþR 4 I x ð:þ divide all terms of (.) byi x V x ¼ R x ¼ V ð þ kr 4 ÞþR 4 : ð:3þ I x I x Replace V =I x by (.8): R x ¼ R ð þ R ÞR 3 ð þ kr 4 ÞþR 4 : ð:4þ R þ R þ R 3 (b) Substituting given component values into (.4) yields R x ¼ ð þ Þ þ 0:5 0:5 þ þ ð Þþ0:5 ¼ :75 X I x ¼ V x ¼ 0:5 R x :75 ¼ A ¼ 0:4857 A: 7 Problem..6 (a) Use node voltage method and find the voltage drop across R (in mv). (b) Verify the solution using SPICE and print SPICE netlist (vccs.cir). (I s =A,R =Ω, R = / Ω, R 3 = /4 Ω) (Fig..6). KCL at : KCL at : V ðv V Þ ¼ 0 V 4V þ ðv V Þ ¼ 0 Fig..6 The circuit of Problem..6
15 . Nodal Analysis 99 3V þ V ¼ V þ V 4V V ¼ 0 3V V ¼ 3V V ¼ 0: ð:5þ ð:6þ From (.6) 3V +V! put into (.5) V V ¼! V ¼ 0:V 3V ¼ :! V ¼ 0:4V V V ¼ 0:4 0: ¼ 0:3V¼ 300 mv: (b) SPICE netlist: Operating point analysis, vccs.cir *gx N+ N NC+ NC VALUE *x Name of the source *N+ Name of positive node *N Name of negative node *NC+ Name of positive controlling node *NC Name of negative controlling node *VALUE Transconductance in S I 0 G *Specifies that the current through G flowing from node to ground *is 0.5 times the potential difference between node and ground. R 0 R 0.5 R Problem..7 A R R ladder circuit is shown in Fig..7. (a) Find the node voltages and shunt branch currents. (b) Find the current supplied by the voltage source. (c) Compute numerical values if V i =5V,R =kω. Fig..7 The circuit for Problem..7
16 00 Analysis Methods Start at the rightmost node of the circuit (node f), looking to the right of each node, R f ¼ R==R ¼ R; R e ¼ R== ðr þ RÞ ¼ R; R d ¼ R== ðr þ RÞ ¼ R R c ¼ R== ðr þ RÞ ¼ R; R b ¼ R== ðr þ RÞ ¼ R: Node voltages: V a ¼ V V b ¼ R b V ¼ R V R þ R b R ¼ V V V c ¼ R c V R þ R c ¼ R R V ¼ V 4 V V d ¼ R d V R þ R d 4 ¼ R R U 4 ¼ V 8 V : V e ¼ R e V R þ R e 8 ¼ R R V 8 ¼ V 6 V V f ¼ R f V R þ R f 6 ¼ R R V 6 ¼ V 3 V The shunt branch currents are calculated by Ohm s law: The right part branch departing from (f) carries a current of V I o ¼ V f R ¼ 3 R ¼ V 64R : This is the same current through the left branch departing from (f). The shunt branch current departing from node (e) is V I eo ¼ V e R ¼ 6 R ¼ V 3 ¼ I o: The shunt branch current departing from node (d) is V I do ¼ V d R ¼ 8 R ¼ V 6R ¼ I ð oþ ¼ 4I o : The shunt branch current departing from node (c) is Similarly, V I co ¼ V c R ¼ 4 R ¼ V 8R ¼ 8I o:
17 . Nodal Analysis 0 V I bo ¼ V b R ¼ R ¼ V 4R ¼ 6I o: Overall current supplied by the voltage source is I ¼ I bo þ I co þ I do þ I eo þ I fo þ I fo ¼ I o ð6 þ 8 þ 4 þ þ þ Þ ¼ 3 I o V I ¼ 3 64R ¼ V R V ¼ 5V; R ¼ kx; V b ¼ :5V; V c ¼ :5 V; V d ¼ 0:65 V; V e ¼ 0:35 V; V f ¼ 0:565 V; : I o ¼ V 64R ¼ 5 ma ¼ 0:0785 ma 64 I ¼ V R ¼ 5 ma ¼ :5mA Problem..8 In the circuit shown in Fig..8, find the voltage at node (=V ). Use node voltage method and Cramer s rule for the solution of matrix equations. R ¼ R ¼ R 3 ¼ R 4 ¼ kx; I ¼ I ¼ I 3 ¼ I 4 ¼ ma: ½¼ I ½G½V; G ¼ R G ¼ G ¼ G 3 ¼ G 4 ¼ 0 3 S I 3 G þ G G 0 V I ði 3 þ I 4 Þ5 ¼ 4 G G þ G 3 G V I þ I 4 0 G 3 G 3 þ G 4 V þ 0 4 ð þ Þ5 ¼ 4 þ 5 V V 4 V 5! 4 5 ¼ V 5: þ 0 þ V 3 0 V 3 Using Cramer s rule for the solution of matrix equation, Fig..8 The circuit for Problem..8
18 0 Analysis Methods 0. V ¼ D D ¼ 4 þ þ 0 ð0þþþ 3 ¼ 0 8 þ 0 þ 0 ð0þþþ ¼ ¼ 0:75 V: Problem..9 In the circuit shown in Fig..9, find the values of node voltages V and V. Use Cramer s rule when necessary R ¼ R 3 ¼ X; R ¼ 4 X; I ¼ I 3 ¼ A; I ¼ I 4 ¼ I 5 ¼ A: V þ V V 5 ¼ 0! V 4 V ¼ 5 V V þ V 4 þ ¼ 0! 4 V þ 3 4 V ¼ V 5 ¼ V D ¼ 3 ¼ ; D 4 ¼ ¼ ; D ¼ ¼ 4 V ¼ D D ¼ 7V; V ¼ D D ¼ V: Problem..0 In the circuit shown in Fig..0, I =A, I = / A, R = / Ω, R = /4 Ω, R 3 = /8 Ω. Fig..9 The circuit for Problem..9
19 . Nodal Analysis 03 Fig..0 The circuit for Problem..0 (a) Find the node voltages, (b) Find the currents flowing in the circuit (Sim_Lin_Eq_Solve.m, matrix_solve. xlsx). ðaþ V R þ V V R þ I I ¼ 0 ð:7þ V V þ V I ¼ 0 R R 3 V þ V ¼ I I R R R V þ V þ ¼ I : R R R 3 ð:8þ ð:9þ ð:30þ Using last two equations, þ 3 6 R R R 4 þ R R R V V V V ¼ I I I ¼ 0:5 0:5 : of this set of simultaneous linear equations by either manually using Cramer s rule or by substitution methods or by employing available software (see, MATLAB m file or EXCEL file referenced in the statement) yields V ¼ 0:4 V; V ¼ 0:09 V:
20 04 Analysis Methods ðbþ i ¼ V R ¼ 0:8 A; i ¼ V V R ¼ 0:A; i 3 ¼ V R 3 ¼ 0:7 A: Problem.. Use node voltages method and find the values of currents and voltages in the circuit shown in Fig... R ¼ X; R ¼ 4 X; R 3 ¼ R 4 ¼ X; I ¼ I ¼ A; I 3 ¼ A: Applying KCL at node, I þ I i i ¼ 0! þ v R v v R ¼ 0 v v v 4 ¼ 0! 6v þ 4v ¼ 0 6v þ 4v ¼: ð:3þ Applying KCL at node, i i 3 i 4 ¼ 0! v v v v v 3 ¼ 0! R R 3 R 4 v v v v v 3 ¼ 0 4 4v 6v þ v 3 ¼ 0: ð:3þ Applying KCL at node 3, I 3 I þ i 4 ¼ 0! þ v v 3 ¼ 0! þ v v 3 ¼ 0 R 4 v v 3 ¼: ð:33þ Fig.. The circuit for Problem..
21 . Nodal Analysis 05 Combining Eqs. (.3) (.33) into matrix form, v v 5 ¼ 4 0 5: 0 v 3 The solution of this matrix equation yields v ¼ V; v ¼ V; v 3 ¼ V i ¼ v ¼ A; i ¼ v v ¼ 0A; i 3 ¼ v ¼ A; i 4 ¼ v v 3 ¼A: R R R 3 R 4 Problem.. Determine the node voltages in the circuit shown in Fig... R ¼ X; I ¼ I ¼ I 3 ¼ A: Analysis by inspection, G ¼ R ¼ S 3 G þ G G 0 4 G Gþ G þ G G 5 V 3 3 I þ I þ I 3 4 V 5 ¼ 4 I 5! 0 G Gþ G V 3 I V V 5 ¼ 4 5: 0 V 3 This matrix equation is solved for unknown voltages and yields the following voltage values: V ¼ :5V; V ¼ 0V; V 3 ¼0:5V: Problem..3 (a) Use node voltages and Cramer s methods to find the values of currents and voltages in the circuit shown in Fig..3. Use SPICE for checking the results. Print the SPICE netlist (cccs5.cir). Fig.. The circuit for Problem..
22 06 Analysis Methods Fig..3 The circuit for Problem..3 R ¼ R 4 ¼ X; R ¼ R 3 ¼ X; I ¼ A; f ¼ 3A=A: (b) Determine the node voltages using the following component values. Use SPICE for checking the results. Print the new SPICE netlist. R ¼ X; R ¼ 4 X; R 3 ¼ 8 X; R 4 ¼ 4 X I ¼ 3A; f ¼ A=A: (a) Applying KCL at node, I i i x ¼ 0! v v 3 v v ¼ 0! v v 3 R 4 R ðv v 3 Þðv v Þ ¼ 0 v v ¼ 0 4v þ v þ v 3 ¼: ð:34þ Applying KCL at node, i x i i 3 ¼ 0! v v v v v 3 ¼ 0! v v v R R R 3 v v 3 ¼ 0 ðv v Þv ðv v 3 Þ ¼ 0 v 4v þ v 3 ¼ 0: ð:35þ Applying KCL at node 3, 3i x þ i þ i 3 ¼ 0!3 v v R þ v v 3 R 4 þ v v 3 R 3 ¼ 0
23 . Nodal Analysis 07 3 v v þ v v 3 þ v v 3 ¼ 0!3v ½ ð v Þþðv v 3 Þþðv v 3 Þ ¼ 0 4v þ 7v 3v 3 ¼ 0: ð:36þ Combining Eqs. (.34) (.36) into matrix form, v v 5 ¼ 4 0 5: v 3 0 The solution of this matrix equation yields v ¼ 0:5V; v ¼ 0:V; v 3 ¼0:V i ¼ v v 3 R 4 ¼ :4A; i ¼ v R ¼ 0:A; i 3 ¼ v v 3 ¼ 0:4A; i x ¼ v v ¼ 0:6A: R 3 R *SPICE Netlist cccs5.cir *Analysis: DC Operating Point current controlled current sourcenodal analysis 5 i0 f 3 0 vref 3 vref 4 0 R R0 R33 R (b) Using new data set, KCL at node, ¼ 5v 4v v 3 KCL at node, 0 ¼ 8v v þ v 3 KCL at node 3, 0 ¼ 4v 7v þ 3v 3 : Using three nodal equations, one obtains the following matrix equation of the circuit:
24 08 Analysis Methods v v 5 ¼ v D ¼ ¼58; D ¼ ¼9 5 D ¼ ¼0; 5 4 D 3 ¼ ¼ 6 v ¼ D D ffi 3:3V; v ¼ D D ffi :06 V; v 3 ¼ D 3 ffi3:7 V: D Renewed SPICE Netlist; * DC operating point analysis cccs5.cir i03 f 3 0 vref vref 4 0 R4 R04 R338 R434 Advantages of using SPICE are apparent here. It can be easily used for many different component variations of a circuit, rather than performing tedious calculations. Problem..4 Determine the ratio of node voltages V =V in the circuit shown in Fig..4. Use Cramer s rule when necessary. Fig..4 The circuit for Problem..4
25 . Nodal Analysis 09 I ¼ A; R ¼ 0 X; R ¼ X; R 3 ¼ 5 X; k ¼ 5A=A; k ¼ A=V: KCL at node : I i 5i i R ¼ I V 5 V V V ¼ I 6 V V þ V ¼ 0 R R R R R R KCL at node : I 7V R þ R þ V R ¼ 0: 5i þ i R i R3 þ ðv V Þ ¼ 5 V þ V V V ðv V Þ ¼ 0 R R R 3 ð:37þ 5 V R þ V R V R V R 3 þ V V ¼ 0: ð:38þ From Eqs. (.37) and (.38), 7 þ 3 R R R 6 5 þ þ þ V ¼ V þ 0 R R R R 3 R ¼ 0 X; R ¼ X; R 3 ¼ 5 X 70: ð þ Þ 0:5 þ þ ð þ 0: þ Þ 7:7 V ¼ 3:5 3: 0 V V V ¼ 0 D V ¼ D D ; V ¼ D D ; V ¼ D ¼ D V D D D D ¼ 0 3: ¼3:; D ¼ 7:7 3:5 0 ¼3:5 V ¼ 3: ¼ 0:943 V=V V 3:5 ð Þ:
26 0 Analysis Methods Fig..5 The circuit for Problem..5 Problem..5 Determine currents flowing through each resistor in the circuit shown in Fig..5 (ladder_node.xlsx). R ¼ R ¼ R 3 ¼ R 4 ¼ R 5 ¼ 0 X; R 6 ¼ R 7 ¼ 5 X; R 8 ¼ R 9 ¼ 0 X; I ¼ I ¼ A: where T stands for transpose operation. ½G½V ¼ ½¼ I ½000 T ð:39þ G ¼ 0 þ 5 ; G ¼ 0 þ 5 ; G 33 ¼ 0 þ 5 þ 0 ; G 44 ¼ 0 þ 0 ; G 55 ¼ 0 þ 0 ; G ¼ G ¼ 5 ; G 3 ¼ G 3 ¼ 5 ; G 43 ¼ G 34 ¼ 0 ; ½G ¼ 6 4 G 45 ¼ G 54 ¼ 0 ; 0:3 0: : 0:5 0: : 0:35 0: :05 0: 0:05 5 : :05 0:5 of Eq. (.39) using these numerical values yields the node voltages: V ¼ 0:545 V; V ¼ 5:88 V; V 3 ¼ 4V; V 4 ¼ 4:77 V; V 5 ¼ 4:909 V: 3 Current values through resistors are obtained using Ohm s Law:
27 . Nodal Analysis i ¼ V ¼ : A; i ¼ V ¼ 0:5888 A R R i 3 ¼ V 3 ¼ 0:4A; i 4 ¼ V 4 ¼ 0:4777 A R 3 R 4 i 5 ¼ V 5 ¼ : A; i 6 ¼ V V ¼ 0: A : R 5 R 6 i 7 ¼ V V 3 ¼ 0: A; i 8 ¼ V 3 V 4 ¼0:03636 A; R 7 R 8 i 9 ¼ V 4 V 5 ¼0:50909 A R 9 Problem..6 Find the node voltage values for the circuit shown in Fig..6. I = I 3 =A,I =A,I 4 = A,R = R = R 3 = R 5 = R 6 = R 7 = R 8 =Ω, R 9 = 0. Ω (Sim_Lin_Eq_Solve.m, matrix_solve.xlsx). The conductance values are G ¼ G ¼ G3 ¼ G4 ¼ G5 ¼ G6 ¼ G7 ¼ G8 ¼ S; G9 ¼ 0 S ½G½V ¼½I: of this matrix equation in MATLAB or EXCEL platform yields V ¼ :795 V; V ¼ 0:590 V; V 3 ¼:08 V; V 4 ¼0:04 V; V 5 ¼0:58 V; V 6 ¼0:39 V: Problem..7 (a) Find the node voltage values in terms of current gain of the CCCS for the circuit shown in Fig..7. (b) Verify the solution using SPICE and print SPICE netlist (cccs.cir). I S ¼ A; R ¼ R ¼ R 3 ¼ X: Fig..6 The circuit for Problem..6
28 Analysis Methods Fig..7 The circuit for Problem..7 KCL at node : (G = G = G 3 =S) I S V ðv V Þfi ¼ 0! V V þ V f V ¼ 0 KCL at node : From (.4), V þ V ð f Þ ¼ 0 V ð f ÞV ¼ : ð:40þ fi þ ðv V Þi ¼ 0! fv þ V V V ¼ 0 fv þ V V ¼ 0 V þ V ðf Þ ¼ 0: ð:4þ V ¼ V ðf Þ ¼ ð f ÞV : ð:4þ
29 . Nodal Analysis 3 Table. Circuit voltages and current as a function of current gain f V (V) V (V) i (A) Substitute into (.40): From (.4), V ð f Þð f ÞV ¼! V ½ ð f Þð f Þ ¼ V ð4 f þ f Þ ¼! V ð3 f Þ ¼ V ¼ 3 f : ð:43þ V ¼ f 3 f : ð:44þ Note that f 6¼ 3. As a check in SPICE, Table. displays the results. Note that f = 3 yields a SPICE error message. The circuit used in SPICE including the CCCS is shown in Fig..7b. SPICE Netlist (cccs.cir): *fx N+ N Vy Value *x Name of the source *N+ : Name of positive node *N : Name of negative node. Current flows from the + node * through the source to the  node *Vref : Name of the voltage source through the controlling *current flows. * The direction of positive control current is * from + node through the source to the  node of Vref=0 *Value: Current gain i0 f Vref 4 Vref R0 R R33 *.op
30 4 Analysis Methods Fig..8 The circuit for Problem..8 Problem..8 Find the node voltage values in the circuit shown in Fig..8. All resistors are Ω and I =4A,I =A,I 3 =A,I 4 = 4 A (Sim_Lin_Eq_Solve. m, matrix_solve.xlsx). I ¼ V V R þ V V 3 R! V V V 3 ¼ 4 I ¼ V V R þ V V 4 R 3! V þ V V 4 ¼ I þ I 3 þ V 3 R 5 þ V 3 V 4 R 4 þ V 3 V R ¼ 0! V þ 3V 3 V 4 ¼5 I þ I 4 þ V 4 R 6 þ V 4 V R 3 þ V 4 V 3 R 4 ¼ 0! V V 3 þ 3V 4 ¼5: From these four equations, the following matrix equation is obtained: ½G½V ¼ ½ I! V V V 3 V ¼ of this matrix equation by employing any available software yields the voltage values as V ¼ :77 V; V ¼ :73 V; V 3 ¼ 0:8 V; V 4 ¼:8 V: 3 7 5:
31 . Nodal Analysis 5 Fig..9 The circuit for Problem..9 Problem..9 In the circuit shown in Fig..9, R ¼ R ¼ R 3 ¼ R 4 ¼ R 5 ¼ R 6 ¼ R 7 ¼ R 8 ¼ kx; R 9 ¼ 00 X; I ¼ I ¼ I 3 ¼ I 4 ¼ ma: (a) Find the conductance matrix for the circuit. (b) Compute the node voltages (Sim_Lin_Eq_Solve.m, matrix_solve.xlsx). (a) ½G½V ¼ ½ I G ¼ G ¼ G 3 ¼ G 4 ¼ G 5 ¼ G 6 ¼ G 7 ¼ G 8 ¼ ms; G 9 ¼ 0 ms: Then, the conductance matrix is ½G ¼ S ð¼ msþ G 7 þ G 8 þ G 9 ¼ 0 3 þ 0 3 þ 0 ¼ 0 3 þ 0 S ¼ 0 3 S: (b) I ¼ ½ 0 0 T ma, (T: transpose operator), ½G½V ¼ ½: I of this matrix equation for voltage vector (e.g., using MATLAB or EXCEL) yields ½V ¼ ½385:5 77: 6:0 97:7 006:0 84:3 T mv:
32 6 Analysis Methods Fig..30 The circuit for Problem..30 Problem..30 Find the node voltage values in the circuit shown in Fig..30. R =R6 =R7 =X, R =R3 =R5 =X, R4 =4X, I =A, I =A, I3 = 3 A (Sim_Lin_Eq_Solve.m, matrix_solve.xlsx). Analysis by inspection, G ¼ S; G ¼ 0:5S; G 3 ¼ 0:5S; G 4 ¼ 0:5 S; G 5 ¼ 0:5S; ½G½V ¼ ½; I G 6 ¼ S; G 7 ¼ S 3 3 :5 0:5 0 0 v 0:5 :5 0: v :5 :75 54 v 3 5 ¼ v 4 3 þ þ : of this matrix equation (e.g., using EXCEL or MATLAB tools) gives node voltages: v ¼ 3:806 V; v ¼ :49 V; v 3 ¼0:56 V; v 4 ¼:58 V: Problem..3 In the circuit shown in Fig..3a, R = R = R 3 =Ω, i S =A. (a) Use SPICE to find the values of the node voltages V, V, and the current i for currentcontrolled current source constants of f =,f = 4, and f =8. (b) Plot i versus i S curve, i S A, if the CCCS constant is 8 A/A. Include net list. (a) The circuit used in SPICE analysis is shown in Fig..3b: (Table.) (b) Fig..3c displays the current sweep.
33 . Nodal Analysis 7 Fig..3 a The circuit for Problem..3, b The circuit for Problem..3 for SPICE analysis, c SPICE analysis result for the Circuit of Problem..3. The current sweep (a) (b) (c) 00.0m 'i(vref)' 50.0m i [A] 0.0m 50.0m 00.0m is sweep [A] Table. Voltage and current values against current gain f V (V) V (V) i (A) *SPICE Netlist for current sweep with f=8: *Analysis: DC Transfer Curves cccs is0dc R3 R0 R30 f 0 vref 8 vref 3 0.dc is .
34 8 Analysis Methods Problem..3 Use node voltages method and determine all currents (ma) and V ðmvþ in the circuit shown in Fig..3. V ¼ V; V 3 ¼ V; R ¼ 5 X; R ¼ 3 X; R 3 ¼ 4 X; R 4 ¼ X: KCL at node : i ¼ V ¼ V; V 3 ¼ V; i 3 ¼ V V 3 ¼ ¼ 500 ma: R 4 i i i 4 ¼ 0! V V V ¼ 0! ð V Þ0ðV Þ5V ¼ 0 0: V ¼ 44 ¼ 0:9367 V ¼ 936:7 mv 47 ¼ :766 ma; i ¼ 0: ¼:77 ma i 4 ¼ 0:9367 ¼ 34:043 ma; i V ¼ i þ i 3 ¼ :766 þ 500 ¼ 7:766 ma 4 i V ¼i ð þ i 3 Þ ¼ ð:77 þ 500Þ ¼ 478:73 ma: Problem..33 Determine the node voltages in the circuit shown in Fig..33. V ¼ V; V ¼ 6V; R ¼ 4 X; R ¼ X; R 3 ¼ X; R 4 ¼ 6 X: There is a voltage source ðv Þ connected between two nonreference nodes (,3). These nodes form a supernode. KCL and KVL can be applied to obtain the node voltages in this circuit. Fig..3 The circuit for Problem..3
35 . Nodal Analysis 9 Fig..33 The circuit for Problem..33 On the other hand, V is connected between node and ground. Thus, the voltage at node equals to v ¼ V ¼ V: At the supernode, i þ i 4 i i 3 ¼ 0 v v R þ v v 3 R 4 v R v 3 R 3 ¼ 0: ð:45þ ð:46þ But constraint equation is v v 4 v v 3 ¼ V ¼ 6V! v ¼ V þ v 3 ¼ 6 þ v 3 ð:47þ þ v v 3 6 Since v ¼ V ¼ V; ¼ ð v þ v 3 Þ ¼ ð 6 þ v 3 þ v 3 Þ ¼ 6 þ v 3 : ð:48þ ð6 þ v 3 Þ 4 þ v 3 6 ¼ 6 þ v 3 ð:49þ v 3 þ 4 v 3 ¼ 36 þ v 3 ð:50þ 7v 3 ¼8! v 3 ¼ 8 7 V: From (.47), v ¼ 6 þ 8 7 ¼ 0 7 V: Summary: v ¼ V; v ¼ 6:47 V; v 3 ¼ 0:47 V: Problem..34 Determine the voltage at node and the current flowing through the voltage source in the circuit shown in Fig..34. Prove the latter result by applying KCL at node 3. I ¼ A; V ¼ V; R ¼ 4 X; R ¼ 8 X; R 3 ¼ 8 X; R 4 ¼ X (supernode.cir).
36 0 Analysis Methods Fig..34 The circuit for Problem..34 Since independent voltage source is connected between nodes (, 3), these nodes form a supernode. Node is included in this supernode. Thus, i i þ 5 ¼ 0! v R v 3 R 4 þ ¼ 0! v 4 v 3 þ ¼ 0 v þ v 3 ¼ 8: ð:5þ The constraint is Substituting (.5) into (.5), From (.5) At node : v 3 v ¼ V! v 3 ¼ þ v : ð:5þ v þ þ ð v Þ ¼ 8! v þ 4 þ v ¼ 8! 3v ¼ 4 v ¼ 4 ¼ :3333 V: 3 v 3 ¼ þ :3333 ¼ 3:3333 V: i 3 þ I i 4 ¼ 0! v v R þ I v v 3 R 3 ¼ 0: ð:53þ Substituting the values for v and v into (.53), :3333 v 8 þ v 3: ¼ 0! 4:6667 v ¼6
37 . Nodal Analysis 6 4:6667 v ¼! v ¼ 0:6667 ¼ 0:3333 V i 3 ¼ v v :3333 0:3333 ¼ ¼:5 A: R 4 The current flowing through the voltage source is calculated by applying KCL at node : i i 5 i 3 ¼ 0! v R i 5 v v R ¼ 0 :3333 :3333 0:3333 i 5 ¼ 0! 0:3333 i 5 þ :5 ¼ i 5 ¼ 0:7967 A: Proof KCL at node 3, i 5 þ i 4 i ¼ 0! 0:7967 þ v v 3 R 3 v 3 R 4 ¼ 0 0:7967 þ 0:3333 3:3333 3:3333 ¼ 0 8 0:7967 þ 7 3:3333 ¼ 0! 0:7967 þ 0:875 :66667 ¼ 0 Q:E:D: 8 SPICE netlist: supernode *OP analysis R04 R8 R338 R440 V3 I0 V3340 Problem..35 Determine the current through dependent source and current though independent voltage source in the circuit shown in Fig..35. Here, v is the node voltage at node.
38 Analysis Methods Fig..35 The circuit for Problem..35 I ¼ 4A; V ¼ 5V; R ¼ R 4 ¼ X; R ¼ 4X; R 3 ¼ X: Consider the supernode consisting of nodes (,). Applying KCL, I v R v v 3 R 4 v v 3 R 3 v R ¼ 0 4 v v v 3 v v 3 v 4 ¼ 0 6 v ðv v 3 Þ4ðv v 3 Þv ¼ 0 4v þ 5v 6v 3 ¼ 6 ð:54þ But, v ¼ v þ 4v ¼ 5v v 3 ¼ V ¼ 5V: ð:55þ ð:56þ Substituting (.55), (.56), into (.54), 45v ð Þþ5v 6 5 ¼ 6! 0v þ 5v 30 ¼ 6! v ¼ :84 V: From (.55), KCL at node : i ¼ I v R v v 3 v ¼ 5 :84 ¼ 9:V: ¼ 4 9: R 4 9: 5 ¼:7A:
39 . Nodal Analysis 3 KCL at node 3: i 30 ¼ v v 3 þ v v 3 :84 5 9: 5 ¼ þ ¼:06 A: R 3 R 4 Following is the SPICE netlist (supernode.cir) for the operating point analysis of this circuit: supernode *OP analysis R0 R04 R33 R43 V305 I04 *VCVS: e{name} {+node} {node} {+cntrl} {cntrl} {gain} e04. Mesh Analysis Problem.. Find the values of V x ; V 0 in the circuit shown in Fig..36. U =35V. KVL around the loop, U V x V x þ V 0 ¼ 0 V x ¼ 0i; V 0 ¼5i 35 0i 0i 5i ¼ 0 i ¼ A; V 0 ¼5i¼5V; V x ¼ 0i ¼ 0 V: SPICE netlist (mesh0): Fig..36 The circuit for Problem..
40 4 Analysis Methods mesh0 *OP ANALYSIS VU 335 R 0 R 305 *VCVS: Ex N+ N NC+ NC VALUE E 0 Problem.. (a) Determine the current i ab in the circuit shown in Fig..37. (b) If U = 0 V, U =6V,R =kxwhat is the value of i ab? (ma) (c) If U =U = 0 V, R =kxwhat is the value of i ab? (ma) (d) If U =U/ = 0 V, R =kxwhat is the value of i ab? (ma) (a) Current through the left mesh (in CW direction), i ¼ U R Current through the right mesh, (in CW direction) (b) (c) If U =U = 0 V, i ¼ U R i ab ¼ i þ i ¼ U U R R ¼ R i ab ¼ U U ¼ ¼ 7mA i ab ¼ ¼ 5mA Fig..37 The circuit for Problem..
41 . Mesh Analysis 5 (d) If U = U/ = 0 V,R = kx, i ab ¼ ¼ 0mA: Problem..3 In the circuit shown in Fig..38, use mesh currents method and find the value of voltage V x. What is the voltage drop across R?. R ¼ R 3 ¼ R ¼ 4 X; V ¼ 3V; V ¼ 5 V (Sim_Lin_Eq_Solve.m, matrix_solve.xlsx). By applying mesh currents and analysis by inspection, the governing equation of the circuit is R þ R R R R þ R 3 i i ¼ V V V or 6 6 i i ¼ 5 : From this matrix equation, i and i can be obtained as i ¼0:065 A; i ¼ 0:85 A: The voltage V x is calculated as V x ¼ i R 3 ¼ 0:85 4 ¼ 3:5 V: The voltage drop across R, V R ¼ V x V ¼ 3:5 5 ¼:75 V: Problem..4 Apply mesh analysis method to find the values of currents i and i and the node voltage in the circuit shown in Fig..39 (V =V, V =V, R =Ω, R =Ω, R 3 =Ω) (Sim_Lin_Eq_Solve.m, matrix_solve.xlsx). Using mesh analysis for the circuit, V þ i R þ ði i ÞR 3 ¼ 0 Fig..38 The circuit for Problem..3
42 6 Analysis Methods Fig..39 The circuit for Problems..4 and..5 ði i ÞR 3 þ i R þ V ¼ 0: Substituting V, V, R, and R values in these equations, þ 3i i ¼ 0 3i i ¼ 0: of this set of simultaneous linear equations for unknown current values yields i ¼ 0:8A; i ¼ 0:A V x ¼ ði i Þ:R 3 ¼ 0:6 ¼ :V: Problem..5 In the circuit shown in Fig..39, use mesh currents method and Cramer s rule to find the values for V x, V R, V R. (R = R =Ω, R 3 =Ω, V =V,V = V) (Sim_Lin_Eq_Solve.m, matrix_solve.xlsx). Using analysis by inspection and mesh currents method, R þ R 3 R 3 R 3 R þ R 3 Substituting given component values, 3 3 i i i i ¼ V : V ¼ D ¼ 3 3 ¼ 8; D ¼ 3 ¼ 6 ¼ 5; i ¼ D D ¼ 5 8 A V R ¼ i R ¼ 5 0 ¼ 8 8 ¼ :5 V
43 . Mesh Analysis 7 V x ¼ V V R ¼ :5 ¼ 0:75 V V R ¼ V x V ¼ 0:75 ¼ 0:5 V: Problem..6 Find the values of mesh currents and the node voltage in the circuit shown in Fig..40. R =0X, R =X, R3 =X, V =4V, V =V, V3 = V (Sim_Lin_Eq_Solve.m, matrix_solve.xlsx). KVL in mesh: 4 þ 0i þ ði i Þþ ¼ 0 i i ¼ : ð:57þ KVL in mesh: þ ði i Þþi þ ¼ 0 i þ 3i ¼ : ð:58þ Using Eqs. (.57) (.58), one obtains the following matrix equation: 3 The solution of this matrix equation gives i ¼ : i i ¼ 0:5 A i ¼ 0:5A v x ¼ V þ ði i ÞR ¼ 0:5 ¼ :5V: Fig..40 The circuit for Problem..6
44 8 Analysis Methods Alternatively, v x ¼ V 3 þ i R 3 ¼ þ 0:5 ¼ :5V: Problem..7 In the circuit shown in Fig..4, determine the voltage drop across R using mesh analysis. R =3X, R =5X, R3 =4X, v =V,I =A. Since a current source exists in the second mesh, i ¼I ¼A: The mesh equation for the other mesh, þ 3i þ ði i Þ5 ¼ 0: Solving this equation for the unknown current, 3i þ 5i þ 0 ¼ 0! 8i þ 8 ¼ 0! i ¼A: The voltage drop across R is V R3 ¼ 5ði i Þ ¼ 5ð þ Þ ¼ 5V: Problem..8 Find the values of numbered (clockwise flowing) mesh currents in the circuit shown in Fig..4. Use Cramer s rule when necessary U ¼ 4V; I ¼ A; R ¼ R ¼ R ¼ R 3 ¼ R 4 ¼ X (mesh.cir). Assume clockwise rotation for mesh currents. Since i ¼ I; it is not necessary to write down KVL equation associated with second mesh. KVL in mesh : U ¼ ði i 3 ÞR þ ði i ÞR 3 ¼ 0 i ðr þ R 3 Þi 3 R ¼ U þ IR 3 : ð:59þ Fig..4 The circuit for Problem..7
45 . Mesh Analysis 9 Fig..4 The circuit for Problem..8 KVL in mesh 3: i 3 R 4 þ ði 3 i ÞR þ ði 3 i ÞR ¼ 0 From (.59) and (.60), R þ R 3 R R R þ R þ R 4 i 3 ðr þ R 3 þ R 4 Þi R ¼ U þ IR : ð:60þ i i 3 ¼ U þ IR 3 IR i ¼ 6 ; D ¼ 6 ¼ 5; D 3 i 3 ¼ 6 3 ¼ 0; D ¼ 6 ¼ 0 i ¼ D D ¼ 0 5 ¼ 4A; i ¼ D D ¼ 0 5 ¼ A: Problem..9 For the circuit shown in Fig..43, write down the circuit equation in matrix form and solve for mesh currents. R =X, V =V =V3 =V (Sim_Lin_Eq_Solve.m, matrix_solve.xlsx). Analysis by inspection yields Fig..43 The circuit for Problem..9
46 30 Analysis Methods I I I 3 I ¼ ðvþ: of this equation (using given EXCEL or MATLAB tools) gives I ¼ 0:88 A; I ¼ 0:68 A; I 3 ¼ :73 A; I 4 ¼ :7 A: Problem..0 In the circuit shown in Fig..44, find the value of current i AB through 3 Ω resistor.ði AB ¼i BA Þ. Use Cramer s rule, when necessary. V =7V, V =6V, R =R5 =X, R =R4 =X, R3 =3X. i AB ¼ i 3 i : The mesh current equations yield the following matrix equation: i i 5 ¼ ¼ i D ¼ ¼ 3 36 þ ð Þþð6Þ½4 þ 7 þ 6 ¼ D ¼ ¼ 0 þ 6 þ ð Þ ¼ 8 ð 60 Þ ¼ 78 Fig..44 The circuit for Problem..0
47 . Mesh Analysis D 3 ¼ ¼ ð6þ ¼ 7 ¼ 08 þ 0 þ 3 ð þ 0 þ 6 Þ i ¼ D D ¼ ¼ A; i 3 ¼ D 3 D ¼ 7 39 ¼ 3A; i AB ¼ i 3 i ¼ 3 ¼ A: Problem.. For the circuit shown in Fig..45, determine the ratio of currents, r ¼ i R ; for k ¼; 0; ; : i R3 Assume mesh currentsði ; i Þ flow clockwise in the left and right meshes, respectively, KVL at mesh : v ¼ ði i ÞR ; i ¼ i R ; i ¼ i R3 : kv þ ðr þ R Þi i R ¼ 0! kði i ÞR þ ðr þ R Þi i R ¼ 0 KVL at mesh : i ðr þ R kr Þþi ðkr R Þ ¼ 0: ð:6þ From Eqs. (.6) and (.6), R þ R ð kþ R ð kþ R R þ R 3 R i þ ðr þ R 3 Þi ¼ U: ð:6þ i i ¼ U 0 D ¼ 0 R ð kþ U R þ R 3 ¼ UR ð kþ Fig..45 The circuit for Problem..
48 3 Analysis Methods D ¼ R þ R ð kþ 0 R U ¼ UR ½ þ R ð kþ i ¼ D D ; i ¼ D D ; r ¼ i ¼ i R ¼ D ¼ R ð kþ i i R3 D R þ R ð kþ ¼ R : R k þ R Values of r for different k parameters are shown in Table.3. Problem.. In the circuit shown in Fig..46, use mesh analysis and calculate the value of current through 0 X internal resistance of the 4 V voltage source, currents through R =Xand R 3 =4X. Find the node voltage. Applying KVL in the left mesh, taking clockwise current directions, (i = i ) 0i þ i i ¼ 4 i 6i ¼ : ð:63þ KVL : i i þ i ¼4V x ¼4ðði i ÞÞ ¼ 48i þ 48i i þ 48i þ 6i 48i ¼ 0 36i 3i ¼ 0 9i 8i ¼ 0 ð:64þ 9i ¼ 8i! i ¼ 8 9 i ð:65þ Table.3 Values of r for different k parameters k 0 r R R 0 R þ R R þ R Fig..46 The circuit for Problem..
49 . Mesh Analysis i 6i ¼! 8 i 9 6 ¼! i ¼ ¼ 08 ¼ 3:76 A; i ¼ i ¼ 8 3:76 9 ð Þ ¼ :84 A i R ¼ i i ¼ :84 3:76 ¼0:35 A V x ¼ i R ¼4:4 V: Problem..3 Find the values of currents i, i, i 3, i AB in the circuit shown in Fig..47. Use Cramer s rule, when necessary. V =4V, R =0X, R =R3 =4X, R4 =6X (Sim_Lin_Eq_Solve.m, matrix_solve.xlsx). i 0 ¼ i i KVL at mesh: 4 þ 0ði i Þþ6ði i 3 Þ ¼ 0! 6i 0i 6i 3 ¼ 4 KVL at mesh: 4i þ 4ði i 3 Þþ0ði i Þ ¼ 0! 0i þ 8i 4i 3 ¼ 0 KVL at mesh3: ði i Þþ6ði 3 i Þþ4ði 3 i Þ ¼ 0! 4i 6i þ 0i 3 ¼ 0: Collecting three equations in a matrix form, i i 5 ¼ i 3 0 Fig..47 The circuit for Problem..3
50 34 Analysis Methods D ¼ ¼ 544; D ¼ ¼ 464; D ¼ ¼ D 3 ¼ ¼ 58 i ¼ D D ffi :5 A; i ¼ D D ffi 0:85 A; i 3 ¼ D 3 D ffi 0:97 A; i ab ¼ i 3 i ffi 0: A: Problem..4 In the circuit shown in Fig..48, R =Ω, V =V,V =V. Find the value of current i X (Sim_Lin_Eq_Solve.m, matrix_solve.xlsx). KVL at mesh : Simplifying, þ ði i Þþði þ i 3 Þþði i 5 Þþi ¼ 0: Similarly, ¼ 4i i i 3 i 5 KVL at mesh : i þ ði i 3 Þþði i Þ ¼ 0! 0 ¼i þ 3i i 3 KVL at mesh3 : i 3 þði 3 i Þþði 3 i Þ¼0! 0 ¼i i þ 4i 3 i 4 Fig..48 The circuit for Problem..4
51 . Mesh Analysis 35 KVL at mesh4 : ði 4 i 3 Þþi 4 þ þði 4 i 5 Þ¼0! 0 ¼i 3 þ 3i 4 i 5 KVL at mesh5 : i 5 þði 5 i Þþði 5 i 4 Þ¼0! 0 ¼i i 4 þ 3i 5 : Collecting these equations in a matrix form, i i i 3 i 4 i ¼ : 0 of this matrix equation by employing available software (see, MATLAB m file or EXCEL file) yields i ¼ 0:646 A; i ¼ 0:73 A; i 3 ¼ 0:7 A; i 4 ¼0:3 A; i 5 ¼ 0:39 A i y ¼ i ¼ 0:73 A i x ¼ i 5 i 4 ¼ 0:39 A þ 0:3 A ¼ 0:369 A: Problem..5 Find the values of mesh currents and the node voltage V X in the circuit shown in Fig..49. R =kx (matrix_solve.xlsx). 0 þ i þ i i þ i i 4 ¼ 0 3i i i 4 ¼ 0 ð:66þ i þ i i 3 þ i i 5 þ i i ¼ 0 i þ 4i i 3 i 5 ¼ 0 ð:67þ Fig..49 The circuit for Problem..5
52 36 Analysis Methods i 3 þ i 3 i 6 þ i 3 i ¼ 0 i þ 3i 3 i 6 ¼ 0 ð:68þ i 4 i þ i 4 i 5 ¼ 0 i þ i 4 i 5 ¼ 0 ð:69þ i 5 i 4 þ i 5 i þ 5 ¼ 0 i i 4 þ i 5 ¼5 ð:70þ 5 þ i 6 i 3 þ i 6 ¼ 0 i 3 þ i 6 ¼ 5: Using Eqs. (.66) (.7), one obtains the following matrix equation: i i i i 4 ¼ : i i 6 5 ð:7þ The solution of this matrix equation gives i ¼ 4: ma; i ¼ :8 ma; i 3 ¼ :447 ma i 4 ¼ :53 ma; i 5 ¼:84 ma; i 6 ¼ 3:4 ma v x ¼ 5 ði 6 i 3 ÞR ¼ 5 :777 ¼ 3:3 V: Problem..6 In the circuit shown in Fig..50, use mesh currents method to determine currents flowing through each resistor (mesh6.xlsx) R ¼ R ¼ R 3 ¼ X; R 4 ¼ R 5 ¼ R 6 ¼ 4 X; R 7 ¼ R 0 ¼ 8 X; R 8 ¼ R 9 ¼ X; V ¼ V ¼ 0 V: Analysis by inspection, ½R½¼ I ½V; I I I ¼ I 4 ¼ : I I 6 0 of this matrix equation (via Excel) gives mesh currents as
53 . Mesh Analysis 37 Fig..50 The circuit for Problem..6 I ¼ 0:790 A; I ¼ 0:47 A; I 3 ¼0:937A I 4 ¼ :473 A; I 5 ¼0:796 A; I 6 ¼:677 A: Individual currents flowing through each resistor are calculated as follows: I R ¼ I ¼ 0:790 A; I R ¼ I ¼ 0:47 A; I R3 ¼ I 3 ¼0:937 I R4 ¼ I I 4 ¼:683 A; I R5 ¼ I I 5 ¼ 0:943 A; I R6 ¼ I 3 I 6 ¼ 0:740 A I R7 ¼ I I ¼ 0:644 A; I R8 ¼ I I 3 ¼ 3:69 A; I R9 ¼ I 4 I 5 ¼ 3:69 A I R0 ¼ I 5 I 6 ¼ 0:880 A: Problem..7 Use mesh analysis and determine the node voltages at nodes and 4 in the circuit shown in Fig..5. V a ¼ V; V b ¼ 6V; I ¼ 8A; R ¼ R ¼ 3R 3 ¼ X (Supermesh.cir). Since there is a current source between two meshes, a supermesh results by excluding the current source and resistor connected in series with it. Thus KVL around supermesh (Fig..5): V a þ i R þ i R þ V b ¼ 0! V b V a ¼i R i R V a V b ¼ i R þ i R! 6 ¼ i þ 6i Fig..5 The circuit for Problem..7
54 38 Analysis Methods Fig..5 The supermesh for the circuit of Problem..7 i þ i ¼ : ð:7þ Constraint equation is obtained by applying KCL at node, i i þ 8 ¼ 0! i i ¼8: ð:73þ From (.7) and (.73), i ¼8 i! i ¼ 8i ð:74þ i þ 8 þ i ¼! 3i ¼7! i ¼ 7 3 A ¼:333 A i ¼ ¼ ¼ A ¼ 5:667 A v ¼ V b þ i R ¼ 6 þ ð5:667þð6þ ¼ 40 V; v 4 ¼ ¼ 8V: Following is the SPICE netlist for the operating point analysis of this circuit: supermesh *OP Analysis R R36 R344 Va0 Vb306 I048 Problem..8 In the circuit shown in Fig..53, determine voltages at nodes, 3, and 4 using mesh current analysis (Supermesh.cir). V a ¼ 0 V; R 3 ¼ 3 X: g ¼ 0:ðA=VÞ; h ¼ 0ðV=AÞ; R ¼ 5 X; R ¼ 4 X; There is a current source between two meshes. For the supermesh circuit, (KVL) (Fig..54),
55 . Mesh Analysis 39 Fig..53 The circuit for Problem..8 Fig..54 The supermesh for the circuit of Problem..8 V a ¼ R i þ R i þ hi ¼ 0 0 ¼ 5i þ 4i þ 0i ¼ 0! 5i þ 4i ¼ 0: ð:75þ KCL at node (constraint equation): þ i i þ V a 0 ¼ 0! þ i i þ 0 0 ¼ 0 i i ¼A: ð:76þ From (.7) and (.7), Alternatively, i ¼ 40 9 ¼ :053 A; i ¼ 9 ¼ 0:053 A V ¼ i R þ 0i ¼ þ 0 9 ¼ 80 ¼ 9:474 V: 9 V ¼ V a i R ¼ 0 5 ¼ 9:474 V 9 V 4 ¼ 9V a R 3 ¼ 0: 0 3 ¼3V V 3 ¼ hi ¼ 0 0:053 ¼ :053 V: Following is the SPICE netlist for the operating point analysis of this circuit:
56 40 Analysis Methods supermesh *OP Analysis Va00 I0 R55 R34 R3403 *CCVS: hxx N+ N VNAME VALUE *Controlling current is through a zero volt voltage source VREF 5 0 h 3 0 VREF 0 *VCCS: gxx N+ N NC+ NC VALUE g Linearity and Superposition Problem.3. In the circuit shown in Fig..55, find the value of the current flowing through R ¼ 9 X resistor using superposition. R ¼ 6X; U ¼ 3V; I ¼ A: Current due to voltage source alone is Current due to current source alone is The sum: i 0 ¼ 3 9 þ 6 ¼ 5 A: i 00 ¼ 6 9 þ 6 ¼ 4 5 A: i ¼ i 0 þ i 00 ¼ 5 þ 4 5 ¼ A: Fig..55 The circuit for Problem.3.
57 .3 Linearity and Superposition 4 Fig..56 The circuit for Problem.3. Problem.3. (a) Determine the node voltage in the circuit shown in Fig..56 (use superposition). (b) Calculate the node voltage if all conductances are S, and current source values are both A. (a) By superposition, I off, I on; Vx 0 ¼ I =G 3 ; I on, I off; Vx 00 ¼ I =G 3 ; V x ¼ V 0 x þ V 00 x ¼ I G 3 þ I G 3 ¼ G 3 ði þ I Þ: ðbþ G ¼ G ¼ G 3 ¼ S; I ¼ I ¼ A V x ¼ ð þ Þ ¼ V: Problem.3.3 (a) In the circuit shown in Fig..57, determine the voltage at node a, if V =V, V =V,R =kω. (b) i =? (Use superposition theorem). Fig..57 The circuit for Problem.3.3
58 4 Analysis Methods ðaþ V 0 a ¼ V 3 V; V 00 a ¼ V 3 V; V a ¼ V 0 a þ V 00 a ¼ V 3 þ V 3 ¼ 3 þ 3 ¼ V ðbþ i ¼ V a R ¼ V kx ¼ ma: Problem.3.4 (a) Assuming a singleinput and singleoutput (SISO) system, state criteria to determine the linearity of such a system. (b) If y is the output and x is the input of a system of the form y = mx + n, what can be said about its linearity? (a) Assuming y is the output and x is the input of a system, three criteria to determine the linearity of such a system are as follows:. Homogeneity: if y = f(x) then k.y = f(k.x) where k is a constant factor (more generally stated, k is any real number for real systems and it is any complex number for complexvalued signals and systems).. Additivity: If y = f(x) and y = f(x), then y + y = f(x + x). 3. For x = 0, then y = f(0) = 0. If a system satisfies all of these criteria stated above, it is a linear system. (b) Let m =,n =, then y =x +, then x x=0 x = x3 = x4 =3 y y= y =3 y3 =5 y4 =7 All criteria are violated. For example, y(x + x3) 6¼ y(x) + y(x3). Therefore, this system is incrementally linear so that the output is a scaled reproduction of the input except for a fixed offset in the output. Problem.3.5 What can be said about the linearity of the modified voltage divider circuit shown in Fig..58?
59 .3 Linearity and Superposition 43 Fig..58 The circuit for Problem.3.5 V a ðuþ ¼ R U þ R U ref! y ¼ mx þ n; where x ¼ U: R þ R R þ R This circuit is incrementally linear so that the output voltage is a scaled reproduction of the input voltage except for a fixed offset in the output voltage. Problem.3.6 In the circuit shown in Fig..59, use linearity principle to find the values for the voltage at node C (=V C ) and the current i through the resistor R 6 (R = R = R 3 =Ω, R 4 = R 5 = R 6 =4Ω, i S = A). Let i =A V C ¼ 4 i ¼ 4 ¼ 4V i R3 ¼ V C R 3 ¼ 4 ¼ 4A i BC ¼ i R3 þ i ¼ 4 þ ¼ 5A V B ¼ i BC 4 þ V c ¼ 5 4 þ 4 ¼ 4 V i R ¼ V B R ¼ 4 ¼ 4 A i AB ¼ i R þ i BC ¼ 4 þ 5 ¼ 9 A V A ¼ 4 i AB þ V B ¼ 4 9 þ 4 ¼ 6 þ 4 ¼ 40 V Fig..59 The circuit for Problem.3.6
60 44 Analysis Methods Fig..60 The circuit for Problem.3.7 i R ¼ V A R ¼ 40 A i 0 s ¼ i R þ i AB ¼ 40 þ 9 ¼ 69 A! i0 s i s ¼ i! i ¼ 69 ¼ 0:0834 A V C ¼ i 4 ¼ 0:04734 V: Problem.3.7 Find the current through resistor R 5 in the circuit shown in Fig..60 (use linearity principle). V s ¼ 0 V; R ¼ 0:5 X; R ¼ 8 X; R 3 ¼ X; R 4 ¼ X; R 5 ¼ X: Let i 5 ¼ A; v 5 ¼ i 5 R 5 ¼ V i 4 ¼ v 5 R 4 ¼ 0:5A i 3 ¼ i 4 þ i 5 ¼ :5A v 3 ¼ i 3 R 3 þ v 5 ¼ ð:5þðþþ ¼ 4V i ¼ v 3 R ¼ 4 8 ¼ 0:5V i ¼ i þ i 3 ¼ 0:5 þ :5 ¼ A V sx ¼ i R þ v 3 ¼ ðþ0:5 ð Þþ4 ¼ 5V When V s ¼ 5V; i 5 ¼ A But v s ¼ 0 V; then i 5 ¼ A: Problem.3.8 Determine the current (I X ) in the circuit shown in Fig..6. Use linearity principle. R =0X, U =0V.
61 .3 Linearity and Superposition 45 Fig..6 The circuit for Problem.3.8 Let I x ¼ A; V b ¼ R ¼ 0 V V a ¼ U ¼ i ab R þ V b ¼ ð þ ÞR þ 0 ¼ 0 þ 0 ¼ 40 V: Since given value of U = 0 V (which is half the calculated value), I x ¼ 0:5A: Problem.3.9 Calculate the value of currents through R 3 and R in the circuit shown in Fig..6. Use superposition. R ¼ R ¼ kx; R 3 ¼ kx; I ¼ 9I ¼ 9mA: By current division rule due to I By current division rule due to I, The sum of the currents: i R3 ¼ i R3 þ i R3 ¼ R i R3 ¼ I : R þ R þ R 3 R i R3 ¼I : R þ R þ R 3 R þ R þ R 3 ði R I R Þ¼ 9 þ þ ð Þ ¼ ma: Fig..6 The circuit for Problem.3.9
62 46 Analysis Methods By Kirchhoff s current law, 9 ¼ i R þ i R3 ¼ i R þ or, i R3 ¼ 7mA: Note that application of superposition principle is somewhat lengthy even though it is straightforward. Problem.3.0 Using superposition theorem in the circuit shown in Fig..63, find the value of (a) Vx. (b) Vx, ifr ¼ 0 X. (c) Vx, ifr ¼ 0 X. Step. (Fig..64), i 0 ¼ 0A!V þ ir þ ir þ kv ¼ 0 Step. V ¼ 0V; ir ð þ R Þ ¼ V kv ¼ V ð kþ;! i ¼ V ð kþ R þ R V x ¼ i R þ kv ¼ V ð kþ ð kþr R þ kv ¼ V þ k : R þ R R þ R Fig..63 The circuit for Problem.3.0 Fig..64 The circuit after killing the current source
63 .3 Linearity and Superposition 47 V x ¼ i R R R þ R ð kþr R R a. V x ¼ V x þ V x ¼ V þ k þ i R þ R R þ R b. R ¼ 0 X; V x ¼ V ½V c. R ¼ 0 X; V x ¼ kv ½V: ½V Problem.3. In the circuit shown in Fig..65, R ¼ R ¼ R 3 ¼ X: Find the values of V a ; V b ; I ¼ I R ; I ¼ I R ; I 3 ¼ I R3 : According to superposition theorem, V a ¼ V a þ V a þ V a3 and V b ¼ V b þ V b þ V b3 : When 3 A current source is closed, I =A,I =I 3 =A. V a ¼ I ¼ ¼ V V b ¼ I 3 ¼ ¼ 4V: When A current source is closed, I =5A,I = A and I 3 =5A. V a ¼ I ¼ 5 ¼ 0 V V b ¼ I 3 ¼ 5 ¼ 0 V When A current source is closed, I =4A,I = A and I 3 =4A. V a3 ¼ I ¼ 4 ¼ 8V; V b3 ¼ I 3 ¼ 4 ¼ 8V V a ¼ V a þ V a þ V a3 ¼ þ 0 þ 8 ¼ 0 V V b ¼ V b þ V b þ V b3 ¼ 4 þ 0 þ 8 ¼ V: Fig..65 The circuit for Problem.3.
64 48 Analysis Methods Fig..66 The circuit for Problem.3. Problem.3. (a) Use superposition theorem and find the value of voltage at node of the circuit shown in Fig..66 (R = R = R 4 = R 5 =Ω, R 3 =Ω). (b) Check your result using SPICE analysis. Print netlist (superposition check.cir). (a) (i) See Fig..67. ðr 4 ==R 5 þ R 3 Þ==R V ¼ ¼ ðr 4 ==R 5 þ R 3 Þ==R þ R þ ¼ 3 V: (ii) See Fig..68. i ¼ 3 þ þ ¼ A; V ¼ ¼ V: Fig..67 The circuit for the calculation of V for Problem.3. Fig..68 The circuit for the calculation of V for Problem.3.
65 .3 Linearity and Superposition 49 Fig..69 The circuit after killing the current source (iii) See Fig..69. ð V 3 ¼ R 3 þ R ==R Þ==R 4 ð þ Þ== ¼ ðr 3 þ R ==R Þþ R 4 ½ð þ Þ==þ ¼ == == þ ¼ þ ¼ 3 V ðivþ V þ V þ V 3 ¼ 3 þ þ 6 ¼ 4 þ 6 þ ¼ ¼ 833 V: 6 6 (b) SPICE netlist, *OP analysis,superposition check v0 v343 v350 R R0 R33 R440 R545 Problem.3.3 Use superposition theorem and find the value of voltage Vx in the circuit shown in Fig..70. First, voltage source is short circuited (Fig..7). The current flow in 4 X branch is calculated by current division, Fig..70 The circuit for Problem.3.3
66 50 Analysis Methods Fig..7 The circuit after voltage source is killed Fig..7 The circuit after current source is killed i ¼ þ 4 ¼ 0:A! V x ¼ 0: ¼ 0:4V: Second, current source is open circuited (Fig..7). In this circuit, one may use the voltage division rule and obtain the unknown voltage as V ¼ 5 5 ¼ V: Finally, superposition results are collected together, V x ¼ V x þ V x ¼ :4V: Problem.3.4 Using superposition theorem, find the values of currents and voltages in the circuit shown in Fig..73 (i =A, V =0V, R =kω, R =kω). Fig..73 The circuit for Problem.3.4
67 .3 Linearity and Superposition 5 First, the voltage source is ignored (short circuited, Fig..74). R eq ¼ ðr ==R Þ ¼ 0:667 kv V x ¼ I R eq ¼ 0:667 ¼ 0:667 kv: Then, the current source is ignored (open circuited, Fig..75). By voltage division, V x ¼ R V ¼ 0 ¼ 3:333 V R þ R þ V x ¼ V x þ V x ¼ 667 V þ 3:33 V ¼ 670:33 V I R ¼ V x ¼ 670:33 V R kx ¼ 0:67033 A I R ¼ I I R ¼ 0:67033 ¼ 0:3967 A: Problem.3.5 In the circuit shown in Fig..76, find the value of i (in ma) by using superposition theorem. Fig..74 The circuit after voltage source is killed Fig..75 The circuit after current source is killed (open circuited) Fig..76 The circuit for Problem.3.5
68 5 Analysis Methods Deactivated voltage source: (Fig..77). Applying current division, i 0 ¼ A V 0 ¼ A V¼ V: Deactivated current source (Fig..78): By voltage division, V 00 3== ¼ 3== þ ¼ 3 3 þ ¼ 3 3 þ þ 6 5 ¼ 6 5 þ V ¼ V 0 þ V 0 ¼ þ 3 4 ¼ :5 V i ¼ V ¼ :5 ¼ 0:65 A ¼ 65 ma: 6 6 ¼ 6 ¼ 3 4 V Fig..77 Deactivated voltage source Fig..78 Deactivated current source Fig..79 The circuit for Problem.3.6
69 .3 Linearity and Superposition 53 Problem.3.6 Use superposition theorem to find the values of voltages at nodes and in the circuit shown in Fig..79. (a) Kill the voltage source as shown in Fig..80: i X ¼ A; V 0 ¼ X A¼ V V 0 ¼ == ½ ð þ Þ ¼ ð==þ ¼ X A¼ V: (b) Kill the current source as shown in Fig..8. By voltage division, R p ¼ ð þ Þ ð þ Þþ ¼ 6 5 X V 00 ¼ R p R p þ ¼ 6 5 ¼ 6 5 þ ¼ 6 V ¼ 3 4 V: Voltage division: V 00 ¼ V 00 þ ¼ ¼ V V ¼ V 0 þ V 00 ¼ þ 0:5 ¼ :5V V ¼ V 0 þ V 00 ¼ þ 3 ¼ :75 V: 4 Fig..80 Deactivated voltage source (short circuited) Fig..8 Deactivated current source (open circuited)
70 54 Analysis Methods Fig..8 The circuit for Problem.3.7 Problem.3.7 The supply voltage v and output current i are mutually transferable in a linear passive circuit. A circuit composed of linear bilateral elements (e.g., R, L, C) is reciprocal. The ratio of v and i is called the transfer resistance (transresistance). This means that if the positions of a voltage source and an ammeter are interchanged, the reading of ammeter remains the same, assuming ideal situation (i.e., internal resistance of both the voltage source and ammeter are null). Alternatively, interchanging a current source and a voltmeter in a linear bilateral circuit does not change the voltmeter reading. Reciprocity is based on the symmetry property of nodal conductance (mesh resistance) matrix. Thus, even a circuit containing dependent sources can be reciprocal for some specific dependent source coefficients, provided that its conductance or resistance matrix is symmetric. Application of reciprocity theorem is limited only to circuits containing a single independent source. (a) Use SPICE and determine the current flowing through 3 X resistor in the circuit shown in Fig..8, assuming that an ammeter is placed in that branch. What is transresistance value? (b) Interchange the ammeter and the voltage source and determine the new ammeter reading, again. What is new transresistance value? (c) If the voltage is 50 V in part (b), determine the new ammeter reading. (a) The current flowing through 3 X resistor (ammeter reading) is A. Transresistance is X. (b) Interchanging the ammeter and the voltage source, the ammeter reading is A, again. Therefore, transresistance is X, as well. (c) If the voltage is 00 V in part (b), (due to linearity) the new ammeter reading is A. This is also verified by SPICE analysis. SPICE netlist (Reciprocity.cir) is given below.
71 .3 Linearity and Superposition 55 Reciprocity *v 0 0 V400 R R04 R33 R430 R5343 *VX VX00.4 Source Transformation Problem.4. In the circuit shown in Fig..83, find the value of node voltage Vx, if V =3V,I =9A,R =X using source transformation. Applying source transformation to the given circuit gives the circuit shown in Fig..84; then, V R I þ V R þ I ¼ 3 V x R ¼ 3V x V ¼ 3V x V x ¼ 3 V ¼ 3 ¼ V: 3 Problem.4. In the circuit shown in Fig..85, use source transformation method and determine the current through resistor R. Fig..83 The circuit for Problem.4.
72 56 Analysis Methods Fig..84 Source transformed circuit of Problem.4. Fig..85 The circuit for Problem.4. Fig..86 Source transformation applied to circuit of Problem.4. By source transformation and KVL, (see Fig..86), V V V 3 ¼ iðr þ R þ R 3 Þ I R V V 3 ¼ iðr þ R þ R 3 Þ i ¼ I R V V 3 R þ R þ R 3 : Problem.4.3 Use source transform to calculate the value of node voltage V a in the circuit shown in Fig..87. R = R =R 3 =8X, I = I =A. i ¼ I R I R 3 4V a R þ R þ R 3 ¼ 8 4 4V a 8 þ 4 þ 4 ¼ 4 V a ¼ ir 3 þ I R 3 : ð 0 V a Þ ¼ V a 5 ð:77þ ð:78þ
73 .4 Source Transformation 57 Fig..87 The circuit for Problem.4.3 Fig..88 The circuit for Problem.4.4 Substitute (.77) in(.78), use given data, V a ¼ V a 5 4 þ 4 ¼ 4 5 4V a 5 þ V a ¼ 4! V a ¼ ¼ 8 ¼ :667 V: 3 Problem.4.4 In the circuit shown in Fig..88, find the value of voltage V x using source transformation. U =3V,U =5V,I =A,R = R 3 =R =4X. Application of source transformation to voltage sources results in with the following circuit equation: 3 4 þ 5 þ ¼ V x 4 : Solving for the unknown voltage yields V x ¼ V: Problem.4.5 Using source transformation, find the node voltage V X in the circuit shown in Fig..89. R = R =4X. Fig..89 The circuit for Problem.4.5
74 58 Analysis Methods By source transformation, (Fig..90), R kr ¼ X: KCL at node x: þ V x 3 V x ¼ 0 V x 3 ¼ V x ¼ ¼ 6V: 6 Problem.4.6 In the circuit shown in Fig..9, find the value of node voltage using superposition theorem, and source transformation (E =0V,R =0Ω, R =0Ω, I = A). First, the current source is deactivated, and voltage source is transformed to current source. By KCL, E R þ V x 0 V x R p ¼ 0; R p ¼ R ==R ¼ 0==0 ¼ 5 X E þ V x R 0 ¼ 0! V x ¼ R p E R 0 R p ¼ ¼ 0 V: 5 Fig..90 Source transformation applied to the circuit of Problem.4.5 Fig..9 The circuit for Problem.4.6
75 .4 Source Transformation 59 Fig..9 Deactivated voltage source (short circuited) Then, the voltage source is deactivated by short circuiting it; and using KCL (Fig..9), I þ V x 0 V x ¼ 0! I þ V x R p 0 ¼ 0! V x ¼ I R p 0 ¼ 0 V: R p Finally, adding these superposition results, V x ¼ V x þ V x ¼ 0 þ 0 ¼ 40 V. Problem.4.7 In the circuit shown in Fig..93, find the values of V X by using source transformation and Kirchhoff s current law. Use Cramer s rule when necessary. R ¼ R ¼ R 3 ¼ R 4 ¼ R 5 ¼ R ¼ X; U ¼ U ¼ U 3 ¼ V; I ¼ A: The voltage sources are transformed into A current sources (Fig..94), and simplified as shown in Fig..95. Fig..93 The circuit for Problem.4.7 Fig..94 The voltage sources are transformed into current sources
76 60 Analysis Methods Fig..95 Simplified circuit for Problem.4.7 Equivalent resistance of X parallel resistors is calculated and the current sources are added: þ ½¼ I ½G½V ¼ þ þ Vx V y D ¼ ¼ 4 ¼ 3 D x ¼ ðþðþ ¼ 5 V x ¼ D x D ¼ 5 V ¼ :667 V: 3 Problem.4.8 Determine the value of voltage at node in the circuit shown in Fig..96. Use source transformation and Cramer s rule, when necessary. R ¼ R ¼ R 3 ¼ kx; R 4 ¼ R 5 ¼ 4kX; I ¼ 4I ¼ 4mA; U ¼ V: Fig..96 The circuit for Problem.4.8
77 .4 Source Transformation 6 4 GV ¼ I =R þ =R =R 0 =R =R þ =R 3 þ =R 4 =R 4 0 =R 4 =R 4 þ =R 5 = þ = = = = þ = þ =4 =4 0 =4 =4 þ =4 5 V 3 4 V 5 ¼ 4 V V 3 4 V 5 6 ¼ 4 V 3 ð4 Þ I I U R 3 I ¼ or terms cancel out; V V V ¼ D ¼ 4 ¼ ; 0 V ¼ D D ; D ¼ ¼ þ 8 ¼ ¼ 7 6 V ¼ D D ¼ ¼ V ¼ 4:57 V: Problem.4.9 In the circuit shown in Fig..97, find the value of current I using source transformation method R ¼ R ¼ R 3 ¼ R 4 ¼ X; I ¼ A; U ¼ V (matrix_solve.xlsx). V voltage source is transformed to A independent current source, VCVS is transformed and circuit is simplified by taking only equivalent resistance of parallel resistors into consideration, see Fig..98.
78 6 Analysis Methods Fig..97 The circuit for Problem.4.9 Fig..98 Simplified circuit for Problem.4.9 Apply Kirchhoff s Current Law: At node : ¼ V þ V V V V ¼ : ð:79þ At node : V þ þ V V V ¼ 0 V V ¼ : ð:80þ Put (.80) and (.8) in matrix form: V V Solving the matrix equation yields ¼ : 0:5 V ¼ :5V; V ¼ V; I ¼ðV V Þ=R ¼ :5 ¼0:5A: Problem.4.0 In the circuit shown in Fig..99, use source transformation to find the value of node voltage V x,ifv = V =0V,I =A,R = R = R 3 =kω.
79 .4 Source Transformation 63 Fig..99 The circuit for Problem.4.0 Fig..00 Source transformation applied to the circuit of Problem.4.0 If source transformation is used for the circuit (Fig..00), V is transformed into I a and V transformed into I b I a ¼ 0 V 0 3 X ¼ 0 A I b ¼ 0 V 0 3 X ¼ 0 A; I þ I a þ I b ¼ :0 A ¼ R t 0 3 þ 0 3 þ 0 3 ¼ 3 0 3! R t ¼ 333:3 X; V x ¼ I R t ¼ :0 333:3 ¼ 340 V: Problem.4. In the circuit shown in Fig..0, find the value of current through resistor R using source transformation method (E =0V, R =Ω, R =3Ω, R 3 =5Ω, I = 8 A). Source transformation is applied on E and R 3 (see, Fig..0). I ¼ E R 3 ¼ 0 V 5 X ¼ 4A Fig..0 The circuit for Problem.4.
80 64 Analysis Methods Fig..0 Source transformation applied to the circuit of Problem.4. I and I added together, I ¼ 8 þ 4 ¼ A R eq ¼ ðr þ R Þ==R 3 ¼ðþ3Þ==5 ¼ :5 X V Req ¼ I R eq ¼ :5 ¼ 30 V I R ¼ V R eq ¼ 30 V R þ R 5 X ¼ 6A: Problem.4. In the circuit shown in Fig..03, V S =V, R = R = R 3 = R 4 =Ω, f = 4 A/A. V =?,V a =?,i =? Use source transformation and node voltage method. Use source transformation and note that V a ¼ V (see, Fig..04), Fig..03 The circuit for Problem.4. Fig..04 Source transformation applied to the circuit of Problem.4.
81 .4 Source Transformation 65 V s R f V R 3 V R V V R ¼ 0 ð:8þ f V R 3 þ V V R V R 3 ¼ 0: ð:8þ Substituting numerical values and rearranging above equations, 3 V ¼ 0 of this matrix equation gives V V ¼ V; V ¼ V a ¼V; i ¼ V R 3 ¼ ¼ A:.5 Thévenin Norton Equivalent Circuits and Maximum Power Transfer Problem.5. A signal source has an opencircuit voltage of mv and a shortcircuit current of 00 na. What is the source resistance? Opencircuit voltage = Thévenin voltage, Shortcircuit current = Norton current R s ¼ V T ¼ ¼ I N ¼ 04 X ¼ 0 kx: Problem.5. For which one of the following circuits in Fig..05, Thévenin s theorem cannot be applied? Fig..05 The circuit for Problem.5.
82 66 Analysis Methods Fig..06 The circuit for Problem.5.3 Thévenin s theorem helps to reduce any oneport linear electrical network to a singlevoltage source and a single impedance. The circuit in Fig..05c contains a nonlineardependent source. Therefore, it does not suit for the application of this theorem. Problem.5.3 A carbon zinc battery can be thought of its Thévenin s equivalent circuit with Thévenin resistance being the internal battery resistance, see Fig..06. In an experiment, the opencircuit voltage of a battery is measured as.596 V. When a resistor of R = 33.0 X is connected across its terminals, the load voltage is measured as.580 V. What is the internal battery resistance? Assuming that measuring equipment probes and battery terminals have no contact resistances, internal resistance of the battery is serially connected to the load. The load current is I L ¼ V Th ¼ V L R Th þ R L R L I L ¼ :596 R Th þ 33 ¼ : ¼ 0:47878 A R Th ¼ ð8:886þð:596þ33 R Th ¼ 0:334 X: Problem.5.4 In the circuit shown in Fig..07, determine the current through R L at maximum power transfer condition if I =4A. Turning off the current source and calculating Thévenin resistance gives R Th ¼ R: Fig..07 The circuit for Problem.5.4
83 .5 Thévenin Norton Equivalent Circuits and Maximum Power Transfer 67 Thévenin voltage is the voltage drop across the grounded resistor, V Th ¼ V oc ¼ IR i RLpmax ¼ V Th ¼ IR R Th þ R L R þ R ¼ I 4 ¼ A: Problem.5.5 (a) Determine the Thévenin and Norton equivalents for the circuit shown in Fig..08, between a and b terminals. (b) Find the power delivered to a load resistance, if R L ¼ 5 X: (c) Determine the value of load resistor for maximum power transfer. R ¼ R ¼ 6 X; R 3 ¼ X; U ¼ 0 V: (a) Thévenin equivalent circuit values are R Th ¼ þ 6k6 ¼ þ 3 ¼ 5 X V Th ¼ V ab ¼ V oc ¼ 6 0 ¼ 5V: 6 þ 6 Note that X resistance has no effect here. Norton equivalent circuit values are (b) I ¼ V Th ¼ 5 R Th þ R L 5 þ 5 ¼ 4 A R N ¼ R Th ¼ 5 X I N ¼ V Th R Th ¼ 5 5 ¼ 3 ¼ 0:333 A P ¼ I R L ¼ 5 ¼ ¼ 0:35 W (c) For maximum power transfer; R L ¼ R Th ¼ 5 X: Fig..08 The circuit for Problem.5.5
84 68 Analysis Methods Problem.5.6 Use Thévenin s theorem to find the value of current, i R6 (Fig..09). ðr ¼ 50 X; R ¼ 5 X; R 3 ¼ R 4 ¼ 0 X; R 5 ¼ 4 X; R 6 ¼ X; V ¼ 0 VÞ: Remove R 6 from circuit: R 3 ==R 4 ¼ 5 X; Voltage division: V ab ¼ V Th V ab ¼ R 3==R 4 R þ R 3 ==R 4 0 V Th ¼ V ab ¼ 5 0 ¼ 0 V: 5 þ 5 Thévenin resistance: When the voltage source is short circuited, V =0V,R is shorted (see, Fig..0): R Th ¼ R 5 þðr kr 3 kr 4 Þ¼4þð5k0k0Þ R Th ¼ 4 þð5k5þ¼4þ:5¼6:5x: The value of the current flowing through resistor R 6 i R6 ¼ V Th ¼ 0 R Th þ R 6 6:5 þ ¼ 0 ¼ :77 A: 8:5 Problem.5.7 In the circuit shown in Fig.., use Thévenin s Theorem and source transformation method to determine the current through the resistor R L. Fig..09 The circuit for Problem.5.6 Fig..0 The circuit for the calculation of Thévenin resistance in Problem.5.6
85 .5 Thévenin Norton Equivalent Circuits and Maximum Power Transfer 69 Fig.. The circuit for Problem.5.7 R ¼ R ¼ X; R 3 ¼ R L ¼ X; V ¼ V; I ¼ 0:5A: First calculate Thévenin resistance for the circuit to the left of R L (Fig..): R Th ¼ðR kr ÞþR 3 : Then, determine Thévenin voltage, noting that R 3 has no current flow at the output terminals (aground), as shown in Fig..3. The node voltage V x becomes the opencircuit voltage. Using KCL at this node, V V þ I ¼ V x þ V þ I x R! V x ¼ V Th ¼ R R R þ : R R The load current through the resistor R L is found using Fig..4: i RL ¼ V Th R Th þ R L ¼ V þ I R þ R R ¼ R kr ÞþR 3 þ R L ð þ þ k ð Þþ þ ¼ 3 A: Fig.. The circuit for the calculation of Thévenin resistance in Problem.5.7 Fig..3 The circuit for the calculation of Thévenin voltage in Problem.5.7
86 70 Analysis Methods Fig..4 The circuit for the calculation of load current in Problem.5.7 Problem.5.8 Determine Thévenin equivalent parameters between a and b terminals of the circuit shown in Fig..5. R =X. (Hint: Apply source transformation for the current source in calculating Thévenin equivalent voltage.) Thévenin equivalent resistance is found by eliminating independent sources (i.e., shortcircuit voltage source and opencircuit current source) in the circuit and finding the resistance between a and b terminals. R Th ¼ R ab ¼ ½ðR þ RÞkRþR ¼ R ¼ 4 X: Thévenin equivalent voltage between (a) and (b) terminals of the circuit can be found by applying source transformation to the current source, and then determining the voltage at node (c). This is due to the fact that the voltage at terminal (a) equals to voltage at node (c), V cb ¼ V ab (Fig..6). By KVL, i ¼ðÞ= ðr þ R þ RÞ ¼ =8 ¼ 0:5 A Fig..5 The circuit for Problem.5.8 Fig..6 The circuit for the calculation of Thévenin voltage in Problem.5.8
87 .5 Thévenin Norton Equivalent Circuits and Maximum Power Transfer 7 V Th ¼ V ab ¼ V cb ¼ R i þ ¼ 4 0:5 þ ¼ :5V: In summary, V Th ¼ :5V; R Th ¼ 4 X: Problem.5.9 For the circuit shown in Fig..7, (a) Find Thévenin s equivalent to the left of terminals a and b. (b) If R =kω, R =kω, R 3 = 00 Ω, k = 0., R L =kω, what is V ab? For Thévenin s equivalent circuit to the left of terminals a and b, the voltage source is removed, the first circuit becomes a short circuit, so the R Th only depends on R 3 (Fig..8). R Th ¼ R 3 ¼ 00 X: The current generated with currentcontrolled current source, CCCS becomes i Th ¼ k i R ¼ k V R þ R ¼ 0: V 000 ¼ V 04 A i L ¼ V Th ¼ R Th i Th V Th ¼ 00 V 04 ¼ V V Th ¼ V ¼ 4:6 0 6 V A R 3 þ R L 00 þ 000 V RL ¼ R L i L ¼ 000 4:6 0 6 V ¼ 4:6 0 3 V : Fig..7 The circuit for Problem.5.9 Fig..8 R Th only depends on R 3
88 7 Analysis Methods Fig..9 The circuit for Problem.5.0 Problem.5.0 For the circuit shown in Fig..9, a. Find the current through R L, and voltage across R L, using Thévenin s method (as function of k, V, R, R, R L ) b. What is the Norton s equivalent circuit to the left of a bifv =V,k = V/V, R =0Ω, R =5Ω. (a) i ¼ V R ; V oc ¼ V T ¼ k i ¼ k V R I RL ¼ V T R þ R L ¼ R T ¼ R ; kv R ðr þ R L Þ ; V R L ¼ I R L ¼ k V R L R ðr þ R L Þ : b. Thévenin equivalent circuit is shown in Fig..0. V T ¼ k V R ¼ 0 ¼ 0:4V; R T ¼ R N ¼ R ¼ 5 X: Norton equivalent circuit is shown in Fig... Fig..0 Thévenin equivalent circuit Fig.. Norton equivalent circuit
89 .5 Thévenin Norton Equivalent Circuits and Maximum Power Transfer 73 Fig.. The circuit described in Problem.5. Problem.5. A DC voltage source with internal resistance of 0 X and with an opencircuit voltage of V feeds a resistive load, R L : Determine the range of load resistance values so that the circuit operates for V L 5 V and i L 0:5A. Figure. shows the equivalent circuit. Using voltage constraint, R Th ¼ R i ¼ 0 X; V oc ¼ V Th ¼ V R L R Th þ R L V Th 5! R L 0 þ R L 5! 7R L 50! R L 7:4 X: Using current constraint, V Th i ¼ 0:5A! R Th þ R L R L 4 X: 0 þ R L 0:5! 5 þ 0:5R L! Therefore, 7:4 X R L 4 X: A proper value of load resistance is to choose the arithmetic mean of limiting values. This allows for component variations in the circuit. In that case, R L ¼ 0 X can be a suitable value of the load resistance. Problem.5. Maximum power delivered by a DC circuit to a resistor of R = 0.5 Ω is W. Find the opencircuit voltage at the output of the circuit. Draw its Thévenin s equivalent circuit. P max ¼ V th 4R L ; V Th ¼ V oc ; R ¼ R Th V oc ¼ 4 ð0:5þ ¼ V oc 8! V oc ¼ 8! V oc ¼ 9V: Figure.3 describes the Thévenin equivalent circuit. V Th ¼ V oc ¼ 9V; R Th ¼ R ¼ 0:5 X: Problem.5.3 Determine Thévenin and Norton equivalent circuit parameters for the circuit between terminal b and ground, as shown in Fig..4.
90 74 Analysis Methods Fig..3 Thévenin equivalent circuit Fig..4 The circuit described in Problem.5.3 First, determine the loop current under opencircuit conditions (i.e., no load is connected between terminal b and ground). By KVL and assuming clockwise current flow in the loop, 8 ð4 þ Þi þ V ab ¼ 8 6i þ i ð Þ ¼ 8 i ¼ 0! i ¼ 4A V oc ¼ V th ¼ V ab ¼ i ð Þ ¼ 4i ¼ 4 4 ¼ 6 V: When dependent source is shorted to ground, one can determine the shortcircuit current at the output port, as Then, R th ¼ R N ¼ V oc i sc ¼ i sc ¼ 8 4 þ ¼ 8 6 A: ¼ X; I N ¼ V th ¼ 6 ¼ :333 A: R th Problem.5.4 For the circuit shown in Fig..5, find Fig..5 The circuit described in Problem.5.4
91 .5 Thévenin Norton Equivalent Circuits and Maximum Power Transfer 75 Fig..6 The circuit for the calculation of Thévenin resistance in Problem.5.4 (a) the value of load at maximum power transfer. (b) the value of current through load resistor at maximum power. (c) the value of maximum power transferred to the load ðu ¼ V; R ¼ 6 X; R ¼ 8 X; R3 ¼ 4 XÞ: (a) Thévenin resistance (see Fig..6) R Th ¼ð6==4Þþ8 ¼ 4 0 þ 8 ¼ 04 0 R L ¼ R Th ¼ 0:4 X: (b) The value of current at maximum power is calculated from Thévenin s equivalent circuit, see Figs..7 and.8. By voltage division, Fig..7 The circuit for the calculation of Thévenin voltage in Problem.5.4 Fig..8 Thévenin equivalent circuit
92 76 Analysis Methods i ¼ 4 V Th ¼ V oc ¼ 4 þ 6 ¼ 8 0 ¼ 0:8V V Th 0:8 ¼ R Th þ R L 0:4 þ 0:4 ¼ 0:8 ¼ 0:038 A: 0:8 (c) P max ¼ i R L ¼ ð0:038þ ð0:4þ ¼ 0:05 W: Problem.5.5 In the circuit shown in Fig..9, all resistors (except R x ) have the same resistance of R =0Ω, and V = 0 V. Find the value of R x for maximum power transfer (matrix_solve.xlsx) (Fig..30). Thévenin s equivalent circuit R x ¼ R Th ¼ v oc i sc : Opencircuit voltage (Fig..3), 3 3R R 0 R 0 R 4R R 0 R 0 R 4R R 0 0 R R 5 : 0 R 0 R 3R 6 4 i i i 3 i 4 i ¼ 6 4 V : Fig..9 The circuit of Problem.5.5 Fig..30 Thévenin s equivalent circuit in Problem.5.5
93 .5 Thévenin Norton Equivalent Circuits and Maximum Power Transfer 77 Fig..3 Circuit for Thévenin voltage calculation in Problem.5.5 Fig..3 Circuit for short circuit current calculation in Problem.5.5 Mesh currents can be found here using an EXCEL spreadsheet as i ¼ 0:509 A; i ¼ 0:85 A; i 3 ¼ 0:046 A; i 4 ¼ 0:343 A; i 5 ¼ 0:76 A V oc ¼ V Th ¼ i 5 R þ i 3 R ¼ 0ð0:046 þ 0:76Þ ¼ 0 0: ¼ : V: Shortcircuit current (Fig..3): 6 4 3R R 0 R 0 0 R 4R R 0 R 0 0 R 4R 0 0 R R 0 0 R R 0 0 R 0 R 3R R 0 0 R 0 R R i sc ¼ 0:95 A i i i 3 i 4 i 5 i sc 3 ¼ R x ¼ R Th ¼ V oc ¼ : ¼ :8 X: i sc 0:95 Problem.5.6 In the circuit shown in Fig..33, (a) Determine Thévenin s equivalent circuit parameters. (b) If R = R = R 3 =Ω, V =V,I =A,R Th =?,V Th =? V
94 78 Analysis Methods Fig..33 The circuit of Problem.5.6 (c) Determine the condition for V ab <0. (d) Norton equivalent circuit parameters in (b)? (a) Deactivate all independent sources (see, Fig..34): R Th ¼ R þ ðr ==R 3 Þ ¼ R R þ R R 3 þ R R 3 : R þ R 3 Because node a is open, R has no effect. By source transformation of the voltage source, following circuit is obtained. Then, applying KCL at node a (Fig..35), V V a I V a ¼ 0! V I ¼ V a þ! R R R 3 R R R 3 V I R V a ¼ þ ¼ V Th : R R 3 Fig..34 The circuit for the calculation of Thévenin resistance in Problem.5.6 Fig..35 The circuit for the calculation of Thévenin voltage in Problem.5.6
95 .5 Thévenin Norton Equivalent Circuits and Maximum Power Transfer 79 (b) R = R = R 3 =Ω, V =V,I = A, substituting the values, R Th ¼ þ þ þ V Th ¼ V a ¼ ¼ :5 X þ ¼ 0:5V (c) If V R \I ; V Th ¼ V a \0: (d) Norton equivalent circuit parameters are I Th ¼ V Th R Th ¼ 0:5 :5 ¼ 0:333 A R N ¼ R Th ¼ :5 X: Problem.5.7 (a) Find Thévenin equivalent circuit for the circuit shown in Fig..36. (b) Find the limiting value of k if R = R =Ω. (c) Norton equivalent circuit? (a) The circuit has no independent sources. Apply source transform to dependent source and A current at terminals a, b (Fig..37). KCL at terminal (a), with i ¼ V a =R Fig..36 The circuit of Problem.5.7 Fig..37 Source transformation of dependent source and application of A current at terminals a, b of the circuit of Problem.5.7
96 80 Analysis Methods Fig..38 Thévenin s (=Norton s) equivalent of the circuit in Problem.5.7 kv a þ V a V a k ¼ 0! V a ¼ R R R R R R R R V Th ¼ V a ¼ k þ R R R R R Th ¼ V a A ¼ þ R R k : R R (b) Denominator of Thévenin resistance must be different than zero, k R R 6¼ R þ R : If R = R =Ω, then k 6¼ (c) Norton equivalent circuit = Thévenin equivalent circuit (Fig..38). Problem.5.8 In the circuit shown in Fig..39, determine the inequality condition on parameter C in terms of known quantities of the circuit so that, R ab <0X. The circuit does not contain an independent source, therefore its Thévenin equivalent circuit has only a Thévenin resistance. We assign a current source at the output, and source transformdependent source (Fig..40a, b), Nodal equation: Fig..39 The circuit of Problem.5.8 I 0 þ I x þ C I x V a ¼ 0; ðv b ¼ 0VÞ:
97 .5 Thévenin Norton Equivalent Circuits and Maximum Power Transfer 8 Fig..40 Assign a current source at the output, and source transformdependent source But, V a ¼8I x! I x ¼ V a 8 I 0 V a 8 þ C V a V a 8 ¼ 0; I 0 ¼ CV a 6 þ V a 8 þ V a ¼ V a Thévenin equivalent of the circuit, R Th ¼ V a I 0 ¼ C 6 þ 8 þ V a ¼ 6I 0 0 þ C : 6I 0 0 þ C ¼ 6 I 0 0 þ C : C þ 0 ¼ V a 6 Therefore, for R ab ¼ R Th \0, C\ 0. Problem.5.9 Find Thévenin equivalent circuit for the circuit shown in Fig..4 (Vs =V,f =4,R = R = R 3 = R 4 =Ω) (matrix_solve.xlsx). R 4 has no influence since node a is open. V Th = V oc = V. Apply source transform to voltage source and write nodal equations at, with i = V /R 3 (see, Fig..4) V s R f V R 3 V R V V R ¼ 0 ð:83þ f V R 3 þ V V R V R 3 ¼ 0: ð:84þ
98 8 Analysis Methods Fig..4 The circuit of Problem.5.9 Fig..4 The circuit of Problem.5.9 after independent source transformation Fig..43 Determining shortcircuit current Substituting numerical values and rearranging equations, 3 V ¼ 0 V ð:85þ Solving for V, V ¼ V Th ¼V¼ V oc R Th ¼ V oc I sc : KCL at nodes, ; with i = V /R 3, (Fig..43)
99 .5 Thévenin Norton Equivalent Circuits and Maximum Power Transfer 83 Simplify (.86) and (.87), 4V V ðv V Þ ¼ 0 ð:86þ 4V þ ðv V ÞV V ¼ 0 ð:87þ V þ 3V ¼ V þ V ¼ 0: ð:89þ ð:89þ Solving these two simultaneous equations for V yields V ¼ V; I sc ¼ V R 4 ¼ ¼ A: Therefore, Thévenin equivalent circuit consists of a negative resistor (Fig..44), with R Th ¼ V oc ¼ ¼X: I sc Problem.5.0 In the circuit shown in Fig..45, determine the maximum power (in Watts) that can be transferred to load resistance R L. Given that, when R L is removed from the circuit, the voltage at node 4 is measured as 4.5 V: R ¼ R 5 ¼ R 6 ¼ R 7 ¼ X; R ¼ R 3 ¼ R 4 ¼ X; I ¼ I = ¼ A: Fig..44 Thévenin equivalent circuit consists of a negative resistor Fig..45 The circuit of Problem.5.
Electric Circuits I. Nodal Analysis. Dr. Firas Obeidat
Electric Circuits I Nodal Analysis Dr. Firas Obeidat 1 Nodal Analysis Without Voltage Source Nodal analysis, which is based on a systematic application of Kirchhoff s current law (KCL). A node is defined
More informationChapter 2. Engr228 Circuit Analysis. Dr Curtis Nelson
Chapter 2 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 2 Objectives Understand symbols and behavior of the following circuit elements: Independent voltage and current sources; Dependent voltage and
More informationmywbut.com Mesh Analysis
Mesh Analysis 1 Objectives Meaning of circuit analysis; distinguish between the terms mesh and loop. To provide more general and powerful circuit analysis tool based on Kirchhoff s voltage law (KVL) only.
More informationChapter 5. Department of Mechanical Engineering
Source Transformation By KVL: V s =ir s + v By KCL: i s =i + v/r p is=v s /R s R s =R p V s /R s =i + v/r s i s =i + v/r p Two circuits have the same terminal voltage and current Source Transformation
More informationSinusoidal Steady State Circuit Analysis
Sinusoidal Steady State Circuit Analysis 9. INTRODUCTION This chapter will concentrate on the steadystate response of circuits driven by sinusoidal sources. The response will also be sinusoidal. For
More information3.1 Superposition theorem
Many electric circuits are complex, but it is an engineer s goal to reduce their complexity to analyze them easily. In the previous chapters, we have mastered the ability to solve networks containing independent
More informationUNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 1.0 Kirchoff s Law Kirchoff s Current Law (KCL) states at any junction in an electric circuit the total current flowing towards that junction is equal
More informationChapter 3 Methods of Analysis: 1) Nodal Analysis
Chapter 3 Methods of Analysis: 1) Nodal Analysis Dr. Waleed AlHanafy waleed alhanafy@yahoo.com Faculty of Electronic Engineering, Menoufia Univ., Egypt MSA Summer Course: Electric Circuit Analysis I (ESE
More information6. MESH ANALYSIS 6.1 INTRODUCTION
6. MESH ANALYSIS INTRODUCTION PASSIVE SIGN CONVENTION PLANAR CIRCUITS FORMATION OF MESHES ANALYSIS OF A SIMPLE CIRCUIT DETERMINANT OF A MATRIX CRAMER S RULE GAUSSIAN ELIMINATION METHOD EXAMPLES FOR MESH
More informationCircuit Theorems Overview Linearity Superposition Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer
Circuit Theorems Overview Linearity Superposition Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer J. McNames Portland State University ECE 221 Circuit Theorems Ver. 1.36 1
More informationDC STEADY STATE CIRCUIT ANALYSIS
DC STEADY STATE CIRCUIT ANALYSIS 1. Introduction The basic quantities in electric circuits are current, voltage and resistance. They are related with Ohm s law. For a passive branch the current is: I=
More informationBasic Electrical Circuits Analysis ECE 221
Basic Electrical Circuits Analysis ECE 221 PhD. Khodr Saaifan http://trsys.faculty.jacobsuniversity.de k.saaifan@jacobsuniversity.de 1 2 Reference: Electric Circuits, 8th Edition James W. Nilsson, and
More information1.7 DeltaStar Transformation
S Electronic ircuits D ircuits 8.7 DeltaStar Transformation Fig..(a) shows three resistors R, R and R connected in a closed delta to three terminals, and, their numerical subscripts,, and, being opposite
More informationModule 2. DC Circuit. Version 2 EE IIT, Kharagpur
Module 2 DC Circuit Lesson 7 Superposition Theorem in the context of dc voltage and current sources acting in a resistive network Objectives Statement of superposition theorem and its application to a
More informationNotes for course EE1.1 Circuit Analysis TOPIC 4 NODAL ANALYSIS
Notes for course EE1.1 Circuit Analysis 200405 TOPIC 4 NODAL ANALYSIS OBJECTIVES 1) To develop Nodal Analysis of Circuits without Voltage Sources 2) To develop Nodal Analysis of Circuits with Voltage
More informationChapter 5 Solution P5.22, 3, 6 P5.33, 5, 8, 15 P5.43, 6, 8, 16 P5.52, 4, 6, 11 P5.62, 4, 9
Chapter 5 Solution P5.22, 3, 6 P5.33, 5, 8, 15 P5.43, 6, 8, 16 P5.52, 4, 6, 11 P5.62, 4, 9 P 5.22 Consider the circuit of Figure P 5.22. Find i a by simplifying the circuit (using source transformations)
More informationmywbut.com Lesson 6 Wye (Y)  Delta ( ) OR Delta ( )Wye (Y) Transformations
Lesson 6 Wye (Y)  Delta ( ) O Delta ( )Wye (Y) Transformations 1 Objectives A part of a larger circuit that is configured with three terminal network Y (or Δ ) to convert into an equivalent Δ (or Y )
More informationPHYSICS 171. Experiment 3. Kirchhoff's Laws. Three resistors (Nominally: 1 Kilohm, 2 Kilohm, 3 Kilohm).
PHYSICS 171 Experiment 3 Kirchhoff's Laws Equipment: Supplies: Digital Multimeter, Power Supply (020 V.). Three resistors (Nominally: 1 Kilohm, 2 Kilohm, 3 Kilohm). A. Kirchhoff's Loop Law Suppose that
More informationNodal and Loop Analysis Techniques
IRW3652.I ALL 522 3:53 Page 65 Nodal and Loop Analysis Techniques LEARNING Goals In Chapter 2 we analyzed the simplest possible circuits, those containing only a singlenode pair or a single loop.
More informationModule 2. DC Circuit. Version 2 EE IIT, Kharagpur
Module 2 DC Circuit esson 8 evenin s and Norton s theorems in the context of dc voltage and current sources acting in a resistive network Objectives To understand the basic philosophy behind the evenin
More informationChapter 4: Methods of Analysis
Chapter 4: Methods of Analysis When SCT are not applicable, it s because the circuit is neither in series or parallel. There exist extremely powerful mathematical methods that use KVL & KCL as its basis
More informationNetwork Topology2 & Dual and Duality Choice of independent branch currents and voltages: The solution of a network involves solving of all branch currents and voltages. We know that the branch current
More informationSystematic Circuit Analysis (T&R Chap 3)
Systematic Circuit Analysis (T&R Chap 3) Nodevoltage analysis Using the voltages of the each node relative to a ground node, write down a set of consistent linear equations for these voltages Solve this
More informationAutomatic Formulation of Circuit Equations
ECE 570 Session 3 IC 752E Computer Aided Engineering for Integrated Circuits Automatic Formulation of Circuit Equations Objective: Basics of computer aided analysis/simulation Outline:. Discussion of
More informationKirchhoff's Laws and Circuit Analysis (EC 2)
Kirchhoff's Laws and Circuit Analysis (EC ) Circuit analysis: solving for I and V at each element Linear circuits: involve resistors, capacitors, inductors Initial analysis uses only resistors Power sources,
More informationLecture #3. Review: Power
Lecture #3 OUTLINE Power calculations Circuit elements Voltage and current sources Electrical resistance (Ohm s law) Kirchhoff s laws Reading Chapter 2 Lecture 3, Slide 1 Review: Power If an element is
More informationElectric Circuit Theory
Electric Circuit Theory Nam Ki Min nkmin@korea.ac.kr 01094192320 Chapter 18 TwoPort Circuits Nam Ki Min nkmin@korea.ac.kr 01094192320 Contents and Objectives 3 Chapter Contents 18.1 The Terminal Equations
More informationNotes on Electric Circuits (Dr. Ramakant Srivastava)
Notes on Electric ircuits (Dr. Ramakant Srivastava) Passive Sign onvention (PS) Passive sign convention deals with the designation of the polarity of the voltage and the direction of the current arrow
More informationOperational amplifiers (Op amps)
Operational amplifiers (Op amps) v R o R i v i Av i v View it as an ideal amp. Take the properties to the extreme: R i, R o 0, A.?!?!?!?! v v i Av i v A Consequences: No voltage dividers at input or output.
More informationE2.2 Analogue Electronics
E2.2 Analogue Electronics Instructor : Christos Papavassiliou Office, email : EE 915, c.papavas@imperial.ac.uk Lectures : Monday 2pm, room 408 (weeks 211) Thursday 3pm, room 509 (weeks 411) Problem,
More informationR 2, R 3, and R 4 are in parallel, R T = R 1 + (R 2 //R 3 //R 4 ) + R 5. CC Tsai
Chapter 07 SeriesParallel Circuits The SeriesParallel Network Complex circuits May be separated both series and/or parallel elements Combinations which are neither series nor parallel To analyze a circuit
More informationLecture 3 BRANCHES AND NODES
Lecture 3 Definitions: Circuits, Nodes, Branches Kirchoff s Voltage Law (KVL) Kirchoff s Current Law (KCL) Examples and generalizations RC Circuit Solution 1 Branch: BRANCHES AND NODES elements connected
More informationSimple Resistive Circuits
German Jordanian University (GJU) Electrical Circuits Laboratory Section 3 Experiment Simple Resistive Circuits Post lab Report Mahmood Hisham Shubbak 7 / / 8 Objectives: To learn how to use the Unitr@in
More informationWriting Circuit Equations
2 C H A P T E R Writing Circuit Equations Objectives By the end of this chapter, you should be able to do the following: 1. Find the complete solution of a circuit using the exhaustive, node, and mesh
More informationDiscussion Question 6A
Discussion Question 6 P212, Week 6 Two Methods for Circuit nalysis Method 1: Progressive collapsing of circuit elements In last week s discussion, we learned how to analyse circuits involving batteries
More informationTWOPORT NETWORKS. Enhancing Your Career. Research is to see what everybody else has seen, and think what nobody has thought.
C H A P T E R TWOPORT NETWORKS 8 Research is to see what everybody else has seen, and think what nobody has thought. Albert SzentGyorgyi Enhancing Your Career Career in Education While two thirds of
More informationQUESTION BANK SUBJECT: NETWORK ANALYSIS (10ES34)
QUESTION BANK SUBJECT: NETWORK ANALYSIS (10ES34) NOTE: FOR NUMERICAL PROBLEMS FOR ALL UNITS EXCEPT UNIT 5 REFER THE EBOOK ENGINEERING CIRCUIT ANALYSIS, 7 th EDITION HAYT AND KIMMERLY. PAGE NUMBERS OF
More informationBasics of Network Theory (PartI)
Basics of Network Theory (PartI). A square waveform as shown in figure is applied across mh ideal inductor. The current through the inductor is a. wave of peak amplitude. V 0 0.5 t (m sec) [Gate 987: Marks]
More informationChapter 2 Resistive Circuits
1. Sole circuits (i.e., find currents and oltages of interest) by combining resistances in series and parallel. 2. Apply the oltagediision and currentdiision principles. 3. Sole circuits by the nodeoltage
More informationBASIC NETWORK ANALYSIS
SECTION 1 BASIC NETWORK ANALYSIS A. Wayne Galli, Ph.D. Project Engineer Newport News Shipbuilding SeriesParallel dc Network Analysis......................... 1.1 BranchCurrent Analysis of a dc Network......................
More informationChapter 5 Objectives
Chapter 5 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 5 Objectives State and apply the property of linearity State and apply the property of superposition Investigate source transformations Define
More informationThevenin equivalent circuits
Thevenin equivalent circuits We have seen the idea of equivalency used in several instances already. 1 2 1 2 same as 1 2 same as 1 2 R 3 same as = 0 V same as 0 A same as same as = EE 201 Thevenin 1 The
More informationEE 321 Analog Electronics, Fall 2013 Homework #3 solution
EE 32 Analog Electronics, Fall 203 Homework #3 solution 2.47. (a) Use superposition to show that the output of the circuit in Fig. P2.47 is given by + [ Rf v N + R f v N2 +... + R ] f v Nn R N R N2 R [
More informationEE40 KVL KCL. Prof. Nathan Cheung 09/01/2009. Reading: Hambley Chapter 1
EE40 KVL KCL Prof. Nathan Cheung 09/01/2009 Reading: Hambley Chapter 1 Slide 1 Terminology: Nodes and Branches Node: A point where two or more circuit elements are connected Branch: A path that connects
More informationScience Olympiad Circuit Lab
Science Olympiad Circuit Lab Key Concepts Circuit Lab Overview Circuit Elements & Tools Basic Relationships (I, V, R, P) Resistor Network Configurations (Series & Parallel) Kirchhoff s Laws Examples Glossary
More informationIntroductory Circuit Analysis
Introductory Circuit Analysis CHAPTER 6 Parallel dc Circuits OBJECTIVES Become familiar with the characteristics of a parallel network and how to solve for the voltage, current, and power to each element.
More informationProblem Set 4 Solutions
University of California, Berkeley Spring 212 EE 42/1 Prof. A. Niknejad Problem Set 4 Solutions Please note that these are merely suggested solutions. Many of these problems can be approached in different
More informationActive Circuits: Life gets interesting
Actie Circuits: Life gets interesting Actie cct elements operational amplifiers (P AMPS) and transistors Deices which can inject power into the cct External power supply normally comes from connection
More informationElectricity & Magnetism
Electricity & Magnetism D.C. Circuits Marline Kurishingal Note : This chapter includes only D.C. In AS syllabus A.C is not included. Recap... Electrical Circuit Symbols : Draw and interpret circuit diagrams
More informationTwoPort Networks Admittance Parameters CHAPTER16 THE LEARNING GOALS FOR THIS CHAPTER ARE THAT STUDENTS SHOULD BE ABLE TO:
CHAPTER16 TwoPort Networks THE LEARNING GOALS FOR THIS CHAPTER ARE THAT STUDENTS SHOULD BE ABLE TO: Calculate the admittance, impedance, hybrid, and transmission parameter for twoport networks. Convert
More informationUnit 2: Modeling in the Frequency Domain. Unit 2, Part 4: Modeling Electrical Systems. First Example: Via DE. Resistors, Inductors, and Capacitors
Unit 2: Modeling in the Frequency Domain Part 4: Modeling Electrical Systems Engineering 582: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland January 20,
More informationUniversity of Alabama Department of Physics and Astronomy. Problem Set 4
University of Alabama Department of Physics and Astronomy PH 26 LeClair Fall 20 Problem Set 4. A battery has an ideal voltage V and an internal resistance r. A variable load resistance R is connected to
More informationENGR 2405 Chapter 8. Second Order Circuits
ENGR 2405 Chapter 8 Second Order Circuits Overview The previous chapter introduced the concept of first order circuits. This chapter will expand on that with second order circuits: those that need a second
More informationFundamentals of Electric Circuits, Second Edition  Alexander/Sadiku
Chapter 3, Problem 9(8). Find V x in the network shown in Fig. 3.78. Figure 3.78 Chapter 3, Solution 9(8). Consider the circuit below. 2 Ω 2 Ω j 8 30 o I j 4 j 4 I 2 j2v For loop, 8 30 = (2 j4)i ji 2
More informationKirchhoff s laws. Figur 1 An electric network.
Kirchhoff s laws. Kirchhoff s laws are most central to the physical systems theory, in which modeling consists in putting simple building blocks together. The laws are commonly known within electric network
More informationChapter 7 DirectCurrent Circuits
Chapter 7 DirectCurrent Circuits 7. Introduction... 7. Electromotive Force... 7.3 Resistors in Series and in Parallel... 4 7.4 Kirchhoff s Circuit Rules... 6 7.5 VoltageCurrent Measurements... 8 7.6
More informationChapter 20 Electric Circuits
Chapter 0 Electric Circuits Chevy olt  Electric vehicle of the future Goals for Chapter 9 To understand the concept of current. To study resistance and Ohm s Law. To observe examples of electromotive
More informationRESISTANCE AND NETWORKS
PURPOSE The purpose of this laboratory is to learn to construct simple circuits; and, to become familiar with the use of power supplies and the digital multimeter. to experimentally find the equivalent
More information... after Norton conversion...
Norton 's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load.
More informationELEC 202 Electric Circuit Analysis II Lecture 10(a) Complex Arithmetic and Rectangular/Polar Forms
Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(a) Complex Arithmetic and Rectangular/Polar Forms THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street,
More informationFundamental of Electrical circuits
Fundamental of Electrical circuits 1 Course Description: Electrical units and definitions: Voltage, current, power, energy, circuit elements: resistors, capacitors, inductors, independent and dependent
More informationSolutions to Systems of Linear Equations
Solutions to Systems of Linear Equations 5 Overview In this chapter we studying the solution of sets of simultaneous linear equations using matrix methods. The first section considers the graphical interpretation
More informationElectric Circuits I. Inductors. Dr. Firas Obeidat
Electric Circuits I Inductors Dr. Firas Obeidat 1 Inductors An inductor is a passive element designed to store energy in its magnetic field. They are used in power supplies, transformers, radios, TVs,
More informationExperiment #6. Thevenin Equivalent Circuits and Power Transfer
Experiment #6 Thevenin Equivalent Circuits and Power Transfer Objective: In this lab you will confirm the equivalence between a complicated resistor circuit and its Thevenin equivalent. You will also learn
More informationELECTRONICS. EE 42/100 Lecture 2: Charge, Current, Voltage, and Circuits. Revised 1/18/2012 (9:04PM) Prof. Ali M. Niknejad
A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 2 p. 1/26 EE 42/100 Lecture 2: Charge, Current, Voltage, and Circuits ELECTRONICS Revised 1/18/2012 (9:04PM) Prof. Ali M. Niknejad
More informationarxiv: v2 [mathph] 23 Jun 2014
Note on homological modeling of the electric circuits Eugen Paal and Märt Umbleja arxiv:1406.3905v2 [mathph] 23 Jun 2014 Abstract Based on a simple example, it is explained how the homological analysis
More informationENGG 1203 Tutorial. Op Amps 10 Oct Learning Objectives. News. Ack.: MIT OCW Analyze circuits with ideal operational amplifiers
ENGG 1203 Tutorial Op Amps 10 Oct Learning Objectives Analyze circuits with ideal operational amplifiers News Mid term Revision tutorial Ack.: MIT OCW 6.01 1 Q1 This circuit is controlled by the charge
More informationNoteARific: Kirchhoff s
NoteARific: Kirchhoff s We sometimes encounter a circuit that is too complicated for simple analysis. Maybe there is a mix of series and parallel, or more than one power source. To deal with such complicated
More informationActive Circuits: Life gets interesting
Actie Circuits: Life gets interesting Actie cct elements operational amplifiers (OP AMPS) and transistors Deices which can inject power into the cct External power supply normally comes from connection
More informationChapter 4: Techniques of Circuit Analysis
Chapter 4: Techniques of Circuit Analysis This chapter gies us many useful tools for soling and simplifying circuits. We saw a few simple tools in the last chapter (reduction of circuits ia series and
More informationPhysics 7B1 (A/B) Professor Cebra. Winter 2010 Lecture 2. Simple Circuits. Slide 1 of 20
Physics 7B1 (A/B) Professor Cebra Winter 2010 Lecture 2 Simple Circuits Slide 1 of 20 Conservation of Energy Density In the First lecture, we started with energy conservation. We divided by volume (making
More informationLet V1=12V, R1=50 ohms, R2=10K ohms, R3=2K ohms, and R4=500 ohms. RL represents the load placed on the circuit between points Aand B.
Questions on Thevenin Equivalent Circuits Fall 2004 2. Thevenin Circuits (25 points) Let V1=12V, R1=50 ohms, R2=10K ohms, R3=2K ohms, and R4=500 ohms. RL represents the load placed on the circuit between
More informationAnalysis of a singleloop circuit using the KVL method
Analysis of a singleloop circuit using the KVL method Figure 1 is our circuit to analyze. We shall attempt to determine the current through each element, the voltage across each element, and the power
More informationChapter 10: Sinusoids and Phasors
Chapter 10: Sinusoids and Phasors 1. Motivation 2. Sinusoid Features 3. Phasors 4. Phasor Relationships for Circuit Elements 5. Impedance and Admittance 6. Kirchhoff s Laws in the Frequency Domain 7. Impedance
More informationChapter 4 Circuit Theorems: Linearity & Superposition
Chapter 4 Circuit Theorems: Linearity & Superposition Dr. Waleed AlHanafy waleed alhanafy@yahoo.com Faculty of Electronic Engineering, Menoufia Univ., Egypt MSA Summer Course: Electric Circuit Analysis
More informationAC Circuit Analysis and Measurement Lab Assignment 8
Electric Circuit Lab Assignments elcirc_lab87.fm  1 AC Circuit Analysis and Measurement Lab Assignment 8 Introduction When analyzing an electric circuit that contains reactive components, inductors and
More informationBFF1303: ELECTRICAL / ELECTRONICS ENGINEERING
BFF1303: ELECTRICAL / ELECTRONICS ENGINEERING Introduction Ismail Mohd Khairuddin, Zulkifil Md Yusof Faculty of Manufacturing Engineering Universiti Malaysia Pahang Introduction BFF1303 ELECTRICAL/ELECTRONICS
More informationChapter 6: Operational Amplifiers
Chapter 6: Operational Amplifiers Circuit symbol and nomenclature: An op amp is a circuit element that behaes as a VCVS: The controlling oltage is in = and the controlled oltage is such that 5 5 A where
More informationPOE Practice Test  Electricity, Power, & Energy
Class: Date: POE Practice Test  Electricity, Power, & Energy Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which of the following forms of energy is
More information3/17/2009 PHYS202 SPRING Lecture notes Electric Circuits
PHYS202 SPRING 2009 Lecture notes Electric Circuits 1 Batteries A battery is a device that provides a potential difference to two terminals. Different metals in an electrolyte will create a potential difference,
More informationCalculation of voltage and current in electric network (circuit)
UNIVERSITY OF LJUBLJANA Calculation of voltage and current in electric network (circuit) Power distribution and Industrial Systems Alba Romero Montero 13/04/2018 Professor: Grega Bizjak Content Background...2...3
More informationDirect Current Circuits
Name: Date: PC1143 Physics III Direct Current Circuits 5 Laboratory Worksheet Part A: SingleLoop Circuits R 1 = I 0 = V 1 = R 2 = I 1 = V 2 = R 3 = I 2 = V 3 = R 12 = I 3 = V 12 = R 23 = V 23 = R 123
More informationor Op Amps for short
or Op Amps for short Objective of Lecture Describe how an ideal operational amplifier (op amp) behaves. Define voltage gain, current gain, transresistance gain, and transconductance gain. Explain the operation
More informationElectricity and Light Pre Lab Questions
Electricity and Light Pre Lab Questions The pre lab questions can be answered by reading the theory and procedure for the related lab. You are strongly encouraged to answers these questions on your own.
More informationCircuits. Fawwaz T. Ulaby, Michel M. Maharbiz, Cynthia M. Furse. Solutions to the Exercises
Circuits by Fawwaz T. Ulaby, Michel M. Maharbiz, Cynthia M. Furse Solutions to the Exercises Chapter 1: Circuit Terminology Chapter 2: Resisitive Circuits Chapter 3: Analysis Techniques Chapter 4: Operational
More information1. Review of Circuit Theory Concepts
1. Review of Circuit Theory Concepts Lecture notes: Section 1 ECE 65, Winter 2013, F. Najmabadi Circuit Theory is an pproximation to Maxwell s Electromagnetic Equations circuit is made of a bunch of elements
More informationElectron Theory. Elements of an Atom
Electron Theory Elements of an Atom All matter is composed of molecules which are made up of a combination of atoms. Atoms have a nucleus with electrons orbiting around it. The nucleus is composed of protons
More informationMicroelectronic Circuit Design 4th Edition Errata  Updated 4/4/14
Chapter Text # Inside back cover: Triode region equation should not be squared! i D = K n v GS "V TN " v & DS % ( v DS $ 2 ' Page 49, first exercise, second answer: 1.35 x 10 6 cm/s Page 58, last exercise,
More information4.2 Graphs of Rational Functions
4.2. Graphs of Rational Functions www.ck12.org 4.2 Graphs of Rational Functions Learning Objectives Compare graphs of inverse variation equations. Graph rational functions. Solve realworld problems using
More informationDesign Engineering MEng EXAMINATIONS 2016
IMPERIAL COLLEGE LONDON Design Engineering MEng EXAMINATIONS 2016 For Internal Students of the Imperial College of Science, Technology and Medicine This paper is also taken for the relevant examination
More informationProject 4: Introduction to Circuits The attached Project was prepared by Professor YihFang Huang from the department of Electrical Engineering.
Project 4: Introduction to Circuits The attached Project was prepared by Professor YihFang Huang from the department of Electrical Engineering. The examples given are example of basic problems that you
More informationVer 3537 E1.1 Analysis of Circuits (2014) E1.1 Circuit Analysis. Problem Sheet 1 (Lectures 1 & 2)
Ver 3537 E. Analysis of Circuits () Key: [A]= easy... [E]=hard E. Circuit Analysis Problem Sheet (Lectures & ). [A] One of the following circuits is a series circuit and the other is a parallel circuit.
More informationExercise 2: The DC Ohmmeter
Exercise 2: The DC Ohmmeter EXERCISE OBJECTIVE When you have completed this exercise, you will be able to measure resistance by using a basic meter movement. You will verify ohmmeter operation by measuring
More informationNetwork Graphs and Tellegen s Theorem
Networ Graphs and Tellegen s Theorem The concepts of a graph Cut sets and Kirchhoff s current laws Loops and Kirchhoff s voltage laws Tellegen s Theorem The concepts of a graph The analysis of a complex
More informationA~(A'~) = i,(t) (1.34)
GENERAL RESISTIVE CIRCUITS 225 Nonlinear branch equation In vector notation, Eq. (1.31) becomes simply Since the independent current sources do not form cut sets (by assumption), Eq. (1.14) remains valid.
More informationLecture 5 Review Current Source Active Load Modified Large / Small Signal Models Channel Length Modulation
Lecture 5 Review Current Source Active Load Modified Large / Small Signal Models Channel Length Modulation Text sec 1.2 pp. 2832; sec 3.2 pp. 128129 Current source Ideal goal Small signal model: Open
More informationCircuit Analysis. by John M. Santiago, Jr., PhD FOR. Professor of Electrical and Systems Engineering, Colonel (Ret) USAF. A Wiley Brand FOR
Circuit Analysis FOR A Wiley Brand by John M. Santiago, Jr., PhD Professor of Electrical and Systems Engineering, Colonel (Ret) USAF FOR A Wiley Brand Table of Contents. ' : '" '! " ' ' '... ',. 1 Introduction
More informationAP1 Electricity. 1. A student wearing shoes stands on a tile floor. The students shoes do not fall into the tile floor due to
1. A student wearing shoes stands on a tile floor. The students shoes do not fall into the tile floor due to (A) a force of repulsion between the shoes and the floor due to macroscopic gravitational forces.
More informationCapacitor of capacitance C; Their schematic representations are shown in Figure below.
UNIT 1 Basic Circuit Concepts In modern life, circuits are everywhere. Without circuits, you wouldn't have indoor lights. Without circuits, you wouldn't have the computer you're using to watch this lesson.
More informationSol: Semiconductor diode.
48 49 1. What is the resistance value of a resistor of colour code Brown, Black, Red and silver? Sol: Brown1, Black0, Red2, Silver 10%. Resistance, R = 10 X 102 ±10Ω. 2. Mention a nonohmic device.
More information