# Chapter 2 Analysis Methods

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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) Parallel-connected 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 current-controlled 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 source-nodal 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 source-nodal 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 voltage-to-current 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 source-nodal 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 current-controlled 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 single-input and single-output (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 complex-valued 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 (trans-resistance). 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 trans-resistance 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. Trans-resistance is X. (b) Interchanging the ammeter and the voltage source, the ammeter reading is A, again. Therefore, trans-resistance 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 open-circuit voltage of mv and a short-circuit current of 00 na. What is the source resistance? Open-circuit voltage = Thévenin voltage, Short-circuit 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 one-port linear electrical network to a single-voltage source and a single impedance. The circuit in Fig..05c contains a nonlinear-dependent 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 open-circuit 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 (a-ground), as shown in Fig..3. The node voltage V x becomes the open-circuit 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., short-circuit voltage source and open-circuit 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 current-controlled 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 open-circuit 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 open-circuit 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 open-circuit 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 short-circuit 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 : Open-circuit 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: Short-circuit 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) De-activate 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 transform-dependent 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 transform-dependent 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 short-circuit 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.

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