Network Analysis by Van Valkenburg chap 5 solution manual

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

1

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+, for element values as follows: V 100 V R 1000 Ω C 1µ F

+

R V

C

-

Switch is closed at t = 0 (reference time) We 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 ≥ 0 Sum of voltage rise = sum of voltage drop 1

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} ∫ 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 kΩ ) 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 of i, di/dt, and d2i/dt2 at t = 0+ if R 10 Ω L 1H V 100 V

K +

R V

-

L

Key closed at t = 0

2


3

iL(0+) = iL(0-) = i(0+) = 0 Amp According to the statement under Q#5.2: Drop Rise Open circuit

Drop

i(0+)

Element and initial condition Applying KVL for t ≥ 0 Sum of voltage rise = sum of voltage 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

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 Ω L 1H C 0.1 µ F V 100 V a

K

b V C

R L

Equivalent circuit at t = 0+

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

4


5

∫ idt + = 0 [eq. 1] C dt Equivalent circuit at t = 0Ri +

a

i(0+) sc

V iL(0+) = iL(0-) = i(0+) =

100 =

= 0.1 Amp

R 1000 Initial condition: VC(0-) = VC(0+) = 0 Also we know for t ≥ 0 VR + VL + VC = 0 iR + VL + VC = 0 At t = 0+ i(0+)R + VL(0+) + VC(0+) = 0 (0.1)(1000) + VL(0+) + 0 = 0 VL(0+) = -100 Volts And di VL = L dt di VL = dt L di(0+) VL(0+) = dt L Putting corresponding values di(0+) = -100 Amp/sec dt Differentiating [eq. 1] with respect to ‘t’ Rdi i Ld2i + + =0 dt C dt2

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

Ld2i(0+)

i(0+) +

6

+

=0 2

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

2

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

V1 V2 R C1 C2

V2

2V 1V 1Ω 1F ½F

After switching: V1

V2

Applying KCL at node 1: V1 – V2

dV1 + C1

R dt V1(0+) – V2(0+)

=0 dV1(0+)


7

+ 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 “Sum of currents entering into the junction must equal to the sum of the currents leaving the junction” 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

V2

C -

R1 R2 C VC(0-) = VC(0+) = 0 V

10 Ω 20 Ω 1/20 F

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Applying KCL at NODE for t ≥ 0 V2 dV2 V2 – V1 +C + = 0 … (1) R2 dt R1 At t = 0+ V2(0+) dV2(0+) V2(0+) – V1(0+) +C + =0 R2 dt R1 V1 = e-t Volts V2 = VC(0+) = 0 Volts 0 dV2(0+) 0 – e-0+ +C + =0 R2 dt 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 d2V2

1

dV2

1

20

dt

20 dt2

1

dV2

1

+

dt

20 dt

3

dV2

1

2

20 dt

10

dt

d(e-t)

1

=0 10

dt

20 dt2 1

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

+ dt

=0

+ 0.1e-t = 0 … (2)

At t = 0+ 3 dV2(0+) 20

d(e-t)

d2V2

1

dt

1 -

2

+ 20

dV2

10 dt

d2V2

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

1

=0 dt

+

20

dt

10 dt

d2V2

d(e-t)

dV2

+

+

20

1

20

dt

2

8


9

Simplifying d2V2(0+) = -8 Volt/sec2 dt2 Differentiating eq. 2 3 d2V2 1 d3V2 + 20 dt2 20 dt3 At t = 0+ 3 d2V2(0+)

1

- 0.1e-t = 0

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

+ 2

3

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

1Ω 1H ½F 2V


i(0+)

V iL(0+) = iL(0-) = i(0+) =

sc

2 =

R

10

= 2 Amp 1

1 Also VK = VC = C dVK

∫ idt

i

= dt C At t = 0+ dVK i(0+) = dt C dVK 2 = dt (1/2) dVK = 4 Volts/sec dt 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 Ω C 1µ F V


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 ≥ 0; according to KCL at V V dV +C = I … (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

d2V

dV +C

=0 2

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

11


12

Simplifying d2V(0+) = -1010 Volts/sec2 dt2 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 R 100 Ω L 1H V

Equivalent circuit before switching:

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


So V(0+) = (I)(R) V(0+) = (1)(100) V(0+) = 100 Volts For t ≥ 0 Applying KCL at node V V 1 + ∫ Vdt = I … (1) R L Differentiating (1) with respect to ‘t’ 1 dV V + = 0 … (2) R 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+ d2V(0+)

1

1

dV(0+)

+ 2

R dt Simplifying

=0 L

dt

d2V(0+) = 1, 000, 000 Volts/sec2 dt

2

13


14

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


sc


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

Req

i(0-) 5V

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

15


16

Va(0-)

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

Vb Va


17

Applying KCL at node Va Va – 5 Va - Vb Va + + =0 10 20 10 After simplification we get Vb = 5Va – 10 … (i) Applying KCL at node Vb Vb - Va Vb – 5 2 + + =0 20 10 3 9Vb – 3Va + 10 = 0 … (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 K open. 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


18

Equivalent circuit before switching

Req

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

+ -

Applying KCL at node Va


19

Va – 5

Va - Vb Va + =0 10 20 10 After simplification we get Vb = 5Va – 10 … (i) At Vb = 5 V 5 = 5Va – 10 +

Va(0+) = 3 Volts 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)

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Req = (10 + 20) + 10 300 Req = 40 Req = 7.5 Ω After simplification

Req

i(0-) 5V

V i(0-) = Req 5 i(0-) = 7.5 i(0-) = 0.667 Amp. i(0-) = iL(0-) iL(0-) = iL(0+) = 0.667 Amp. Equivalent network at t = ∞

Va

20


Applying KCL at node Va Va – 5 Va Va + + =0 10 20 10 After simplification we get

iL(∞ )

Va(∞ ) = 2 Volts V

Va

iL(∞ ) =

+ 10 20 After simplification we get iL(∞ ) = 0.6 Amp. Q#5.12: In the network given in figure p5-10, determine Vb(0+) and Vb(∞ ) for the conditions stated in Prob. 5-10. At t = 0-, equivalent network is

21


22

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


23

After simplification we get Va = 0.2Vb + 2 … (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 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) V1 and 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


24

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 – iR1R1 VR1 VC(∞ ) = V – R1 + R2 After simplification we get VR2

0V 0V 0A 0A


25

VC(∞ ) = R1 + R2 Since VC(∞ ) = V1(∞ ) + V2(∞ ) VR2 V2(∞ ) = R1 + R2 V1(∞ ) = VC(∞ ) - V2(∞ ) VR2 VR2 V1(∞ ) = R1 + R2 R1 + R2 V1(∞ ) = 0 Volts

Self justified

(c)

V1

V2

According to KCL at node ‘V1’ V – V1 dV1 1 +C + ∫ V1 – V2)dt = 0 … (i) R1 dt L Differentiating with respect to ‘t’ 1

dV

dV1 -

d2V1 1 +C +

(V1 – V2) = 0 … (iii)

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

dt

dt2

dt

L

According to KCL at node ‘V2’ V2 1 + ∫ V2 – V1)dt = 0 … (ii) R2 L Differentiating with respect to ‘t’ 1

dV2

1 +

R2

dt

(V2 – V1) = 0 … (iV) L

As V1 = V1 + V2 By putting the value of V1 in (iii) & (iv) 1

dV

R1

dt

1

dV

d2(V1 + V2)

d(V1 + V2) -

+C dt2

dt dV1 -

-

R1

dt

dt

1

dV2

R2

dt

L

1

dV2

1

R2

dt

1

dV2

d2V1

dV2 +C dt

dt

2

1 + (V1 + V2 – V2) = 0 L

d2V2 + dt2

1 +

(V1) = 0 … (V) L

1 +

(V2 – V1) = 0 … (iV)

+

(V2 – (V1 + V2)) = 0 … (iV) L 1

+

(V2 – V1 - V2) = 0

R2

dt

L

1

dV2

1

R2

dt

+

(–V1) = 0 … (Vi) L

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

26


27

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 -


28

R1 + R2 VR2 VC(0-) = R1 + R2 V2 = iR2R2 V iR2(0-) = iR1(0-) = R1 + R2 VR2 V2 = R1 + R2

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+)


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+) =

VR2 -

R1 + R2

R1 + R2

V1(0+) = 0 Volts (b) Equivalent network at t = ∞ At t = ∞ capacitor will be fully discharged and acts as an open circuit.

Hence VC(∞ ) = 0 V iL(∞ ) = 0 A

29


30

V2(∞ ) = iL(∞ )R2 V2(∞ ) = (0)R2 V2(∞ ) = 0 Volt (c) For t ≥ 0, the equivalent network is

V1

V2 V2 Applying KCL at node ‘V1’ 1 dV1 ∫ (V1 – V2)dt + C = 0 … (i) L dt d C

-V2 (V1 + V2) =

dt

… (ii) R2

As V1 = V1 + V2 1 d(V1 + V2) ∫ (V1 + V2 – V2)dt + C =0 L dt 1 d(V1 + V2) ∫ V1dt + C =0

V1


31

L dt Differentiating (i) with respect to ‘t’ d2V1

1 V1 + C

dt2

L

d2V2 +C dt2

= 0 … (iii)

Differentiating (ii) with respect to ‘t’ d2 C

-1

dV2

R2

dt

(V1 + V2) = dt2

… (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


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 … (i) dt At t = 0+ di(0+) L + i(0+)R2 = V0 … (i) dt By putting corresponding values we get di(0+)

V0 =

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+)

-RV0 =

dt2

L2

Referring to the network at t = 0+ V1(0+) = VR1(0+) = iR1(0+)(R1) V0 iR1(0+) =

32


33

R1 V0R1 V1(0+) = R1 V1(0+) = V0 (b) Also V1 = V0 for all t ≥ 0

d2V1(0+)

dV1(0+) = dt

=0 dt

2

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 circuit before 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


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 –

R1 R2 R2 – R1

V1(0+) = V0 – R2

34


35

Equivalent circuit for t ≥ 0

R1 L i C R2 Applying KVL around the loop 1 di i(R1 + R2) + ∫ idt + L = 0 … (i) C dt Also 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 R1 R2

R2 – (-V0)

V0 VL(0+) = -

R1 – V0 + V0 R2

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

R1 R2

And

di VL = L dt di(0+) VL(0+) = dt L Putting corresponding value di(0+)

-V0R1 =

dt

R2L

Differentiating eq. (i) with respect to ‘t’ 1 di i(R1 + R2) + ∫ idt + L = 0 … (i) C dt di (R1 + R2)

d2i

i +

dt

+L C

=0 dt

2

At t = 0+ di(0+) (R1 + R2)

+ dt

d2i(0+)

i(0+) +L

=0 dt2

C

Here di(0+)

-V0R1 =

dt

R2L V0

i(0+) = R2 Putting corresponding values and simplifying d2i(0+)

V0

R1(R1 + R2)

= dt2

1 -

R2

L

C

36


37

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

We know -1 V1 = ∫ idt - iR1 C Differentiating with respect to ‘t’ dV1

-i

di

= - R1 dt C 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+)

V0R1 2 =

dt2

R1(R1 + R2) -

R2L C

L

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


38

i1 = i2 = R1 + R2 + R3 i3 = 0

a

C3

L2

b V

R2 +

R3 i1

i2

-

i3 R1

L1 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 capacitors also 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

-


39

+ -

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. When R1 2 MΩ


40

V0 R2 C1 C2 2 2 solve for d i2/dt at t = 0+.

1000 V 1 MΩ 10 µ F 20 µ F

+ C1

i1

V0 -

C2 i2 R2

R1


i2

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


41

K + V0 -

i1

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

i2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} V0(R1 + R2) i1 =

… (b)

R1R2 Putting corresponding values we get i1(0+) i2(0+) Substituting value of R2(i2 – i1) in eq. (ii) 1 1 ∫ i1dt + ∫ i2dt + R1i2 = 0 C1 C2 Differentiating with respect to ‘t’ i1 di2 i2 + R1 + = 0 … (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 i1 R2 - 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+)

1.5(10-3) Amp. 5(10-4) Amp.

42


43

+ R1 + =0 C1 dt dt2 C2 dt Putting corresponding values and simplifying d2i2(0+) = 1.40625(10-5) Amp/sec2. dt2 Q#5.19: In the circuit shown in the figure, the switch K is closed at t = 0 connecting a voltage, V0sin ω t, to the parallel RL-RC circuit. Find (a) di1/dt and (b) di2/dt at t = 0+.

+ -

Equivalent circuit after switching

+ -

At t ≥ 0 Applying KVL around outside loop di2 Ri2 + L = V0sin ω t … (i) dt Applying KVL around inside loop

i1

i2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} 1 Ri1 +

∫ i1dt = V0sin ω t … (ii)

C Equivalent circuit at t = 0+

+ -

iL(0+) = iL(0-) = 0 A VC(0+) = 0 V V0sin ω t i1 = R At t = 0+ V0sin ω (0+) i1(0+) = R V0sin 0 i1(0+) = R i1(0+) = 0 A From (i) At t = 0+ di2(0+) Ri2(0+) + L = V0sin ω (0+) … (i) dt By putting corresponding values we get di2(0+) = 0 Amp/sec dt Differentiating eq. (ii) with respect to ‘t’ 1 Ri1 + ∫ i1dt = V0sin ω t … (ii) C

44

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} di1 R

45

i1 +

= V0ω cos ω t

dt C At t = 0+ di1(0+) i1(0+) R + = V0ω cos ω (0+) dt C By putting corresponding values & simplifying we get di1(0+) = dt

V0ω R

Q#5.20: In the network shown, a steady state is reached with the switch K open with V 100 V R1 10 Ω R2 20 Ω R3 20 Ω L 1H C 1µ F . At time t = 0, the switch is closed. (a) Write the integrodifferential equations for the network after the switch is closed. (b) What is the voltage V0 across C before the switch is closed? What is its 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) What is the value of di1/dt at ∞ ?

Circuit diagram: i


46

i2

i1


R2 R1 R3

Simplifying

i

VC(0-)

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


47

By putting corresponding values we get VC(0-) = VR2 = 66.667 V V iL(0-) = R1 + R2 iL(0-) = 3.334 A For t ≥ 0

i2

Applying KVL around outside loop di1 R2i1 + L = V … (i) dt Applying KVL around inside loop 1 R3i2 + ∫ i2dt = V … (ii) C Since 10 iL(0+) = iL(0-) = iR2(0-) = i1(0-) = i1(0+) =

Amp. 3

V – VC(0+) i2(0+) = R3 Here VC(0+) = 6.667 Volts Putting corresponding values & simplifying i2(0+) = 1.667 Amp. From eq. (i) At t = 0+ di1(0+) R2i1(0+) + L = V … (i) dt

i1


48

Putting corresponding values we get di1(0+) = 33.334 Amp/sec. dt Differentiating eq. 2: di2 R3

i2 +

=0 C

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 … (i) dt Here di1(∞ ) = 0 Amp/sec dt i1(∞ ) = 5 Amp.

Equivalent circuit after switching:

+ -


49

iL(0+) VC(0+) Q#5.21: The network shown in the figure has two independent node pairs. If the switch K is opened at t = 0, find the following quantities at t = 0+: (a) V1 (b) V2 (c) dV1/dt (d) dV2/dt Circuit diagram: V1 V2 L

i(t)

K

R1

C R2

Initial conditions: iL(0-) = iL(0+) = 0 A VC(0-) = VC(0+) = 0 V Applying KCL at node ‘V1’ 1 V1 ∫ (V1 – V2)dt + = i(t) … (i) L R1 Differentiating with 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 we get dV1(0+)

di(t)(0+)

V1(0+) R1

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

50

dt

L

Applying KCL at node ‘V2’ 1 V2 dV2 ∫ (V2 – V1)dt + +C = 0 … (ii) L R2 dt At t = 0+ (V2(0+) – V1(0+)) 1 dV2(0+) dV2(0+) + +C =0 L R2 dt dt Putting corresponding values we get dV2(0+) = 0 V/sec. dt Equivalent circuit at t = 0+

V1(0+)

V2(0+)

Q#5.22: In the network shown in the figure, the switch K is closed at the instant t = 0, connecting an unenergized system to a voltage source. Show that if V(0) = V, then: di1(0+)/dt, di2(0+)/dt =? L1 L2

R1 L3

+ -

R3 i1

i2


51

R2

iL1(0-) = iL1(0+) = 0 A iL2(0-) = iL2(0+) = 0 A For t ≥ 0 According to KVL Loop 1: di1

d(i1 – i2)

+ L3 dt After simplification

d(i1 – i2) di1 di2 + R2(i1 – i2) + M13 + M31 + (-M32) = V(t) dt dt dt

R1i1 + L1

i(R1 + R2) – i2R2 +

dt

di1 (L1 + L3 + 2M13) dt

di2 (L3 + M13 + M23) = V(t) … (i) dt

i1(0+) = i2(0+) = 0 At t = 0+ di1(0+) i(0+)(R1 + R2) – i2(0+)R2 + (L1 + L3 + 2M13) dt Putting corresponding values we get di1(0+) (L1 + L3 + 2M13) dt

di2(0+) (L3 + M13 + M23) = V(t) dt

di2(0+) (L3 + M13 + M23) = V … (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 … (iii) dt dt At t = 0+ i2(0+)(R3 + R2) – i1(0+)R2 +

di2(0+) (L2 + L3 + 2M23) -

di1(0+) (L3 + M13 + M23) = 0

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} dt Putting corresponding values we get di2(0+) dt

52

dt

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

From (ii) & (iv) we can determine the values of di1(0+)/dt & di2(0+)/dt. MASHAALLAH BHAI’S REFERENCE: In order to indicate the physical relationship of the coils and, therefore, simplify the sign convention for the mutual terms, we employ what 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 will add. If one current enters a dot and the other current leaves a dot, the mutual induced voltage and self-induced voltage terms will have opposite signs. Q#5.23: For the network of the figure, show 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 switching

+ -

C

i2

R2


V(t) i1(t) = R1 At t = 0+ V(0+) i1(0+) = R1 For t ≥ 0 Loop 1: 1 R1i1 + ∫ (i1 – i2)dt = V(t) … (i) C Differentiating (i) with respect to ‘t’ di1 (i1 – i2) dV(t) R1 + = … (ii) dt C dt At t = 0+ di1(0+) (i1(0+) – i2(0+)) dV(0+) R1 + = dt C dt Putting corresponding values we get

di1(0+)

dV(0) =

dt

V(0)

1

R1C

R1

dt

Differentiating (ii) with 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+) =

53

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

C

dt

dt

54

dt2

From here we can determine the value of d2i1(0+)/dt2. Q#5.24: The given network consists of two coupled coils and a capacitor. At t = 0, the switch K is closed connecting a generator of voltage, V(t) = V sin (t/(MC)1/2). Show that Va(0+) = 0, dVa(0+)/dt = (V/L)(M/C)1/2, and d2Va(0+)/dt2 = 0. CIRCUIT DIAGRAM: M K a + -

Va V(t)

Equivalent circuit at t = 0+

+ -


After simplification

diL(0+) Va(0+) = VC(0+) + M dt We know for t ≥ 0, according to KVL di 1 L + ∫ idt = V(t) … (a) dt C At t = 0+ di(0+) L + VC(0+) = V(0+) dt Here VC(0+) = 0 V V(t) = V sin (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 we get di(0+) = 0 Amp/sec. dt iL(0-) = iL(0+) = i(0+) = 0 A Now di(0+) Va(0+) = VC(0+) + M … (i) dt Putting corresponding values we get Va(0+) = 0 Volt Now Differentiating (i) with respect to ‘t’ dVa dVC d2i

55

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} = +M … (b) dt dt dt2 Differentiating (a) with respect to ‘t’ d2i i dV(t) L + = … (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 + = … (c) 2 dt C dt Putting corresponding values we get d2i(0+) V = dt2 L(MC)1/2 At node a, apply KCL 1 dVC ∫ (VC – V(t)) + C

= 0 … (c)

L dt Rearranging dVC(0+) iL(0+) + C =0 dt dVC(0+) 0+C =0 dt dVC(0+) =0 dt d2i = +M … (b) dt dt dt2 At t = 0+ dVa(0+) dVC(0+) d2i(0+) = +M … (b) dt dt dt2 Putting corresponding values we get dVa

dVC

dVa(0+)

V =

M

56

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

L

C

Differentiating (b) with respect to ‘t’ d2Va d2VC d3i = +M 2 2 dt dt dt3 Differentiating (c) with respect to ‘t’ d3i(0+) 1 di(0+) d2V(0+) L + = … (c) 3 2 dt C dt dt d2V(0+) =? dt2 V cos (t/(MC)1/2)

d(V(t)) = (MC)1/2 -V 2

sin (t/(MC)1/2)

d (V(t)) = (MC) -V d2(V(0+)) =

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

d2(V(0+)) =

sin (0) (MC) -V

d2(V(0+)) =

(0) (MC)

d2V(0+) =0 dt

2

d3i(0+) = 0 Amp/sec3 dt3 Differentiating (c) with respect to ‘t’ (VC – V(t)) d2VC +C

= 0 … (c)

L dt2 At t = 0+ d2VC(0+)

(VC(0+) – V(0+)) +C

= 0 … (c)

57


58

L dt2 Putting corresponding values d2VC(0+) = 0 V/sec2 dt

2

d2Va

d2VC

d3i +M dt3

= dt2 dt2 At t = 0+ d2Va(0+) d2VC(0+) d3i(0+) = +M dt2 dt2 dt3 d2Va(0+) = 0 Volt/sec2 2

dt Q#5.25: In the network of the figure, the switch K is opened at t = 0 after the network has attained a steady state with the switch closed. (a) Find an expression for the voltage across the switch at t = 0+. (b) If the parameters are adjusted such that i(0+) = 1 and di/dt (0+) = -1, what is the value of the derivative of the voltage across the switch, dVK/dt (0+) ? CIRCUIT DIAGRAM: +

VK

-

R2 R1

C

V i L

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


59

Equivalent network after switching: 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 we get


60

V VR1(0+) =

R1 R2

For t ≥ 0

1

∫ idt C Differentiating with 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 we get VK = iR1 +

dVK(0+)

1 =

dt

-R C

Q#5.26: In the network shown in the figure, the switch K is closed at t = 0 connecting the battery with 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

C2 R2

V

L


61

Equivalent network at t = 0+

After simplification:

R2

Va(0+) = V Equivalent network at t = ∞


62

VC1(∞ ) = V. Q#5.27: In the network of the figure, the switch 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 ≥ 0 (V1 – V(t)) R1

dV1 (V1 – V2) 1 + C1 + + ∫ (V1 – V3)dt = 0 … (i) dt R2 L

Using KCL at node ‘V2’ (V2 – V1)

dV2 + C2

1 +

∫ V2dt = 0 … (ii)


dt

L2

Using KCL at node ‘V3’ 1 L1

dV3 ∫ (V3 – V1)dt + C3 dt

= 0 … (iii)

At t = 0+, capacitor C1 becomes short circuit as a result of which V1(0+) 0V V2(0+) 0V V3(0+) 0V At t = 0+ 1 dV3 ∫ (V3 – V1)dt + C3 = 0 … (iii) L1 dt dV3(0+) =0 dt After simplification we 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 … (ii) R2 dt iL2(0-) = iL2(0+) = 0 A Putting corresponding values we 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 + + ∫ (V1(0+) – V3(0+))dt = 0 R1 dt R2 L Putting corresponding values we get

63


dV1(0+) = 0 Volt/sec. dt Q#5.28: In the network of the figure, a steady state is reached, and at t = 0, the switch K is opened. (a) Find the voltage across the switch, VK at t = 0+. (b) Find dVK/dt at t = 0+. CIRCUIT DIAGRAM:

VK +

Equivalent network before switching

-

64


65

R2 i(0-)

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 network after switching

V1

K


66

At t ≥ 0 At node K, according to KCL d dVk VK – V1 C1 (VK – V1) + C3 + =0 dt dt R2 After simplification we get dV1 1 dVK (VK – V1) = (C3 + C1) + dt C1 dt R2

… (i)

At node ‘V1’ according to KCL (V1 - V) dV1 d(V1 - VK) (V1 - VK) + C2 + C1 + =0 R1 dt dt R2 After simplification we get dV1 1 dVK (VK – V1) (V – V1) = C1 + + dt (C1 + C2) dt R2 R1

… (ii)

Equating (i) & (ii) we 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 network of the accompanying figure, a steady state is reached with the switch K closed and with i = I0, a constant. At t = 0, switch K, is opened. Find: (a) V2(0-) =? (b) V2(0+) =? (c) (dV2/dt)(0+). CIRCUIT DIAGRAM: 1 2 R2

+


67 V2

C I0

R1

R3

L

Equivalent network at t = 0-

I0

After simplification we get

I0 R1

R2


68

Now According to current divider rule: R1I0 iR2 = R1 + R2 We know V2(0-) = VL(0-) = 0 Equivalent network at t = 0+

+ I0

R1 +

+ I0R1

VC(0+)

iL(0+)


69

After simplification

V2(0+) + VC(0+) + -

I0R1

At node ‘V1’ V1

d(V1 – V2) +C

= I0 … (i)

R1

dt

At node ‘V2’ V2

d(V2 – V1) +C

R3

1 +

dt

L

∫ V2dt … (ii)

From eq. (ii) d(V1 – V2) C

V2 =

dt

1 +

R3 L

∫ V2dt


Substituting the value of Cd(V1 – V2)/dt in (i) we get V2 1 V1 + ∫ V2dt + = I0 R3 L R1 At t = 0+ V2(0+) 1 V1(0+) + ∫ V2(0+)dt + = I0 … (iii) R3 L R1 Putting corresponding values we get I0R1R2 V1(0+) = R1 + R2 Differentiating eq. (iii) with respect to ‘t’ and from here putting the value of dV1(0+)/dt in eq. (i) we get dV2(0+)/dt. Hint: In eq. (i) V1(0+) I0R1R2 = R1 (R1 + R2)R1 dV2(0+)

-I0R1R3 =

dt

C(R1 + R2)(R1 + R3)

dV1(0+)

-R1 dV2(0+) =

dt

R3

dt

THE END.

70

Network Analysis by Van Valkenburg chap 5 solution manual

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