vtusolution.in Initial conditions Necessity and advantages: Initial conditions assist

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1 Necessity and advantages: Initial conditions assist Initial conditions To evaluate the arbitrary constants of differential equations Knowledge of the behavior of the elements at the time of switching Knowledge of the initial value of one or more devivatives of a response is helpful in anticipating the form of the response It is useful in getting to know the elements individually and in combined networks Gives a better understanding of non linear switching circuits The general solution of a first order differential equation contains an arbitrary constant, the order of differential equation increases the no of arbitrary constants increase Solution of differential equation= CF + PI Geometrical interpretation of Derivatives Normally i(0+), di/dt, and d 2 i/dt 2 at t=0+ will be evaluated i(0+) indicates where the current starts di/dt at t=0+ gives the slope fo the curve and the second derivative gives the rate of change of slope Ex: 1. i(0+)=0, di(0+)/dt = 0 and d 2 i(0+)/dt 2 =k >0 2. i(0+)=0, di(0+)/dt = K>0 and d 2 i(0+)/dt 2 =>0

2 3. i(0+)=k>0, di(0+)/dt =0 and d 2 i(0+)/dt 2 =0 Behaviour of the elements at t=0+ and at t= Element Equivalent circuit at t=0+ At t= Resistor Resistor Resistor Inductor Open circuit Short circuit Capacitor Short circuit Open circuit Inductor with initial current Capacitor with initial charge Procedure to find initial conditions: Current source Io Finally short Voltage source + - Finally open circuit 1. History of the network, at t=0- find i(0-), v(0-), preferably current through inductor and voltage across the capacitor before switching, 2. Write the circuit at t=0+ by looking at the table given above 3. Find i(0+), and v(0+) 4. Write general circuit after the switching operation 5. Write general integro differential equation 6. Obtain an expression for di/dt 7. Apply initial conditions like i(0+) find di/dt at t=0+ 8. obtain an expression for d 2 i(0+)/dt 2 9. Apply initial conditions like i(0+), di/dt at t=0+ find out d 2 i(0+)/dt 2. and the process is repeated.

3 Problems1. R-L Circuit (series) If K is closed at t=0, find the values of i, di/dt and d 2 i/dt 2 at t=0+ if R=10 Ω, L=1H and V=100V As per the steps indicated : 1. Previous history before the switch is closed i(0-) =0. 2. Write the general circuit after switching : Inductor acts as open, as inductor wont allow current to change instantaneously Hence i(0+) = 0 3. Write general network after switching : V= Ri + L di/dt 4. Obtain an expression for the first derivative: di/dt = (V R(i) )/L, substituting the values we get di/dt = 100 A/s 5. Obtain an expression for the second derivative: d 2 i(0+)/dt 2 = - R/L di/dt, substituting the known values we get -1000A/s 2. Ans: 0, 100 A/sec, 1000 A/sec 2

4 Problem 2. RL (parallel) If K is opened at t=0, Find v, dv/dt and d 2 v/dt 2 at t=0+, If I=1 amp, R=100 Ω, and L=1H As per the steps indicated : 6. Previous history before the switch is opened indicates that the switch Is closed i L (0-) =0. 7. Write the general circuit after switching : Inductor acts as open, as inductor will not allow current to change instantaneously Hencev(0+) =i*r=100 V 8. Write general network after switching : I= v/r + 1/L 9. Obtain an expression for the first derivative: Diff we get dv/dt at to0+ = V/sec 10. Obtain an expression for the second derivative: d 2 v(0+)/dt 2 = - R/L dv/dt, substituting the known values we get 10*10 6 V/s 2. Ans: 100, v/sec, 10 6 v/s 2

5 Problem 3: R-C Circuit(series) If K is closed at t=0, find the values of i, di/dt and d 2 i/dt 2 at t=0+ if R=1000 ohms, C= 1μf and V=100V History of the network: Vc(0-) = 0, i(0-)=0 Capacitor acts as a short Hence network after switching, Voltage source, switch, resistor, and capacitor as s/c makes a closed path Therefore i(0+)=v/r = 0.1 A. Write the general network after the switch is closed: V= Ri + 1/c Differentiate : 0 = R di/dt + i/c di(0+)/dt= - i(0+)/rc, substituting the values we get 100 A/s differentiate again to obtain second derivative: d 2 i(0+)/dt 2 = - (1/RC) di(0+)/dt, substitutiting we get A/s 2. Ans: 0.1, -100 A/s, 100,000 A/s 2

6 Problem 4: RC circuit(parallel) If K is opened at t=0, Find v, dv/dt and d 2 v/dt 2 at t=0+, if I=10A,R=1000 Ω and C=1μf History of the network, before the switch is opened: Vc(0-) = 0, Network after the switch is opened. Cap acts as a short circuit hence v(0+) = 0; General network after the switch is opened: I= V/R + C dv/dt Obtain an expression for dv(0+)/dt = (I V(0+)/R )/c, substituting we get dv(0+) / dt = 10 5 V/s Expression for second derivative: d 2 v(0+)/dt 2 = (-1/RC)*dv(0+)/dt=-10 8 V/s 2 Ans: 0, 10 5 v/s, v/s 2

7 Problem5: In the network shown, switch K is changed from position a to b at t=0, steady state being established at position a. V=100, R= 1000Ω, L=1H, C=1μf. Find : i, di/dt and d 2 i/dt 2 at t=0+ When the switch is at position a, SS being established, L acts as a short circuit, (t= ) state: Hence : i(0-)= V/R = 0.1A, Vc(0-)=0V. When the switch is closed to position b, Cap acts as short, R behaves as R and L acts as a current source, there fore i(0+)=0.1a Write general network after switching: Write integro diff equation: 0= Ri+Ldi/dt + 1/c, (a) Rearranging the above, we get the expression for di/dt = (-Ri/L) V c /L substituting the values di(0+)/dt = A/s differentiating equation (a), we get d 2 i(0+)/dt 2 = -R/L (di/dt) + i/rc substituting the values we get d 2 i(0+)/dt 2 = 0.

8 Problem6: In the network shown, switch K is changed from position a to b at t=0, steady state being established at position a. Show that i 1 (0+)=i 2 (0+)= -V/(R 1 +R 2 +R 3 ), i 3 (0+)=0. Solution: Steady state being established at position a, Means inductors act as short circuited, capacitors act as open circuit Switch being at position a, we can see that i 1 (0-)=0, i 2 (0-)=0 i 3 (0-)=0 V c1 (0-)=0, V c2 (0-)=0 and V c3 (0-)=V. When switch is moved to position b, Only C3 will act as a voltage source, No current i 1, i 2 and i 3 hence inductor L 1 and L 2 acts as open And Vc3 = V We get i 3 (0+)=0 and i 1 =i 2 = -V/ R 1 +R 2 +R 3

9 Problem7: Steady state reached with switch k open, Switch is closed at t=0. V=100V, R1=10 Ω, R2=R3=20Ω, L=1H, C=1μ f Find: v 0 across C and indicate polarities,initial values of i 1, i 2, di 1 /dt, di 2 /dt at t=0+ find di/dt at t= Solution: Switch k is opened : SS being established : Inductor acts as short circuit: capacitor acts as open : I 1 (0-)= V/R1+R2 = 3.33 A I 2 (0-)=0; Vc(0-)= i1*r2 = 66.67V At t=0 switch is closed : hence R1 becomes redundant< current will not flow through R1 as switch is acting as short across the resistor R1. Inductor acts as a current source of value 3.33 A, Vc(0+)= V I 2 (0+) = ( )/20 = 1.67A I 1 (0+)= 3.33 A General network after switching : V=i 1 R 2 + L di 1 /dt (a) V=i 2 R 3 + 1/c (b) di 1 (0+)/dt = (V-i 1 R 2 ) /L = 33.3A/s diff eqn (b) we get di 2 (0+)/dt=- i 2 (0+)/R 3 C = A/s. Ans: 66.7 volts 3.33A, 1.67A 33.4 A/s, A/s 0

10 Problem8: Switch K is closed at t=0, with zero capacitor voltage and zero inductor curent. Solve for: v 1 and v 2 at t=0+, v 1 and v 2 at t=, dv 1 /dt, dv 2 /dt at t=0+, d 2 v 2 /dt 2 at t=0+ History: at t=0- V c (0-)=0, i L (0-)=0, Network at t=0+: Capacitor acts as short circuit, inductor acts as open circuit. Hence v 1 (0+) =0, V 2 (0+)=0, i 1 (0+)= V/R 1, i 2 (0+)=0, V c (0+)=0 General network after switching: V=Ri+ 1/c (a) 0= 1/c + L di 2 /dt + R 2 i 2. (b) diff eqn (a) we get di 1 (0+)/dt = - V/CR 1 2 di 2 (0+)/dt=0, d 2 i 2 /dt 2 = V/R 1 LC dv 1 /dt= L (d 2 i 2 /dt 2 )= V/R 1 C dv 2 (0+)/dt=0 Solution: 0,0 0, vr 2 / (R 1 +R 2 ) L d 2 i 2 /dt 2 = L V / (LCR 1 ), 0 Rv/(R 1 CL)

11 Problem 9: In the network shown, switch K is opened at t=0 after the network has attained a steady state with switch closed. Find: Expression for voltage across the switch at t=0+ If the parameters are adjusted such that i(0+)=1 and di/dt at t=(0+) = -1, What is the value of the derivative of the voltage across the switch dv k /dt at t = 0+ Solution: History: switch is closed indicates that the voltage across R1 and C is zero, inductor acts as short ckt as steady state having been reached, There fore i(0-) = V/R 2, voltage across cap =0, Switch is opened: Inductor acts as a current source of value V/R 2, capacitor acts as short Hence: i(0+)= V/R 2 only as inductor doesnot allow any sudden change in current. Now general network after switching: V= R 1 i+ 1/c + R 2 i + L di/dt but R 1 i+ 1/c is V k V k (0+) = R 1 V / R 2 dv k /dt = R 1 di 1 /dt + i1/c = -R1 + 1/C Solution vr 1 /R 2 1/c R 1

12 LAPLACE TRANSFORMATION AND APPLICATIONS Laplace transformation It s a transformation method used for solving differential equation. Advantages The solution of differential equation using LT, progresses systematically. Initial conditions are automatically specified in transformed equation. The method gives complete solution in one operation. (Both complementary function and particular Integral in one operation) The Laplace Transform of a function, f(t), is defined as Where S is the complex frequency Condition for Laplace transform to exist is Unit step function,

13 Delta function Ramp function ( ) [ ] Laplace Transform of exponential function

14 ( ) [ ] [ ] ( ) [ ] [ ]

15 Laplace transform of derivative Consider a function f(t) W K T Let * + In general [ ] ( ) [ ] [ ] Laplace Transform of Integration [ ] [ ] [ ]

16 [ ] [ ] [ ] [ ] [ ] Laplace transform of some important functions

17 These equations tell us that transform of any function delayed to begin at time t=a, is times transform of the function when it begins at t=0. This is known as shifting theorem. Initial value Theorem It states that Proof Substituting in integration we have

18 Final value theorem Since s is not a function of t Letting s Hence on LHS Wave form synthesis Unit step function

19 Delayed unit step Delayed ve unit step

20 Waveform synthesis involving unit step function Rectangular pulse Laplace transform of periodic function Let f(t) be a periodic function with period T. Let f1(t),f2(t), f3(t).be the functions describing the

21 first cycle, second cycle, third cycle. Therefore by shifting theorem Rectangular wave of time period 2T [ ] [ ] [ ] [ ]

22 [ ] Half cycle of sine wave * + = * + ( ) ( ) [ ] Show that the transform of the square wave is

23 Ramp Function Ramp with slope A/T

24 Shifted ramp Shifted ramp with negative slope Addition of two ramp function Saw tooth waveform with slope A/T

25 For the waveform shown, show that the transform of this function is

26 Hence proved Triangular Waveform Trapezoidal wave [ ]

27 Find the Laplace transform of the waveform shown in figure Solution of networks using Laplace Transform 1) Consider a series RC network as shown in figure. It is assumed that the switch K is closed at t=0. Find the current flowing through the network. Solution Applying KVL, the equation for the circuit is The transform of the equation is [ ] If the capacitor is initially uncharged, the above equation reduces to the form Therefore [ ]

28 * + [ ] 2) Consider a series RL network as shown in figure. It is assumed that the switch K is closed at t=0. Find the current flowing through the network. Solution Applying KVL, the equation for the circuit is The corresponding transformed equation is Since i(0-)= 0, we have

29 To bring I(s) expression to the standard form, to take Laplace Inverse, let us apply partial fraction expansion for I(s) It can be found that A=V/L and B= Therefore [ ] ( ) 3) Consider a series RLC circuit with the capacitor initially charged to voltage V 0 =1 volt Solution By applying KVL, the differential equation of the circuit can be written as The corresponding Transformation equation is

30 Therefore * + Substituting R=1ohm, L=1H, C=1/2 F and V 0 =1 volt we have 4) In the circuit shown in figure, steady state is reached with switch K open. Obtain the expression for current when switch K is closed at t=0. Assume R1=1Ω,R2=1 Ω, L=1H V=10V. Ω Solution Applying KVL, with the switch is closed Taking Laplace transform of the above equation yields

31 Substituting the values of R1, L and i(0-) Applying Partial fraction Expansion Solving for A and B A=10 and B= Therefore 5) Derive the expression for current i(t) for the series RLC circuit shown. Assume zero initial conditions. Solution

32 The transformed equation is * + [ ] [ ] Where [( ) ] CASE 1: The roots are real and unequal CASE2: CASE 3 Therefore S1=S2 [ ] [ ] [ ]

33 [ ] The Transformed Networks The voltage current relation of network elements can also be represented in frequency domain. Resistor For a resistor, the voltage-current relationship is The Laplace transform of the above equation is Inductor For an inductor, the voltage-current relationship is The Laplace transform of the above equation is For an inductor, the voltage-current relationship can also be written as The Laplace transform of the above equation is

34 Equations (1) and (2) can be represented by the following circuits Capacitor For a capacitor the voltage-current relationship is Laplace transform of above equation voltage-current relationship can also be written as The Laplace transform of the above equation is Equations (3) and (4) can be represented by the following circuits

35 Determine the current in the inductor L1and L2 for the circuit shown below. The switch is closed at t=0 and the circuit has attained steady state before closing the switch. V1= 1 Volt, L1=2 H, L2=3 H, R1=R2=2Ω. Solution Before closing the switch the circuit has reached steady state. Hence the current through inductor L1 is = Hence the transformed network is shown below Therefore the loop equations are

36 [ ] [ ] [ ] By applying Cramer s rule By applying partial fraction expansion ( ) ( ) ( ) ( ) ( ) [ ]

37 Similarly By applying partial fraction expansion Therefore [ ] [ ] [ ] [ ]

38 Solution of networks with AC Excitation For the network shown in figure find the voltage across the capacitor when the switch is closed at t=0. Let R=2Ω, C=0.25 F and V(t)=0.5cost u(t) The transformed network is shown below. Since ( ) Voltage across the capacitor is given by ( )

39 Equating the numerators Equating the coefficients of, we ve A=0.4 B=0.2 and C=-0.4 2)Determine the current in the network when the switch is closed at t=0. Assume v(t)= 50 sin 25t, R=10 ohms, and L=5 H. Solution The transformed network is shown. Hence Since ( )

40 By applying Partial fraction expansion 3)For the network shown in figure find v(t), if the switch is closed at t=0. The transformed network is

41 [ ] By applying partial fraction expansion Evaluating the constants we have [ ] [ ] [ ] [ ] [ ]

42 For the network shown the switch has been in open position for long time and it is closed at t=0. Find the voltage across the capacitor. The transformed network at t=0- is Let us find the solution of the circuit with switch K open. Therefore V(0+)=5 V When the switch is closed at t=0 the transformed network is

43 By applying KCL we have We find In the network shown the switch is opened at t=0. Steady state is reached before t=0. Find i(t)

44 Solution At t=0- the transformed network is Applying KVL for outer loop When the switch is opened the transformed network is ( )

45 The network shown in figure has attained steady state with switch K open. The switch is closed at t=0. Determine v(t) The transformed network before closing the switch is Writing the nodal equation for V(s) [ ]

46 Now writing the nodal equation for V(s) [ ] [ ]

47 Mesh Analysis Mesh analysis is another technique of solving a given circuit. Mesh analysis uses KVL for solving the circuit. Mesh analysis can be applied only for planar circuits. Planar circuits are circuits that can be drawn on a plane surface with no wires crossing each other. A mesh is group of branches within a network joined so as to form a complete closed path such that there is no other closed path inside it. For the circuit shown below in Fig.1 there are two meshes, one mesh is formed with E 1, R 1, R 3, E 1 & the second mesh is formed with E 2, R 2, R 3, E 3. Within each mesh there is no closed path. Fig.1 Loop analysis also uses KVL but there is a difference between loop and a mesh. A loop can have a closed path within it but a mesh cannot. For example consider Fig.1 circuit, E 1,R 1,R 2,E 2,E 1 forms a loop. Mesh is a special case of loop which does not have a closed path within it. Kirchhoff s voltage law: The algebraic sum of the voltages around any closed path is zero. Procedure for Mesh analysis: 1. Indentify number of meshes in the given circuit. 2. For each mesh, give (name) one mesh current. Assign one current direction (clock wise or anti clockwise) for each mesh. 3. These mesh currents are the independent variables. Mesh Analysis Page 1

48 4. Apply KVL for each mesh and write the voltage equation. 5. When KVL is applied for 1 st mesh with mesh current named as I 1, then assume that I 1 >I 2, I 1 >I 3, I 1 > I 4 and so on. Similarly, when KVL is applied for 2 nd mesh assume I 2 >I 1,I 2 >I 3,I 2 >I 4 & when KVL is applied for 3 rd mesh I 3 >I 1,I 3 >I 2,I 3 >I 4 6. Solve the voltage equations to get the values of mesh currents. 7. Using mesh current values calculate the current through each element and voltage across each element. Example1: Find current through 10Ω resistor 4Ω resistor using mesh analysis for the circuit shown in Fig1a. Fig.1a Solution: Identify number of meshes; assign one mesh current for each mesh with direction. Apply KVL to each mesh. Fib.1b Mesh Analysis Page 2

49 Mesh-1 equation 4( i1 i2) i1 4i (1) Mesh-2 equation 5i2 9i2 10( i2 i3) 4( i2 i1) 0 4i2 28i2 10i (2) Mesh-3 equation 1i3 7i3 3 10( i3 i2) 0 0i1 10i2 18i (3) Solving all three above equations i1 2.22A i2 0.47A i A Current through 10Ω resistor i2 i i2 i A Example2: SUPER MESH Concept Current through 4Ω resistor i1 i i1 i2 1.75A Find the current flowing through 4Ω resistor for the circuit shown in Fig. 2a Mesh Analysis Page 3

50 Fig.2a Solution: Identify number of meshes; assign one mesh current for each mesh with direction. Apply KVL to each mesh. Fig.2b There are 3 meshes in the given circuit. In the 1 st mesh there is one current source, which is common between mesh1 & mesh3. Voltage across the current source is unknown; KVL cannot be applied in that case, hence assume that current source is removed. A new mesh will be formed, this new mesh is called super mesh, this has been shown in the Fig.2c. In the super mesh in few components I 1 will be flowing and in few components I 3 will be flowing. Write the super mesh equation. Super mesh equation: 4( i1 i2) 10( i3i2) 1i3 7i i1 14i2 18i (1) Mesh Analysis Page 4

51 Mesh-2 equation 5i2 9i2 10( i2 i3) 4( i2 i1) 0 Fig.2c 4i1 28i2 10i (2) Now there are 2 equations with 3 unknown variables. A third equation is required to get a unique solution. The 3 rd equation will be an equation related to the current source. Equation3 Actual current source direction is upwards. I 3 is flowing upwards & I 1 is flowing down wards. The net current flowing through current source in terms of mesh currents will be the difference of I 3 and I 1, resultant current should flow in the direction of current source. Hence the equation will be i3i ( 3) Solving the above three equations: i A Current through 4Ω resistor is i2. 105A i A Example3: With Dependent Sources i2i1.105 ( 1.933) A Find the value of Vx of the circuit shown in the Fig.3a. Mesh Analysis Page 5

52 Solution: Fig.3a Indentify the number of meshes and assign mesh current directions. There are 2 meshes, mesh current directions shown in the Fig. 3b. Fig.3b There is a current source between mesh 1 and mesh2. Hence a super mesh will be formed here. Super mesh circuit is shown in Fig.3c. Super mesh equation 2i1 4i i1 4i ( 1) Fig.3c Mesh Analysis Page 6

53 Equation related to current source i2 i1 0.1( Vx) Vx 4( i2) (2) Solving above 2 equations Example4: Calculate the mesh currents i1, i2, i3 for the circuit shown in Fig.4a Solution: Fig.4a In mesh 1 there is one current source- 15A, which is exclusively for only mesh1. Hence through that branch only mesh current i1 will flow, therefore i1 will be the magnitude of the current source. There is no need to write KVL equation for mesh1. i (1) Mesh 2 equation 2i2 3( i2 i3) 1( i2 i1) 0 1i1 6i2 3i3 i A i A Vx 4( 9.615) 38.46V (2) Mesh Analysis Page 7

54 Mesh3: There is a dependent current source common to mesh3 & mesh1. KVL cannot be applied for mesh3. Super mesh will not be formed because there won t be any closed path, if dependent current source is removed and also 15A current source is removed from the circuit. The 3 rd equation will be related to mesh currents i3, i1 & the dependent current source. Solving above 4 equations 1 i3 i1 Vx ( 3) 9 Vx 3( i3 i2) (4) i1 15A i2 11A i3 17A Mesh Analysis Page 8

55 Node voltage analysis The node voltage method or nodal analysis is a very power powerful approach for circuit analysis and it is based on the application of KCL and Ohm s law. Equilibrium equations are written using KCL at independent nodes of the circuit. A node refers to a point where two or more circuit elements meet. Procedure for nodal analysis: 1. Choose one node of the circuit as the common node (or the datum or reference) node. 2. Number (or label) the remaining nodes (V 1, V 2. Or V a,v b.). 3. Each node voltage (eg. V 1 or Va) is with respect to the reference node. These node voltages are the independent variables. 4. Apply KCL at each node (except reference node) and write equilibrium equation. If we have four nodes (except reference node) then we get 4 equilibrium equations. 5. When KCL is applied at node1 (V 1 ) it is assumed that V 1 > V 2, V 1 >V 3 and so on. The same rule applies to all nodes, i.e, V 2 >V 1, V 2 >V 3.. when KCL is applied at node2. 6. Assume current leaving the node as positive and current entering the node as negative or vice versa. Same assumption is to be followed throughout. 7. Solve the equilibrium equations using Cramer s rule or any other method to get node voltages. Use these node voltages to calculate current flowing through each branch of the circuit. Example1: Compute the voltage across 7A current source using nodal analysis for the circuit shown in Fig.1 Fig.1 Nodal analysis Page 1

56 Solution: Select one node as reference node and number the remaining nodes. Assume current leaving the node as +ve and current entering the node as ve. Applying KCL at node1 Equation at node1 Assuming v1 v2, v1 v3 Fig.1a v1 v2 v1 v Rearranging the terms 3 1 v 1( ) v2 v3( ) (1) Applying KCL at node2 Equation at node2 v2 v1 Assuming v2 v3 v2 v1 v2 v2 v Rearranging the terms v 1( 1) v2(1.58) v3(0.25) (2) Applying KCL at node3 Equation at node3 Assuming v3 v1 v3 v2 Nodal analysis Page 2

57 v3 v3 v2 v3 v Rearranging the terms v 1(.5) v2(.25) v3(.95) (3) All three equations v1(1.5) v2 v3(.5) 3 v1 v2(1.58) v3(.25) 0 v1(.5) v2(.25) v3(.95) 7 Arranging in the matrix form Using Cramer s rule v v3 is the voltage across 7A current source. v1 3 v2 = 0 v3 7 = 11.49V Similarly, solving for V 1 &V 2 we get V 1 =5.27V, V 2 = 5.15V Example2: Find the voltage across 1Ω resistor and 5Ω resistor using nodal analysis for the circuit shown below in Fig.2. Fig.2 Solution: Nodal analysis Page 3

58 Redraw the circuit to remove the jump Consider one node as reference node and number the remaining nodes Equation at node1 v1 v2 v1 v v1(.583) v2(.33) v3(.25) 11 Equation at node 2 v2 v2 v1 v2 v v1(.33) v2(1.476) v3(.143) 3 Equation at node3 v3 v2 v3 v3 v v1(.25) v2(.143) v3(.593) 25 All three equations v v v3 25 Solving V 1 =5.33V, V 2 =7.7V, V 3 =46.27V Voltage across 1Ω resistor is V 2 =7.7V, Example3: Fig.2a Voltage across 5Ω resistor is V 3 =46.27V Compute the voltage across each current source for the circuit in the Fig.3. Nodal analysis Page 4

59 Solution: Fig.3 Identify the nodes and select one node as reference node. Fig.3a There is one independent voltage source between node1 & node2. If we try to apply KCL at node1, it is not possible to write expression for current flowing through the branch where 5V voltage source is connected. In such cases, short node1 & node2. Shorting node1 & node2 a Super Node will be formed. Apply KCL at the Super node. Fig.3b Nodal analysis Page 5

60 Equation(KCL) at Super node v1 v1 v v2 v2 v v 1 6v (1) There are two variables in the above equation and to solve we need one more independent equation. The other equation will be relation between node1, node2 & the voltage source. Observing the polarities of the voltage source connected between node1 & node2, we can tell that voltage difference between node1 and node2 is 5V. Same thing is put in the equation form Then v 1 v (2) This is the second equation. There are two equations with 2 unknowns, if solved unique solution will be obtained. Solving the two equations V 1 =5.375V; V 2 = 0.375V Voltage across 4A current source is 5.375V and voltage across 9A current source is 0.375V. Example4: Find the value of Vx using nodal analysis for the circuit shown in the Fig. 4 Solution: Fig.4 Mark nodes and select one node as reference node. Nodal analysis Page 6

61 Super node (Node 1 & Node 4 ) equation-1 Super node ( Node 2 & Node 3 ) equation-2 Fig.4a Equation3 & equation4 related to independent voltage sources Solving above equations Example 5: v1 v1 v2 v4 v v1(0.15) v2(0.05) v3(0.08) v4(0.08) 5...; (1) v2 v1 v2 v3 v v1( 0.05) v2(0.09) v3(0.08) v4(0.08) 5...( 2) v3 v ( 3) v4 v ( 4) substituting v3 150 v2 v V, v V v V, v3 150 v V vx v V Determine the power supplied by 2A source for the circuit shown in Fig.5 using nodal analysis. Nodal analysis Page 7

62 Solution: Fig.5 Mark the nodes and select one node as reference node. Node-1 equation Fig.5a v1 1 v v1( 1 ) v2( 1 ) (1) 4 4 node2 v (2) node3 v3 v3 3 v (0) v1 v2( ) v3( ) (3) 2 2 Solving above three equations Nodal analysis Page 8

63 v1 21V, v2 4V v3 1.67V v3 v ( 21) v3 v V Power Supplied by 2A source is =22.67 X 2 =45.34W Example6: With dependent sources Determine i 1 using nodal analysis for network shown in Fig.6. Solution: Select reference node and name the other nodes. Fig.6 Fig.6a Nodal analysis Page 9

64 Equation at node A Va 4 Va 0.5i rearrangin g i1 3 Va( )...( 1) Equation related to Dependent source Solving above two equations Example 7: 4 i1 Va......( 2) 2 Va 0.727V Va i1 2 2 i A Determine all node voltages with respect to reference voltage for the circuit shown in the Fig.7. Fig.7 Nodal analysis Page 10

65 Solution: Mark the super node and write nodal equations. node2 Solving above equations Fig.7a v1 12V ( 1) v2 v1 v2 v (2) supernode v3 v2 v4 v4 v1 0.5Vx...(3) v3 v4 0.2Vy ( 4) 0.2Vy 0.2( v4 v1) (5) & Vx v2 v (6) v1 12V v2 4V v3 0V v4 2V Nodal analysis Page 11

66 Example8: (Nodal analysis using AC sources) Find the value of Vx using nodal analysis for the circuit shown in the Fig.8. Solution: Mark the nodes. Node1 equation Node2 equation Fig.8 Fig.8a v1 v1 v ( 1) j2 j1 v2 v2 v1 Vx ( 2) 2 j1 Dependent source equation Vx v (3) Nodal analysis Page 12

67 Rearranging the above 3 equations v1( ) v2( ) 160 j2 j1 j1 v1( ) v2( j1 2 Solving the above two equations 1 ) 0 j1 v Nodal analysis Page 13

68 TWO POTR NETWORK

69 Two Port Network Overview The concept of a two-port network. The relationship between input and output current and voltages. Combinations of networks in series, parallel, and cascaded. Two Port Network A pair of terminals through which a current may enter or leave a network is known as a port. Two terminal devices or elements (such as resistors, capacitors, and inductors) results in one port network. Most of the circuits we have dealt with so far are two terminal or one port circuits. (Fig. 1(a)) A two port network is an electrical network with two separate ports for input and output. It has two terminal pairs acting as access points. The current entering one terminal of a pair leaves the other terminal in the pair. (Fig. 1(b)) Fig. 1(a) Fig. 1(b) To characterize a two-port network requires that we relate the terminal quantities V 1, V 2, I 1, and I 2. Out of these four, only two are independent. The terms that relate to these voltages and currents are called parameters. Impedance and admittance parameters are commonly used in the synthesis of filters. They are also important in the design and analysis of impedance-matching networks and power distribution networks. Three types of two-port parameters are examined here: impedance, admittance & transmission. Z PARAMETER Z parameter is also called impedance parameter and the unit of Z parameters is ohm (Ω) The black box replaced with Z-parameter is as shown below. 2

70 A two-port network may be either voltage driven or current driven The terminal voltages can be related to the terminal currents as: V Z I Z I V Z I Z I The values of the parameters can be evaluated by setting the input or output port open circuits (i.e. set the current to zero). z z V V I1 I I 0 2 I 0 V I1 I I 0 2 I 0 z z V 2 1 3

71 These are referred to as the open-circuit impedance parameters. These parameters are as follows: z 11 Open circuit input impedance z 12 Open circuit transfer impedance from port 1 to port 2 z 21 Open circuit transfer impedance from port 2 to port 1 z 22 Open circuit output impedance When z 11 =z 22, the network is said to be symmetrical. It should be noted that an ideal transformer has no Z - parameters. The equivalent circuit for two port networks is shown below: 1. Find the Z parameter of the circuit below. Solution: When I 2 = 0(open circuit port 2). Redraw the circuit. 4

72 V 120 I...(1) 1 b V I...(3) a Ib I1...(2) Ia I1...(4) sub (1) (2) sub (4) (3) Z V I1 Z V I1 When I 1 = 0 (open circuit port 1). Redraw the circuit. V I...(1) 160 Ix I2...(2) 400 sub (1) (2) V 2 Z22 96 I2 In matrix form: 2. Find the Z parameter of the circuit below x Z V 120 I...(3) Iy I2...(4) 400 sub (4) (3) V 72 1 Z12 I2 y 5

73 Solution: i) I 2 = 0 (open circuit port 2). Redraw the circuit. ii) I 1 = 0 (open circuit port 1). Redraw the circuit. In matrix form; Z (2 j4) 0 10 (16 - j8) V I (2 j4) 1 1 V 1 Z 11 (2 j4) I1 Y PARAMETER Y Parameter also called admittance parameter and the unit is Siemens (S). The black box that we want to replace with the Y-parameter is shown below. V 2 Z 0 V (short circuit) 10I V I 2I 1 Z I2 2 2 V V - 10I j20 10 V j V 2 Z 22 (16-j8) I2 6

74 I y V y V I y V y V I1 y 11 y12 V1 V1 y I y y V V

75 I I y and y V1 V V 0 1 V y 11 = Short-circuit input admittance y 21 = Short-circuit transfer admittance from port 1 to port 2 I y and y V2 V V 0 2 V y 12 = Short-circuit transfer admittance from port 2 to port 1 y 22 = Short-circuit output admittance 1. Find the Y parameter of the circuit shown below. Solution: i) V 2 = 0 V 20 I...(1) 1 a 5 Ia I1...(2) 25 sub (1) (2) Y I 1 11 V1 1 4 S I V 5I 1 2 I 2 Y21 V1 1 S 5 8

76 ii) V 1 = 0 In matrix form, Y S Find the Y parameters of the circuit shown. Solution: i) V 2 = 0 (short circuit port 2). Redraw the circuit. V 2 15 I...(3) 5 Ix I2...(4) 25 sub (3) (4) Y 2 22 V2 x I 4 S 15 V 5I 2 1 I 1 Y12 V2 1 S 5 9

77 Applying KVL to the loop consisting of dependent source 10I 2 and 10 Ω resistor we get, 10I 2+ 10I 2 = 0 or I 0 2 V (2 j4)i 1 1 I 1 1 Y 11 V1 2 j4 2 Y 21 V1 I 0 S ii) V 1 = 0 (short circuit port 1). Redraw the circuit. 2 I 1...(1) I 2-10I 2 j4 V V - 10I -j (0.1 - j0.2) S 1 1 2I 2 V 2...(2) 10 -j20 I2 Y 22 (0.05 j0.025) S V sub (2) (1) I 2 1 Y 12 (-0.1 j0.075) S V2 In matrix form; j j0.075 Y S j0.025 T (ABCD) PARAMETER T parameter or also ABCD parameter is a another set of parameters relates the variables at the input port to those at the output port. T parameter also called transmission parameters because this parameter are useful in the analysis of transmission lines because they express sending end variables (V 1 and I 1 ) in terms of the receiving end variables (V 2 and -I 2 ). 10

78 The black box replaced with T parameter is as shown below. V AV BI I CV DI V1 A B V2 V2 T I C D I I T terms are called the transmission parameters or simply T or ABCD parameters, and each parameter has different units. V 1 A V 2 I2 0 I 1 C V 2 I2 0 V1 B I I D I 2 V V2 0 A=open-circuit voltage ratio C= open-circuit transfer admittance (S) B= negative short-circuit transfer impedance () D=negative short-circuit current ratio Find the ABCD parameter of the circuit shown below. 11

79 Solution: i) I 2 = 0, ii) V 2 = 0, In matrix form; T S 1.4 Conversion from Y to Z Parameters: For the Y parameters we have, I = Y V..(a) For the Z parameters we have, V = Z I..(b) From (a), V = Y -1 I..(c) Comparing (b) & (c) we have, Z = Y -1 Therefore, Where ΔY = det Y Y 22 -Y 12-1 Z11 Z12 Z= Y = = ΔY ΔY Z21 Z22 -Y 21 Y 22 ΔY V 10I C ΔY I V 2 V 2I V A 2 0.1S V 6 V 2 V V 10 5 V 1.2 V I 10 I 14 I D I B V 2I 10 I I V 12I 10I 14 V 12 I 10I 10 V I 12

80 Conversion Table Network Functions Contents: Network functions of one port and two port networks Properties of poles and zeros of network functions. V-I and I-V relations of basic elements: Component Symbol V-I relation I-V relation Resistor Capacitor v ( t) i ( t) R R R 1 vc( t) ic( t) dt C () v () R t ir t R dvc () t ic () t C dt () di () L t vl t L dt 1 il( t) vl( t) dt L 13

81 Inductor Transform Impedance (Resistor) Transform Impedance (Inductor) Transform Impedance (Capacitor) A network function is the Laplace transform of an impulse response. Its format is a ratio of two polynomials of the complex frequencies. Consider the general two-port network shown in Figure (a). The terminal voltages and currents of the two-port can be related by two classes of network functions, namely, the driving point functions and the transfer functions. 14

82 Network Functions: Driving point impedance & admittance functions I. The driving point functions relate the voltage at a port to the current at the same port. II. These functions are a property of a single port. III. For the input port the driving point impedance function Z IN (s) is defined as: Z IN (s) = V IN (s)/i IN (s) Transfer function This function can be measured by observing the current I IN when the input port is driven by a voltage source V IN (Figure (b)). The driving point admittance function Y IN (s) is the reciprocal of the impedance function, and is given by: Y IN (s) = I IN (s)/v IN (s) The output port driving point functions are defined in a similar way. The transfer functions of the two-port relate the voltage (or current) at one port to the voltage (or current) at the other port. The possible forms of transfer functions are: a) The voltage transfer function, which is a ratio of one voltage to another voltage. G 21 (s) = V 2 (s)/v 1 (s) b) The current transfer function, which is a ratio of one current to another current. α 21 (s) = I 2 (s)/i 1 (s) c) The transfer impedance function, which is the ratio of a voltage to a current. Z 21 (s) = V 2 (s)/i 1 (s) d) The transfer admittance function, which is the ratio of a current to a voltage. Y 21 (s) = I 2 (s)/v 1 (s) The general form of a network function is: 15

83 All the coefficients a i and b i are real An alternate form of H(s) In the above expression z 1, z 2,..., z n are called the zeros of H(s), because H(s) = 0 when s = z i. The roots of the denominator p 1, p 2,..., p m are called the poles of H(s). It can be seen that H(s) = at the poles, s = p i. The poles and zeros can be plotted on the complex s plane (s = σ + jω), which has the real part σ for the abscissa, and the imaginary part jω for the ordinate. Properties of all Network Functions: 1. Network functions are ratios of polynomials in s with real coefficients. A consequence of this property is that complex poles (and zeros) must occur in conjugate pairs. Consider a complex root at (s = -a jb) which leads to the factor (s + a + jb) in the network function. The jb term will make some of the coefficients complex in the polynomial, unless the conjugate of the complex root at (s = -a + jb) is also present in the polynomial. The product of a complex factor and its conjugate is which can be seen to have real coefficients. 2. The networks must be stable: 16

84 A bounded input excitation to the network must yield a bounded response. Or The output of a stable network cannot be made to increase indefinitely by the application of a bounded input excitation. Stability of the general network function H(s) 1. If the network function has a simple pole on the real axis, the impulse response due to it (for t >= 0) will have the form: For p1 positive, the impulse response is seen to increase exponentially with time, representing an unstable circuit. Thus, H(s) cannot have poles on the positive real axis. Suppose H(s) has a pair of complex conjugate poles at s = a +/- jb. The contribution to the impulse response due to this pair of poles is If a is positive, corresponding to poles in the right half s plane, the response is seen to be an exponentially increasing sinusoid Therefore, H(s) cannot have poles in the right half s plane. An additional restriction on the poles of H(s) is that any poles on the imaginary axis must be simple. Higher order poles on the jω axis will also cause the network to be unstable. 17

85 Summary: The network functions of all passive networks and all stable active network must be rational functions in s with real coefficients. May not have poles in the right half s plane. May not have multiple poles on the jω axis. Check to see whether the following are stable network functions: The first function cannot be realized by a stable network because one of the coefficients in the denominator polynomial is negative. It can easily be verified that the poles are in the right half s plane. The second function is stable. The poles are on the jω axis (at s = +/- 2j) and are simple. Note that the function has a zero in the right half s plane; however, this does not violate any of the requirements on network functions. 18

86 1 FORMAT-1B Subject: Electric Circuit Analysis Unbalanced Three Phase Systems Review of A.C three phase system Three phase voltages are generated by the alternator(ac generator), when the rotating magnetic field sweeps across the stator conductors, hence emf s are induced in all the three phases, which are separated by 120 degrees. E A = E m Sinωt E B = E m Sin(ωt ) E C = E m Sin(ωt ) or E c = E m Sin(ωt )

87 2 Phase Sequence It is the order in which the maximum voltages in 3-phase are in sequence that is A, B, C. Balanced 3-Phase Supply When all the three voltages in 3-phase supply having same magnitude but differs in phase by 120 degrees with respect to one another is called balanced 3-phase supply. OR Also in all the three lines the same and equal magnitude of current flows n balanced 3-Phase Supply The magnitude and phase angle in three phase supply are not similar that system is called un balanced 3- phase supply OR In all the three lines the magnitude of currents are different (un equal). Unbalanced 3-Phase Load Suppose among the three phase impedances, if any One of the phase impedance is different then it is called as unbalanced load.

88 3 Star connected 3-phase system In star connected three phase system, all the three ends of coils are joined at one point N (Neutral) other 3 ends being free. Consider E AB, E BC, E CA are the line voltages called as E L. E AN, E BN, E CN are the phase voltages called as E ph. In star connection Line current is equal to phase current I L = I Ph Delta connected 3-phase system All the three coils are connected end to end to form delta connected three phase system Consider I A, I B, I C, are the line Currents called as I L.

89 4 I AB I BC I CA are the phase currents called as I ph. In Delta connection Line Voltage is equal to phase Voltage E L = E Ph In delta system line current is 17.32A and phase current is10a. the reason for this difference in current is that current flows through different windings at different times in a 3-phase circuit. During some periods of time current will flow between two lines only. At other times current will flow from two lines to the third. Delta connection is similar to parallel connection because there is always more than one path for current flow. UnBalanced 3-Phase System : The loads in all three phases of either Star OR Delta connection are not identical to each other in all respects are called unbalanced load. In Unbalanced load the Line current in star and Phase current in Delta will be different. The line or phase current giving rise to the flow of neutral current in this context supply neutral and the star point is inter connected then it is called 3-phase 4-wire system. In unbalanced loading, due to the flow of unequal currents in each of the phase and voltage appears at the neutral. Advantages of 3-phase system over 1-phase: 1.3-phase system is more efficient than 1-phase system 2.Cost of 3-phase equipment is less than 1-phase comparatively 3.3-phase system is 1.5 times more than 1-phase. 4.Harmonics in 3-phase system is minimum 5.Economy of 3-phase power transmission is cheaper than 1-phase 6. 3-phase system produces uniform torque but 1-phase gives pulsating torque 7.All 3-phase motors are self starting but 1-phase motors are not self starting 8.3-phase apparatus are compact in size, requires less material as compared to 1-phase.

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92 7 Example:1 Find the real power and neutral current for the given 3-phase star connected unbalanced load which is connected to 3-phase balanced supply voltage of 400V.

93 8 Example:2 Find the line currents and power drawn for the given 3-phase balanced star connected supply voltage of 50Volts and having unbalanced load impedances.

94 9 Ex.3-Find the neutral current which flows in the 3-phase unbalanced star connected load having balanced 3-phase supply voltage of 100Volts in ACB sequence, 50Hz.

95 10 Example:4. Find the currents and power drawn in the given unbalanced Delta connected load supplied with balanced voltage of 100V.

96 11 Example:5. Find currents in the star connected unbalanced load.

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98 13 Example:6. Find line currents using loop analysis.

99 14 Example:7. Find currents using loop analysis.

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