72560322 Network Analysis by Van Valkenburg Chap 5 Solution Manual.pdf

8/18/2019 72560322 Network Analysis by Van Valkenburg Chap 5 Solution Manual.pdf

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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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PROBLEMS Q#5.1: In the network of the figure, the switch K is closed at t = 0 with the capacitor uncharged. Find values for i, di/dt and d2i/dt2 at t = 0+, forelement values as follows: V 100 V R 1000 W C 1µ F

+ V -

R C

Switch is closed at t = 0 (reference time)

e know Voltage across capacitor before switching = VC(0-) = 0 V According to the statement under Q#5.1. VC(0+) = VC(0-) = 0 V V 100 iC(0+) = i(0+) = = = 0.1 Amp. R 1000 Element and initial condition Equivalent circuit at t = 0+ Sc C Switch Drop Rise i(0+) Short circuit Drop

Applying KVL for t ³ 0 Sum of voltage rise = sum of voltage drop 1


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} ò idt C Differentiating with respect to `t' di i R + =0 dt C di(0+) -i(0+) = [eq. 1] dt CR By putting the values of i(0+), C & R di(0+) -(0.1) = dt (1 µ F)(1 kW ) V = iR + di(0+) = -100 Amp/sec dt Differentiating eq. 1 with respect to `t' d2i(0+) -di(0+) 1 = dt2 dt CR Putting the corresponding values d2i(0+) = 100, 000 amp/sec2 2 dt Q#5.2: In the given network, K is closed at t = 0 with zero current in the inductor. Find the values ofi, di/dt, and d2i/dt2 at t = 0+ if R 10 W L 1H V 100 V

2

K + V L R

Key closed at t = 0


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} iL(0+) = iL(0-) = i(0+) = 0 Amp According to the statement under Q#5.2: Drop Rise Open circuit i(0+) Drop

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Element and initial condition Applying KVL for t ³ 0 Sum of voltage rise = sum ofvoltage drop Ldi V = iR + dt Ldi = V ± iR dt di V - iR = [eq. 1] dt L di(0+) V ± i(0+)R = dt L Putting corresponding values di(0+) V ± (0)R = dt L di(0+) V = dt L di(0+) 100 = dt 1 di(0+) = 100 Amp/sec dt Differentiating [eq. 1] d2i d V iR = -

Equivalent circuit at t = 0+ oc


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} dt2 dt L L d2i -Rdi = dt2 Ldt d2i(0+) -Rdi(0+) = dt2 Ldt Putting corresponding values d2i(0+) = -1, 000 Amp/sec2 dt2 Q#5.3: In the network of the figure, K is changed from position a to b at t = 0. Solve for i, di/dt, and d2i/dt2 at t = 0+ if R 1000 W L 1H C 0.1 µ F V100 V a b V C L K R

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Equivalent circuit at t = 0+

b sc Applying KVL for t ³ 0 Sum of voltage rise = sum of voltage drop 1 Ldi


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Rdi(0+) + i(0+) +2

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Ld2i(0+) =0

dt C dt Putting corresponding values d2i(0+) = 9, 00000 Amp/sec2 dt2

Q#5.4: For the network and the conditions stated in problem 4-3, determine the values of dv1/dt and dv2/dt at t = 0+. R K V1 C1 C2 V2

V1 V2 R C1 C2 After switching: V1

2V 1V 1W 1F ½F V2

Applying KCL at node 1: V1 ± V2 + C1 R dt V1(0+) ± V2(0+) dV1 =0 dV1(0+)


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} + C1 =0 R dt Putting corresponding values 2V±1V dV1(0+) + (1 F) =0 1 dt Simplifying dV1(0+) = -1 Volt/sec dt At node 2: V2 ± V1 dV2 + C2 =0 R dt V2(0+) ± V1(0+) dV2(0+) + C2 =0 R dt Putting corresponding values 1V±2V dV2(0+) + (1/2) =0 1 dt Simplifying dV2(0+) = 2 Volt/sec dt According to KCL ªSum of currents entering into the junction must equal to the sum of the currents leaving the junctionº Q#5.5: For the network described in problem 4.7, determine values of d2v2/dt2 and d3v2/dt3 at t = 0+. NODE R1 R2 V1 R1R2 C VC(0-) = VC(0+) = 0 V C 10 W 20 W 1/20 F + V2

7


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Applying KCL at NODE fort ³ 0 V2 dV2 V2 ± V1 +C + = 0 ¼ (1) R2 dt R1 At t = 0+ V2(0+) dV2(0+) V2(0+) ± V1(0+) + + =0 R2 dt R1 V1 = e-t Volts V2 = VC(0+) = 0 Volts 0 dV2(0+) 0 ± e-0+ +C + =0 R2dt R1 Simplifying dV2(0+) = 2 Volt/sec dt Differentiating eq. 1 with respect to `t'1 dV2 d2V2 1 dV2 dV1 + C + =0 R2 dt dt2 R1 dt dt 1 20 1 20 3 20 dV2 + dt dV2 + dt dV2 + dt 20 dt 1 + 20 dt 20 dt2 1 + 20 dt 20 dt2 2

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1

d2V2 +

1

dV2 -

d(e-t) =0 dt 1 10 =0 d(e-t) =0 dt

20 dt2 1 d2V2 + 20 dt 12

10 dt 1 dV2

10 dt 1 10 dt d(e-t)

d2V2

d(e-t) = -e-t 3 dV2

d2V2 + 0.1e-t = 0 ¼ (2)

At t = 0+ 3 dV2(0+)

d2V2(0+) + 0.1e-t(0+) = 0


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Simplifying d2V2(0+) = -8 Volt/sec2 dt2 Differentiating eq. 2 3 d2V2 1 d3V2 + 20 dt2 20 dt3 At t = 0+ 3 d2V2(0+) +2 3

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- 0.1e-t = 0

1

d3V2(0+) - 0.1e-t(0+) = 0

20 dt 20 dt Putting corresponding values and simplifying d3V2(0+) = 26 Volts/sec3 dt3 Q#5.6: The network shown in the accompanying figure is in the steady state with the switch k closed. At t = 0, the switch is opened. Determine the voltage across the switch, VK, and dVK/dt at t = 0+. SOLUTION: VK

R L C V Equivalent network before switching

1W 1H ½F 2V


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i(0+)

sc

V iL(0+) = iL(0-) = i(0+) = R 1 Also VK = VC = C dVK i = dt C At t = 0+ dVK i(0+) = dt C dVK 2 = dt (1/2) dVK = 4 Volts/sec dt ò idt =

2 = 2 Amp 1

Q#5.7: In the given network, the switch K is opened at t = 0. At t = 0+, solve for the values of v, dv/dt, and d2v/dt2 if I 10 A R 1000 W C 1µ F V


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Switch is opened at t = 0 Equivalent circuit at t = 0-

No current flows through R so VC(0-) = VC(0+) = i(0-)R = V(0+) Here i(0-) = 0 VC(0-) = VC(0+) = (0)R VC(0-) = VC(0+) = 0 Volts For t ³ 0; according to KCL at V VdV +C = I ¼ (1) R dt At t = 0+ V(0+) dV(0+) +C =I R dt Simplifying dV(0+) = 107 Volts/sec dt Differentiating (1) with respect to `t'

1

dV +C

d2V =02

R dt dt At t = 0+ 1 dV(0+) d2V(0+) +C =0 2 R dt dt


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Simplifying d2V(0+) = -1010 Volts/sec2 dt2

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Q#5.8: The network shown in the figure has the switch K opened at t = 0. Solve for V, dV/dt, and d2V/dt2 at t = 0+ if I 1A Rhhfisejjdfir 100 W L 1H V

Equivalent circuit before switching:

Because iL(0-) = 0 A iL(0-) = iL(0+) = 0 A Therefore After switching (t = 0+) V


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So V(0+) = (I)(R) V(0+) = (1)(100) V(0+) = 100 Volts For t ³ 0 Applying KCL at node V V 1 + ò Vdt = I ¼ (1) R L Differentiating (1) with respect to `t' 1 dV V + = 0 ¼ (2 dt L At t = 0+ 1 dV(0+) V(0+) + =0 R dt L Simplifying dV(0+) = -10, 000 Volts/sec dt Differentiating (2) with respect to `t' 1 d2V 1 dV + =0 R dt2 L dt At t = 0+ 1 d2V(0+) + R dt Simplifying d2V(0+) = 1, 000, 000 Volts/sec2 dt2 2

1 L

dV(0+) =0 dt


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Q#5.9: In the network shown in the figure, a steady state is reached with the switch K open. At t = 0, the switch is closed. For the element values given, determine the value of Va(0-) and Va(0+). Circuit diagram: Vb Va

Equivalent circuit before switching:

sc


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Req = (10 + 20) 10 (10 + 20)(10) Req = (10 + 20) + 10 300 Req = 40 Req = 7.5 W After simplification Req

i(0-) 5V

V i(0-) = Req 5 i(0-) = 7.5 i(0-) = 0.667 Amp. i(0-) = iL(0-)


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Va(0-)

20 Va(0-) = (5) (10 + 20) Va(0-) = 3.334 Volts Also equivalent network at t = 0+ t = 0+

Vb Va


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Applying KCL at node Va Va ± 5 Va - Vb Va + + =0 10 20 10 After simplification weget Vb = 5Va ± 10 ¼ (i) Applying KCL at node Vb Vb - Va Vb ± 5 2 + + =0 20 10 3 9Vb ± 3a + 10 = 0 ¼ (ii) Substituting value of Vb from (i) into (ii) 9[5Va ± 10] ± 3Va + 10 = 0 45Va ± 90 ± 3Va + 10 = 0 42Va ± 80 = 0 Va(0+) = 1.905 Volts Q#5.10: In the accompanying figure is shown a network in which a steady state is reached with switch Kopen. At t = 0, the switch is closed. For the element values given, determine the values of Va(0-) and Va(0+).

CIRCUIT DIAGRAM: Vb Va K


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Equivalent circuit before switching

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Req

VC(0-) = Va(0-) = 0 V Equivalent network at t = 0+ Vb Va

+ -

Applying KCL at node Va


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Va ± 5 Va - Vb Va + =0 10 20 10 After simplification we get Vb = 5Va ± 10 ¼ (i) At Vb = 5 V 5 = 5Va ± 10 + Va(0+) = 3 Volts

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Q#5.11: In the network of figure P5-9, determine iL(0+) and iL(¥ ) for the conditions stated in problem 5-9. Equivalent circuit before switching:

sc

Req = (10 + 20) 10 (10 + 20)(10)


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Applying KCL at node Va Va ± 5 Va Va + + =0 10 20 10 After simplification we get Va(¥ ) = 2 Volts V iL(¥ ) = Va + 10 20 After simplification we get iL(¥ ) = 0.6 Amp.

iL(¥ )

Q#5.12: In the network given in figure p5-10, determine Vb(0+) and Vb(¥ ) for theconditions stated in Prob. 5-10. At t = 0-, equivalent network is


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VC(0-) = VC(0+) = 5 Volts Also equivalent network at t = ¥ is

Vb Va

According to KCL at Va Va ± 5 Va - Vb Va + + =0 10 20 10


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} After simplification we get Va = 0.2Vb + 2 ¼ (i) According to KCL at Vb Vb ± 5 Vb ± Va + =0 10 20 After simplification we get 3Vb ± Va ± 10 = 0 Putting the value of Va we get

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Vb(¥ ) = 4.286 Volts Q#5.13: In the accompanying network, the switch K is closed at t = 0 with zero capacitor voltage and zero inductor current. Solve for (a) V1and V2 at t = 0+, (b) V1 and V2 at t = ¥ , (c) dV1/dt and dV2/dt at t = 0+, (d) d2V2/dt2 at t = 0+. CIRCUIT DIAGRAM

R1 V1 L

C V2 R2

Equivalent network after switching


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According to statement under Q#5.13 At t = 0VC(0-) VC(0+) iL(0-) iL(0+) Equivalent network at t = 0+ V2(0+) = iL(0+)(R2) V2(0+) = (0)(R2) V2(0+) = 0 V V1(0+) +V2(0+) = VC(0+) Putting corresponding values V1(0+) + 0 = 0 V1(0+) = 0 Equivalent circuit at t = ¥ V1(¥ ) = iR2R2 V iR2 = R1 + R2 VR2 V1(¥ ) = R1 + R2 VC(¥ ) = V ± iR1VR1 VC(¥ ) = V ± R1 + R2 After simplification we get VR2 0V 0V 0A 0A


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} VC(¥ ) = R1 + R2 Since VC(¥ ) = V1(¥ ) + V2(¥ ) VR2 V2(¥ ) = R1 + R2 V1(¥ ) = VC(¥ ) - V2(¥ ) VR2 VR2 V1(¥ ) = R1 + R2 V1(¥ ) = 0 Volts (c) Self justified

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V1

V2

According to KCL at node `V1' V ± V1 dV1 1 +C + ò V1 ± V2)dt = 0 ¼ (i) R1 dt L Differenng with respect to `t' 1 dV dV1 d2V1 1 +C + (V1 ± V2) = 0 ¼ (iii)


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} R1 dt dt dt2 L

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According to KCL at node `V2' V2 1 + ò V2 ± V1)dt = 0 ¼ (ii) R2 L Differentiating with pect to `t' 1 R2 As V1 = V1 + V2 By putting the value of V1 in (iii) & (iv) 1 R1 1 R1 1 R2 1 R2 1 R2 1 R2 dV dt dV dt dV2 + dt dV2 + dt dV2 + dt dV2 + dt L L 1 (±V1)= 0 ¼ (Vi) L 1 (V2 ± V1 - V2) = 0 L 1 (V2 ± (V1 + V2)) = 0 ¼ (iV) dt 1 (V2 ± V1) = 0 ¼ dt dV1 dt dV2 +C dt2

dV2 + dt

1 (V2 ± V1) = 0 ¼ (iV) L

d(V1 + V2) +C

d2(V1 + V2) dt2 d2V1

1 + (V1 + V2 ± V2) = 0 L 1 + L (V1) = 0 ¼ (V)

d2V2 + dt2

From (V) & (Vi) we can find the values of dV1/dt & dV2/dt. (d) Refer part C.


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Q#5.14: The network of Prob. 5-13 reaches a steady state with the switch K closed. At a new reference time, t = 0, the switch K is opened. Solve for the quantities specified in the four parts of Prob. 5-13. (a) Equivalent circuit before switching

At t = 0V iL(0-) = R1 + R2 VC(0-) = V ± iR1(0-)R1 Here iR1(0-) = iL(0-) VC(0-) = V ± iL(0-)R1 VR1 VC(0-) = V -


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} R1 + R2 VR2 VC(0-) = R1 + R2 V2 = iR2R2 V iR2(0-) = iR1(0-) = R1 + R2 VR2 V2 = R1 + R2

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VR2 V2(0-) = R1 + R2 V1(0-) + V2(0-) = VC(0-) V1(0-) = VC(0-) - V2(0-) VR2 VR2 V1(0-) = R1 + R2 R1 + R2 V1(0-) = 0 Volts Equivalent network after switching

V1(0+)

+ VC(0+) V2(0+)


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VC(0-) = VC(0+) VR2 VC(0+) = R1 + R2 V iL(0-) = iL(0+) = R1 + R2 V2(0+) = iL(0+)R2 VR2 V2(0+) = R1 + R2 V1(0+) + V2(0+) = VC(0+) V1(0+) = VC(0+) - V2(0+) VR2 V1(0+) = R1 + R2 V1(0+) = 0 Volts (b) Equivalent network at t = ¥ At t = ¥ capacitor will be fully discharged and acts as an open circuit. R1 + R2 VR2

Hence VC(¥ ) = 0 V iL(¥ ) = 0 A


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} V2(¥ ) = iL(¥ )R2 V2(¥ ) = (0)R2 V2(¥ ) = 0 Volt (c) For t ³ 0, the equivalent network is

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V1

V2 V2 Applying KCL at node `V1' 1 dV1 ò (V1 ± V2)dt + C = 0 ¼ (i) L dt d C dt (V1 + V2)R2 -V2 ¼ (ii) V1

As V1 = V1 + V2 1 d(V1 + V2) ò (V1 + V2 ± V2)dt + C =0 L dt 1 d(V1 + V2) ò V1dt + C =0


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} L dt Differentiating (i)with respect to `t' 1 V1 + C L dt2 d2V1 d2V2 +C dt2 = 0 ¼ (iii)

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Differentiating (ii) with respect to `t' d2 C dt2 (V1 + V2) = R2 dt -1 dV2 ¼ (iV)

After simplification we get the values of dV1/dt & dV2/dt. (d) Refer part c. Q#5.15: The switch K in the network of the figure is closed at t = 0 connecting the battery to an unenergized network. (a) Determine i, di/dt, and d2i/dt2 at t = 0+. (b) Determine V1, dV1/dt, and d2V1/dt2 at t = 0+.

K V0 i

V1

Equivalent circuit after switching


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Here VC(0-) = VC(0+) = 0 Volt iL(0-) = iL(0+) = 0 A i(0+) = 0 A iL(0+) = i(0+) Applying KVL around outside loop di L + iR2 = V0 ¼ (i) dt At t = 0+ di(0+) L + i(0+)R2 = V0 ¼ (i) dt By putting corresponding values we get di(0+) = dt L Differentiating (i) with respect to `t' d2i di L + R2 =0 dt2 dt At t = 0+ d2i(0+) di(0+) L + R2 =0 2 dt dt Simplifying we get d2i(0+) = dt2 L2 Referring to the network at t =0+ V1(0+) = VR1(0+) = iR1(0+)(R1) V0 iR1(0+) = -RV0 V0


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} R1 V0R1 V1(0+) = R1 V1(0+) = V0 (b) Also V1 = V0 for all t ³ 0

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dV1(0+) = dt

d2V1(0+) =0 dt2

Q#5.16: The network of Prob. 5.15 reaches a steady state under the conditions specified in that problem. At a new reference time, t = 0, the switch K is opened. Solve for the quantities specified in Prob. 5.15 at t = 0+. Equivalent circuitbefore switching

V0

R2

V0 iL(0-) = R2 VC(0-) = VR2(0-) = iL(0-)(R2) V0 VC(0-) = VR2(0-) = (R2) R2 VC(0-) = VR2(0-) = V0 As


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VC(0-) = VC(0+) = V0 Equivalent network at t = 0+

+ i(0+)

V0 i(0+) = iL(0-) = iL(0+) = R2 V1(0+) = V0 ± VR1(0+) V1(0+) = V0 ± iL(0+)R1 V0 V1(0+) = V0 ± R2 R2 ± R1 V1(0+) = V0 ± R2 R1


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Equivalent circuit for t ³0

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R1 L i C R2 Applying KVL around the loop 1 di i(R1 + R2) + ò idt + L = 0 ¼ (i) C dtAlso VR1 + VR2 + VL + VC = 0 i(0+)R1 + i(0+)R2 + VL(0+) + VC(0+) = 0 VL(0+) = -i(0+)R1 - i(0+)R2 - VC(0+) Here VC(0+) = -V0 V0 i(0+) = R2 V0 VL(0+) = R2 V0 VL(0+) = R2 R1 ± V0 + V0 V0 R1 R2 R2 ± (-V0)


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} V0 VL(0+) = R2 And R1

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di VL = L dt di(0+) VL(0+) = dt L Putting corresponding value di(0+) = dt R2L Differentiating eq. (i) with respect to `t' 1 di i(R1 + R2) + ò idt + L = 0 ¼ (i) C dt di(R1 + R2) dt At t = 0+ di(0+) (R1 + R2) dt Here di(0+) = dt V0 i(0+) = R2 Putting corresponding values and simplifying d2i(0+) = dt2 R2 L V0 R1(R1 + R2) C 1 R2L -V0R1 + C i(0+) +L dt2 d2i(0+) =0 + C i +L dt2

-V0R1

d2i =0


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Also di V1 = L + iR2 dt Differentiating with respect to `t' d2i di =L + R2 dt dt2 dt At t = 0+ dV1(0+) d2i(0+) di(0+) =L + R2 dt dt2 dt By putting corresponding values and simplifying dV1 dV1(0+) dt V0 R12 1 = R2 L C

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e know -1 V1 = ò idt - iR1 C Differentiating with respect to `t' dV1 -i di = - R1 dtC dt Differentiating with respect to `t' d2V1 -di 1 d2i= - R1

dt2 dt C dt2 At t = 0+ d2V1(0+) -di(0+) 1 d2i(0+) = - R1 dt2 dt C dt2 Putting corresponding values and simplifying d2V1(0+) = dt2 R2L C V0R1 2 L R1(R1 + R2)

Q#5.17: In the network shown in the accompanying figure, the switch K is changed from a to b at t = 0. Show that at t = 0+, V


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} i1 = i2 = R1 + R2 + R3 i3 = 0

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a b V +

C3

L2 R2 R3

i1 R1 Equivalent circuit before switching L1

i2 i3 C1 C2

+

-

At t = 0-, capacitor C3 is fully charged to voltage V that is VC3(0-) = V and behaves as an open circuit, so current in L1, L2 becomes and other two capacitorsalso fully charged. iL1(0-) = iL1(0+) = 0 A iL2(0-) = iL2(0+) = 0 A VC1(0-) = VC1(0+) = 0 V VC2(0-) = VC2(0+) = 0 V Equivalent circuit after switching


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+ -

After simplification we get

+ i2 i1

Hence -V i1 = i2 = R1 + R2 + R3 C1 behaves short circuit being uncharged at t =0- & L1 behaving open circuit since iL(0-) = iL(0+) = 0 A and i3 = 0 [L2 behaving open circuit]. Q#5.18: In the given network, the capacitor C1 is charged to voltage V0 and the switch K is closed at t = 0.

hen R1 2 MW


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} V0 R2 C1 C2 2 2 solve for d i2/dt at t = 0+. 1000 V 1 MW 10 µ F 20 µ F

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+ C1 V0 i1 C2 i2 R2

R1

Equivalent circuit before switching

i2

VC1(0-) = V0 VC2(0-) = 0 V Equivalent circuit after switching


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K + V0 i1 i2

VC1(0+) = V0 VC2(0+) = 0 V For t ³ 0 For loop 1: 1 R2(i1 ± i2) + ò i1dt = 0 ¼ (i) C1 Fo loop 2: 1 R2(i2 ± i1) + ò i2dt + R1i2 = 0 ¼ (ii) C2 1 R2(i1 ± i2) = ò i1dt C1 1 R2(i2 = ò i1dt ¼ (iii) C1 Taking loop around outside i2R1 = V0 V0 i2 = ¼ (a) R1 In loop 1 According to KVL: R2(i1 ± i2) = V0 R2i1 ± R2i2 = V0 Putting the value of i2 and simplifying


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} V0(R1 + R2) i1 = ¼ (b) 1.5(10-3) Amp. 5(10-4) Amp. R1R2 Putting corresponding values we get i1(0+) i2(0+) Substituting value of R2(i2 ± i1) in eq. (ii) 1 1 ò i1dt + ò i2dt + R1i2 = 0 C1 C2 Differentiating with respect to `t' i1 di2 i2 + R1 + = 0 ¼ (iv) C1 dt C2 At t = 0+ i1(0+)di2(0+) i2(0+) + R1 + =0 C1 dt C2 By putting corresponding values we get di2(0+) = -8.75(10-5) Amp/sec. dt Differentiating eq. (iii) with respect to `t' di2 di1 i1R2 - R2 = dt dt C1 At t = 0+ di2(0+) di1(0+) i1(0+) R2 - R2 = dt dt C1 By putting corresponding values and simplifying di1(0+) = -2.375(10-4) Amp/sec. dt Differentiating eq. (iv) with respect to `t' 1 di1 d2i2 1 di2 + R1 + =0 C1 dt dt2 C2 dt At t = 0+ 1 di1(0+) d2i2(0+) 1 di2(0+)

42


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} + R1 + =0 C1 dt dt2 C2 dt Putting corresponding values and simplifying d2i2(0+) = 1.40625(10-5) Amp/sec2. dt2

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Q#5.19: In the circuit shown in the figure, the switch K is closed at t = 0 connecting a voltage, V0sin w t, to the parallel RL-RC circuit. Find (a) di1/dt and (b) di2/dt at t = 0+.

+ -

i1

i2

Equivalent circuit after s

itching

+ -

At t ³ 0 Applying KVL around outside loop di2 Ri2 + L = V0sin w t ¼ (i) dt Applying KVL around inside loop


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} 1 Ri1 + C Equivalent circuit at t = 0+ ò i1dt = V0sin w t ¼ (ii)

44

+ -

iL(0+) = iL(0-) = 0 A VC(0+) = 0 V V0sin w t i1 = R At t = 0+ V0sin w (0+) i1(0+) =R V0sin 0 i1(0+) = R i1(0+) = 0 A From (i) At t = 0+ di2(0+) Ri2(0+) + L = V0sin w (0+) ¼ (i) dt By putting corresponding values

e get di2(0+) = 0 Amp/sec dt Differentiating eq. (ii)

ith respect to `t' 1 Ri1 + ò i1dt = V0sin w t ¼ (ii) C


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} di1 R + i1 = V0w cos w t

45

dt C At t = 0+ di1(0+) i1(0+) R + = V0w cos w (0+) dt C By putting corresponding values & simplifying

e get di1(0+) = dt V0w R

Q#5.20: In the net

ork sho

n, a steady state is reached

ith the s

itch K open

ith V 100 V R1 10 W R2 20 W R3 20 W L 1H C 1µ F . At time t = 0, the s

itch is closed. a)

rite the integrodifferential equations for the net

ork after the s

itch is closed. (b)

hat is the voltage V0 across C before the s

itch is closed?

hat isits polarity? (c) Solve for the initial value of i1 and i2(t = 0+). (d) Solve for the values of di1/dt and di2/dt at t = 0+. (e)

hat is the value of di1/dt at ¥?

Circuit diagram: i


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46

i2

i1

Equivalent circuit before s

itching:

R2 R1 R3

Simplifying

i

VC(0-)

VC(0-) = iR2(R2) = VR2 Here V iR2 = R1 + R2 VR2 VR2 = R1 + R2


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} By putting correspondingvalues

e get VC(0-) = VR2 = 66.667 V V iL(0-) = R1 + R2 iL(0-) = 3.334 A For t ³0

47

i2

i1

Applying KVL around outside loop di1 R2i1 + L = V ¼ (i) dt Applying KVL around inside loop 1 R3i2 + ò i2dt = V ¼ (ii) C Since 10 iL(0+) = iL(0-) = iR2(0-) = i1(0-) =i1(0+) = 3 V ± VC(0+) i2(0+) = R3 Here VC(0+) = 6.667 Volts Putting correspondingvalues & simplifying i2(0+) = 1.667 Amp. From eq. (i) At t = 0+ di1(0+) R2i1(0+) + L = V ¼ (i) dt Amp.


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Putting corresponding values

e get di1(0+) = 33.334 Amp/sec. dt Differentiating eq. 2: di2 R3 + i2 =0 C

48

dt At t = 0+ di2(0+) i2(0+) R3 + =0 dt C Putting corresponding values di2(0+) =83.334(104) Amp/sec. dt From eq. (i) At t = ¥ di1(¥ ) R2i1(¥ ) + L = V ¼ (i) dt Here di(¥ ) = 0 Amp/sec dt i1(¥ ) = 5 Amp.

Equivalent circuit after s

itching:

+ -


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49

iL(0+) VC(0+) Q#5.21: The net

ork sho

n in the figure has t

o independent node pairs. If the s

itch K is opened at t = 0, find the follo

ing quantities at t = 0+: (a) V1 (b) V2 (c) dV1/dt (d) dV2/dt Circuit diagram: V1 V2 L

i(t)

K

R1 R2

C

Initial conditions: iL(0-) = iL(0+) = 0 A VC(0-) = VC(0+) = 0 V Applying KCL atnode `V1' 1 V1 ò (V1 ± V2)dt + = i(t) ¼ (i) L R1 Differentiating

ith respect to `t'

(V1 ± V2)

1

dV1

di(t)

+ = L R1 dt dt At t = 0+ (V1(0+) ± V2(0+)) 1 dV1(0+) di(0+) + = L R1 dt dt Putting corresponding values

e get dV1(0+) di(t)(0+) V1(0+) R1


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} = dt dt L

50

Applying KCL at node `V2' 1 V2 dV2 ò (V2 ± V1)dt + +C = 0 ¼ (ii) L R2 dt At t = 0+ (V2 ± V1(0+)) 1 dV2(0+) dV2(0+) + +C =0 L R2 dt dt Putting corresponding values

e get dV2(0+) = 0 V/sec. dt Equivalent circuit at t = 0+ V1(0+) V2(0+)

Q#5.22: In the net

ork sho

n in the figure, the s

itch K is closed at the instant t = 0, connecting an unenergized system to a voltage source. Sho

that if V(0) = V, then: di1(0+)/dt, di2(0+)/dt =? L1 L2

R1 + i1 L3 R3 i2


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51

R2

iL1(0-) = iL1(0+) = 0 A iL2(0-) = iL2(0+) = 0 A For t ³ 0 According to KVL Loop 1: di1 R1i1 + L1 d(i1 ± i2) dt + L3 dt After simplification i(R1 + R2) ± i2R2 + i1(0+) = i2(0+) = 0 At t = 0+ di1(0+) i(0+)(R1 + R2) ± i2(0+)R2 + (L1 + L3 + 2M13) dt Putting corresponding values

e get di1(0+) (L1 + L3 + 2M13) dt di2(0+) (L3 + M13+ M23) = V(t) dt d(i1 ± i2) di1 di2 + R2(i1 ± i2) + M13 + M31 + (-M32) = V(t) dt dtdt di2 (L3 + M13 + M23) = V(t) ¼ (i) dt

di1 (L1 + L3 + 2M13) dt

di2(0+) (L3 + M13 + M23) = V ¼ (ii) dt

According to KVL Loop 2: di2 d(i2 ± i1) d(i2 ± i1) R3i2 + L2 + L3 + R2(i2 ± i1) + M23- M31 dt dt dt After simplifying

di1

di2 + M32 =0 dt dt

di2 di1 i2(R3 + R2) ± i1R2 + (L2 + L3 + 2M23) (L3 + M13 + M23) = 0 ¼ (iii) dt dt Att = 0+ i2(0+)(R3 + R2) ± i1(0+)R2 + di2(0+) (L2 + L3 + 2M23) di1(0+) (L3 + M13 + M23) = 0


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} dt Putting correspondingvalues

e get di2(0+) dt dt

52

di1(0+) (L2 + L3 + 2M23) (L3 + M13 + M23) = 0 ¼ (iv) dt

From (ii) & (iv)

e can determine the values of di1(0+)/dt & di2(0+)/dt. MASHAALLAH BHAI'S REFERENCE: In order to indicate the physical relationship of the coilsand, therefore, simplify the sign convention for the mutual terms,

e employ

hat is commonly called the dot convention. Dots are placed beside each coil so that if currents are entering both dotted terminals or leaving both dotted terminals, the fluxes produced by these currents

ill add. If one current enters a dot and the other current leaves a dot, the mutual induced voltage and self-induced voltage terms

ill have opposite signs. Q#5.23: For the net

ork of the figure, sho

that if K is closed at t = 0, d2i1(0+)/dt2? CIRCUIT DIAGRAM: L R1 i1 V(t) + Initial conditions: iL(0-) = iL(0+) = 0 A = i2(0+) VC(0-) = VC(0+) = 0 V Equivalent circuit after s

itching C i2 R2

+ -


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53

V(t) i1(t) = R1 At t = 0+ V(0+) i1(0+) = R1 For t ³ 0 Loop 1: 1 R1i1 + ò (i1 ± i2)dt = V(t) ¼ (i) C Differentiating (i)

ith respect to `t' di1 (i1 ± i2) dV(t) R1 + = ¼ (ii) C dt At t = 0+ di1(0+) (i1(0+) ± i2(0+)) dV(0+) R1 + = dt C dt Putting corresponding values

e get

di1(0+) = dt

dV(0) dt

V(0) R1C

1 R1

Differentiating (ii)

ith respect to `t' d2i1 1 di1 di2 d2V(t) R1 + = 2 dt C dt dt dt2 At t = 0+ d2i1(0+) R1 + 1 di1(0+) di2(0+) d2V(0+) =


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} dt2 C dt dt dt2

54

From here

e can determine the value of d2i1(0+)/dt2. Q#5.24: The given net

orkconsists of t

o coupled coils and a capacitor. At t = 0, the s

itch K is closedconnecting a generator of voltage, V(t) = V sin (t/(MC)1/2). Sho

that Va(0+) =0, dVa(0+)/dt = (V/L)(M/C)1/2, and d2Va(0+)/dt2 = 0. CIRCUIT DIAGRAM: M K a + V(t) Va

Equivalent circuit at t = 0+

+ -


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55

After simplification

diL(0+) Va(0+) = VC(0+) + M dt

e kno

for t ³ 0, according to KVL di 1 L + ò idt =V(t) ¼ (a) dt C At t = 0+ di(0+) L + VC(0+) = V(0+) dt Here VC(0+) = 0 V V(t) = Vsin (t/(MC)1/2) V(0+) = V sin ((0+)/(MC)1/2) V(0+) = V sin (0) V(0+) = V (0) V(0+) = 0 V Putting corresponding values

e get di(0+) = 0 Amp/sec. dt iL(0-) = iL(0+) = i(0+) = 0 A No

di(0+) Va(0+) = VC(0+) + M ¼ (i) dt Putting corresponding values

e get Va(0+) = 0 Volt No

Differentiating (i)

ith respect to `t' dVa dVC d2i


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} = +M ¼ (b) dt dt dt2 Differentiating (a)

ith respect to `t' d2i i dV(t) L + = ¼ (a) 2 dt C dt Here V d(V(t)) = cos (t/(MC)1/2) (MC)1/2 At t = 0+ d2i(0+) i(0+) dV(0+) L + = ¼ (c) 2 dt C dt Putting corresponding values e get d2i(0+) V = dt2 L(MC)1/2 At node a, apply KCL 1 dVC L dt Rearranging dVC(0+) iL(0+) + C =0 dt dVC(0+) 0+C =0 dt dVC(0+) =0 dt d2i= +M ¼ (b) dt dt dt2 At t = 0+ dVa(0+) dVC(0+) d2i(0+) = +M ¼ (b) dt dt dt2 Puttingcorresponding values

e get dVa dVC dVa(0+) = V Mò (VC ± V(t)) + C = 0 ¼ (c)

56


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} dt L C

57

Differentiating (b)

ith respect to `t' d2Va d2VC d3i = +M 2 2 dt dt dt3 Differentiating (c)

ith respect to `t' d3i(0+) 1 di(0+) d2V(0+) L + = ¼ (c) 3 2 dt C dt dt d2V(0+) =? dt2 V d(V(t)) = (MC)1/2 -V d (V(t)) = (MC) -V d2(V(0+)) = (MC) -V d2(V(0+)) = (MC) -V d2(V(0+)) = (MC) d2V(0+) =0 dt2 2

cos (t/(MC)1/2)

sin (t/(MC)1/2)

sin (0+/(MC)1/2) sin (0) (0)

d3i(0+) = 0 Amp/sec3 dt3 Differentiating (c)

ith respect to `t' (VC ± V(t)) d2VC+C = 0 ¼ (c)

L dt2 At t = 0+(VC(0+) ± V(0+)) +C

d2VC(0+)= 0 ¼ (c)


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} L dt2 Putting corresponding values d2VC(0+) = 0 V/sec2 dt d2Va2

58

d2VC

= dt2 dt2 At t = 0+ d2Va(0+) d2VC(0+) d3i(0+) = +M dt2 dt2 dt3 d2Va(0+) = 0 Volt/sec2 dt Q#5.25: In the net

ork of the figure, the s

itch K is opened at t = 0 after the net

ork has attained a steady state

ith the s

itch closed. (a) Find an expression for the voltage across the s

itch at t = 0+. (b) If the parameters are adjusted such that i(0+) = 1 and di/dt (0+) = -1,

hat is the value of the derivative of the voltage across the s

itch, dVK/dt (0+) ? CIRCUIT DIAGRAM: + VK 2

d3i +M dt3

R2 R1 V i L C

Initial conditions: i(0+) = 1 di(0+) = -1 dt


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59

Equivalent net

ork after s

itching: VK

sc

V iL = R2 V iL(0-) = iL(0+) = R2 At t = 0+ VK(0+) = VR1(0+) VR1(0+) = iL(0+)(R1) Putting corresponding value

e get


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60

V VR1(0+) = R2 For t ³ 0 VK = iR1 + 1 R1

ò idt C Differentiating

ith respect to `t' dVK di i = R1 + dt dt C At t = 0+ dVK(0+)di(0+) i(0+) = R1 + dt dt C Putting corresponding value

e get

dVK(0+) = dt

1 -R C

Q#5.26: In the net

ork sho

n in the figure, the s

itch K is closed at t = 0 connecting the battery

ith an unenergized system. (a) Find the voltage Va at t = 0+. (b) Find the voltage across capacitor C1 at t = ¥ . CIRCUIT DIAGRAM: R1

Va K C1 V C2 R2 L


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61

Equivalent net

ork at t = 0+

After simplification:

R2

Va(0+) = V Equivalent net

ork at t = ¥


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62

VC1(¥ ) = V. Q#5.27: In the net

ork of the figure, the s

itch K is closed at t = 0. At t = 0-, all capacitor voltages and inductor currents are zero. Three node to datum voltages are identified as V1, V2, and V3. (a) Find V1 and dV1/dt at t = 0+. (b) Find V2 and dV2/dt at t = 0+. (c) Find V3 and dV3/dt at t = 0+. CIRCUIT DIAGRAM:

V1

V3

+ V2

Using KCL at node `V1' For t ³ 0 (V1 ± V(t)) R1 dV1 (V1 ± V2) 1 + C1 + + ò (V1 ± V3)dt dt R2 L

Using KCL at node `V2' (V2 ± V1) + C2 dV2 + 1 ò V2dt = 0 ¼ (ii)


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} R2 dt L2

63

Using KCL at node `V3' 1 L1 dV3 ò (V3 ± V1)dt + C3 dt = 0 ¼ (iii)

At t = 0+, capacitor C1 becomes short circuit as a result of

hich V1(0+) 0V V2(0+) 0V V3(0+) 0V At t = 0+ 1 dV3 ò (V3 ± V1)dt + C3 = 0 ¼ (iii) L1 dt dV3(0+) =0 dt After simplification

e get iL1(0+) + C3 dV3(0+) = 0 Volt/sec dt Here iL1(0-) = iL1(0+) = 0 A At t = 0+ (V2(0+) ± V1(0+)) dV2(0+) + C2 + iL2(0+) = 0 ¼ (ii) R2 dt iL2(0-) = iL2(0+) = 0 A Putting corresponding values

e get dV2(0+) =0 dt iL3 (0+) At t = 0+ eq. (i) reveals (V1(0+) ± V(0+)) dV1(0+) (V1(0+) ± V2(0+)) 1 + C1 + + ò (V1(0+) ± V3(0+))dt = 0 R1 dt R2 L Putting corresponding values

e get


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64

dV1(0+) = 0 Volt/sec. dt Q#5.28: In the net

ork of the figure, a steady state is reached, and at t = 0, the s

itch K is opened. (a) Find the voltage across thes

itch, VK at t = 0+. (b) Find dVK/dt at t = 0+. CIRCUIT DIAGRAM:

VK + -

Equivalent net

ork before s

itching


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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} R1 R2 i(0-)

65

R3

At t = 0VC1(0-) = i(o-) R2 V i(0-) = R1 + R2 + R3 VR2 VC1(0-) = R1 + R2 + R3 VC1(0-) = i(o-) R2 VR3 VC1(0-) = R1 + R2 + R3 VC2(0-) = V ± VR1(0-) Equivalent net

ork after s

itching

V1

K


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66

At t ³ 0 At node K, according to KCL d dVk VK ± V1 C1 (VK ± V1) + C3 + =0 dt dt R2 After simplification

e get dV1 1 dVK (VK ± V1) = (C3 + C1) + dt C1 dt R2 ¼ (i)

At node `V1' according to KCL (V1 - V) dV1 d(V1 - VK) (V1 - VK) + C2 + C1 + =0 R1 dt dt R2 After simplification

e get dV1 1 dVK (VK ± V1) (V ± V1) = C1 + + dt (C1 + C2) dt R2 R1 Equating (i) & (ii)

e get dVK/dt at t = 0+. Hint: V1(0+) = VC2(0-) = V ± VR1(0-) Here VR1 VR1(0-) = R1 + R2 + R3 VR2 + VR3 V ± VR1(0-) = R1 + R2 + R3 Q#5.29: In the net

ork of the accompanying figure, a steady state is reached

iththe s

itch K closed and

ith i = I0, a constant. At t = 0, s

itch K, is opened.Find: (a) V2(0-) =? (b) V2(0+) =? (c) (dV2/dt)(0+). CIRCUIT DIAGRAM: 1 2 R2 + ¼ (ii)


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67 V2

C I0 R1 R3 L

Equivalent net

ork at t = 0-

I0


e get

I0 R1 R2


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68

No

According to current divider rule: R1I0 iR2 = R1 + R2

e kno

V2(0-) = VL(0-) = 0 Equivalent net

ork at t = 0+

+ I0

R1 + iL(0+) + I0R1 VC(0+)


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69


V2(0+) + VC(0+) + I0R1

At node `V1' V1 +C R1 dt At node `V2' V2 +C R3 dt From eq. (ii) d(V1 ± V2) C dt = V2 + L 1 ò V2dt d(V2 ± V1) + L 1 ò V2dt ¼ (ii) d(V1 ± V2) = I0 ¼ (i)


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70

Substituting the value of Cd(V1 ± V2)/dt in (i)

e get V2 1 V1 + ò V2dt + = I0 R3 LR1 At t = 0+ V2(0+) 1 V1(0+) + ò V2(0+)dt + = I0 ¼ (iii) R3 L R1 Putting corresponding values

e get I0R1R2 V1(0+) = R1 + R2 Differentiating eq. (iii)

ith respectto `t' and from here putting the value of dV1(0+)/dt in eq. (i)

e get dV2(0+)/dt. Hint: In eq. (i) V1(0+) I0R1R2 = R1 (R1 + R2)R1 dV2(0+) = dt dV1(0+) = dt R3 dt C(R1 + R2)(R1 + R3) -R1 dV2(0+) -I0R1R3

THE END.


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72560322 Network Analysis by Van Valkenburg Chap 5 Solution Manual.pdf

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