Network Analysis by Van Valkenburg chap 5 solution manual

Muhammad Irfan Yousuf Peon of Holy Prophet (P.B.U.H) 1

PROBLEMSQ#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 VR 1000 ΩC 1 µ F

+ R

V C-

Switch is closed at t = 0 (reference time)We knowVoltage across capacitor before switching = VC(0-) = 0 VAccording to the statement under Q#5.1.VC(0+) = VC(0-) = 0 V V 100iC(0+) = i(0+) = = = 0.1 Amp. R 1000

Element and initial condition Equivalent circuit at t = 0+

CSc

Switch

Drop

Rise i(0+) Short circuit

Drop

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

1


V = iR + ∫ idt CDifferentiating with respect to ‘t’ di iR + = 0 dt Cdi(0+) -i(0+) = [eq. 1] dt CRBy putting the values of i(0+), C & Rdi(0+) -(0.1) = dt (1 µ F)(1 kΩ )

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 dt2

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 ΩL 1 HV 100 V

K+ R

V- L

Key closed at t = 0


iL(0+) = iL(0-) = i(0+) = 0 AmpAccording to the statement under Q#5.2:

Drop

Rise Open circuit Drop

i(0+)

Element and initial condition Equivalent circuit at t = 0+ oc

Applying KVL for t ≥ 0Sum of voltage rise = sum of voltage drop LdiV = iR + dt Ldi = V – iR dt di V - iR = [eq. 1] dt Ldi(0+) V – i(0+)R = dt LPutting corresponding valuesdi(0+) V – (0)R = dt Ldi(0+) V = dt Ldi(0+) 100 = dt 1di(0+) = 100 Amp/sec dt Differentiating [eq. 1] d2i d V iR = -


dt2 dt L Ld2i -Rdi = dt2 Ldtd2i(0+) -Rdi(0+) = dt2 LdtPutting 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 ΩL 1 HC 0.1 µ FV 100 V

a K

b R

V C L

Equivalent circuit at t = 0+

b

sc

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


Ri + ∫ idt + = 0 [eq. 1] C dtEquivalent circuit at t = 0-

a

i(0+) sc

V 100iL(0+) = iL(0-) = i(0+) = = = 0.1 Amp R 1000Initial condition:VC(0-) = VC(0+) = 0Also we know for t ≥ 0VR + VL + VC = 0iR + VL + VC = 0 At t = 0+i(0+)R + VL(0+) + VC(0+) = 0(0.1)(1000) + VL(0+) + 0 = 0VL(0+) = -100 VoltsAnd diVL = L dtdi VL

= dt Ldi(0+) VL(0+) = dt LPutting corresponding valuesdi(0+) = -100 Amp/sec dt Differentiating [eq. 1] with respect to ‘t’Rdi i Ld2i + + = 0 dt C dt2


Rdi(0+) i(0+) Ld2i(0+) + + = 0 dt C dt2

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 2 VV2 1 VR 1 ΩC1 1 FC2 ½ F

After switching:

V1 V2

Applying KCL at node 1:

V1 – V2 dV1

+ C1 = 0 R dtV1(0+) – V2(0+) dV1(0+)


+ C1 = 0 R dtPutting corresponding values2 V – 1 V dV1(0+) + (1 F) = 0 1 dtSimplifying

dV1(0+) = -1 Volt/sec dt

At node 2:V2 – V1 dV2

+ C2 = 0 R dtV2(0+) – V1(0+) dV2(0+) + C2 = 0 R dtPutting corresponding values1 V – 2 V dV2(0+) + (1/2) = 0 1 dtSimplifying

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 V2

V1 C-

R1 10 ΩR2 20 ΩC 1/20 F

VC(0-) = VC(0+) = 0 V


Applying KCL at NODE for t ≥ 0V2 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 VoltsV2 = 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 + - = 0R2 dt dt2 R1 dt dt

1 dV2 1 d2V2 1 dV2 d(e-t) + + - = 020 dt 20 dt2 10 dt dt

1 dV2 1 d2V2 1 dV2 1 d(e-t) + + - = 020 dt 20 dt2 10 dt 10 dt

3 dV2 1 d2V2 1 d(e-t) + - = 020 dt 20 dt2 10 dt

d(e-t) = -e-t

3 dV2 1 d2V2

+ + 0.1e-t = 0 … (2)20 dt 20 dt2

At t = 0+3 dV2(0+) 1 d2V2(0+)

+ + 0.1e-t(0+) = 020 dt 20 dt2


Simplifying

d2V2(0+)

= -8 Volt/sec2

dt2

Differentiating eq. 23 d2V2 1 d3V2

+ - 0.1e-t = 020 dt2 20 dt3

At t = 0+3 d2V2(0+) 1 d3V2(0+)

+ - 0.1e-t(0+) = 020 dt2 20 dt3

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 1 ΩL 1 HC ½ FV 2 V

Equivalent network before switching


i(0+) sc

V 2iL(0+) = iL(0-) = i(0+) = = = 2 Amp R 1 1Also VK = VC = ∫ idt CdVK i = dt CAt t = 0+dVK i(0+) = dt CdVK 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 AR 1000 ΩC 1 µ F

V


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

No current flows through R soVC(0-) = VC(0+) = i(0-)R = V(0+)Herei(0-) = 0VC(0-) = VC(0+) = (0)RVC(0-) = VC(0+) = 0 VoltsFor t ≥ 0; according to KCL at VV dV + C = I … (1) R dtAt t = 0+V(0+) dV(0+) + C = I R dtSimplifying

dV(0+) = 107 Volts/sec dt

Differentiating (1) with respect to ‘t’

1 dV d2V+ C = 0

R dt dt2

At t = 0+1 dV(0+) d2V(0+)

+ C = 0R dt dt2


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 1 AR 100 ΩL 1 H

V

Equivalent circuit before switching:

BecauseiL(0-) = 0 AiL(0-) = iL(0+) = 0 AThereforeAfter switching (t = 0+)

V


SoV(0+) = (I)(R) V(0+) = (1)(100)

V(0+) = 100 Volts

For t ≥ 0Applying KCL at node VV 1 + ∫ Vdt = I … (1)R LDifferentiating (1) with respect to ‘t’1 dV V

+ = 0 … (2) R dt LAt t = 0+1 dV(0+) V(0+)

+ = 0R dt LSimplifying

dV(0+)= -10, 000 Volts/sec

dt

Differentiating (2) with respect to ‘t’1 d2V 1 dV

+ = 0 R dt2 L dtAt t = 0+

1 d2V(0+) 1 dV(0+) + = 0

R dt2 L dtSimplifying

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

dt2


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 300Req = 40

Req = 7.5 Ω

After simplification Req

i(0-)5 V

Vi(0-) = Req

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


Va(0-)

20Va(0-) = (5) (10 + 20)

Va(0-) = 3.334 Volts

Also equivalent network at t = 0+t = 0+

Vb

Va


Applying KCL at node Va

Va – 5 Va - Vb Va

+ + = 0 10 20 10After simplification we getVb = 5Va – 10 … (i)Applying KCL at node Vb

Vb - Va Vb – 5 2+ + = 0

20 10 39Vb – 3Va + 10 = 0 … (ii) Substituting value of Vb from (i) into (ii)9[5Va – 10] – 3Va + 10 = 045Va – 90 – 3Va + 10 = 042Va – 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


Equivalent circuit before switching

Req

VC(0-) = Va(0-) = 0 VEquivalent network at t = 0+

Vb

Va

Applying KCL at node Va

+ -


Va – 5 Va - Vb Va

+ + = 0 10 20 10After simplification we getVb = 5Va – 10 … (i)At Vb = 5 V5 = 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)


Req = (10 + 20) + 10 300Req = 40

Req = 7.5 Ω

After simplification Req

i(0-)5 V

Vi(0-) = Req

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

iL(0-) = iL(0+) = 0.667 Amp.

Equivalent network at t = ∞

Va


Applying KCL at node Va iL(∞ )Va – 5 Va Va

+ + = 0 10 20 10After simplification we get

Va(∞ ) = 2 Volts

V Va

iL(∞ ) = + 10 20After 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


VC(0-) = VC(0+) = 5 VoltsAlso equivalent network at t = ∞ is

Vb

Va

According to KCL at Va

Va – 5 Va - Vb Va

+ + = 0 10 20 10


After simplification we getVa = 0.2Vb + 2 … (i)According to KCL at Vb

Vb – 5 Vb – Va

+ = 0 10 20 After simplification we get3Vb – Va – 10 = 0Putting the value of Va we get

Vb(∞ ) = 4.286 VoltsQ#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


According to statement under Q#5.13

At t = 0-VC(0-) 0 VVC(0+) 0 ViL(0-) 0 AiL(0+) 0 A

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

ViR2 = R1 + R2 VR2

V1(∞ ) = R1 + R2

VC(∞ ) = V – iR1R1 VR1

VC(∞ ) = V – R1 + R2

After simplification we get VR2


VC(∞ ) = R1 + R2

SinceVC(∞ ) = 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 LDifferentiating with respect to ‘t’

1 dV dV1 d2V1 1- + C + (V1 – V2) = 0 … (iii)


R1 dt dt dt2 L

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

1 dV2 1

+ (V2 – V1) = 0 … (iV) R2 dt L

As

V1 = V1 + V2

By putting the value of V1 in (iii) & (iv)

1 dV d(V1 + V2) d2(V1 + V2) 1- + C + (V1 + V2 – V2) = 0

R1 dt dt dt2 L

1 dV dV1 dV2 d2V1 d2V2 1- - + C + + (V1) = 0 … (V)

R1 dt dt dt dt2 dt2 L

1 dV2 1

+ (V2 – V1) = 0 … (iV) R2 dt L

1 dV2 1

+ (V2 – (V1 + V2)) = 0 … (iV) R2 dt L

1 dV2 1

+ (V2 – V1 - V2) = 0 R2 dt L

1 dV2 1

+ (–V1) = 0 … (Vi) R2 dt L

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


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 = 0- ViL(0-) = R1 + R2

VC(0-) = V – iR1(0-)R1

HereiR1(0-) = iL(0-)VC(0-) = V – iL(0-)R1

VR1

VC(0-) = V -


R1 + R2

VR2

VC(0-) = R1 + R2

V2 = iR2R2

ViR2(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

ViL(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 VR2

V1(0+) = - 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.

HenceVC(∞ ) = 0 ViL(∞ ) = 0 A


V2(∞ ) = iL(∞ )R2

V2(∞ ) = (0)R2

V2(∞ ) = 0 Volt

(c)For t ≥ 0, the equivalent network is

V1

V2

V2 V1

Applying KCL at node ‘V1’ 1 dV1

∫ (V1 – V2)dt + C = 0 … (i) L dt

d -V2

C (V1 + V2) = … (ii) dt R2

AsV1 = V1 + V2

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


L dtDifferentiating (i) with respect to ‘t’

1 d2V1 d2V2

V1 + C + C = 0 … (iii) L dt2 dt2

Differentiating (ii) with respect to ‘t’

d2 -1 dV2

C (V1 + V2) = … (iV) dt2 R2 dt

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


HereVC(0-) = VC(0+) = 0 VoltiL(0-) = iL(0+) = 0 Ai(0+) = 0 AiL(0+) = i(0+)Applying KVL around outside loop diL + iR2 = V0 … (i) dtAt t = 0+ di(0+)L + i(0+)R2 = V0 … (i) dtBy putting corresponding values we get

di(0+) V0

= dt L

Differentiating (i) with respect to ‘t’ d2i di L + R2 = 0 dt2 dtAt t = 0+ d2i(0+) di(0+) L + R2 = 0 dt2 dtSimplifying we get

d2i(0+) -RV0

= dt2 L2

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

iR1(0+) =


R1

V0R1

V1(0+) = R1

V1(0+) = V0

(b)AlsoV1 = V0 for all t ≥ 0

dV1(0+) d2V1(0+) = = 0 dt 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 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


Equivalent circuit for t ≥ 0

R1

Li

C

R2

Applying KVL around the loop 1 dii(R1 + R2) + ∫ idt + L = 0 … (i) C dtAlsoVR1 + 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+) HereVC(0+) = -V0

V0

i(0+) = R2

V0 V0

VL(0+) = - R1 - R2 – (-V0) R2 R2

V0 VL(0+) = - R1 – V0 + V0 R2


V0 VL(0+) = - R1 R2

And

diVL = L dtdi(0+) VL(0+) = dt LPutting corresponding value

di(0+) -V0R1

= dt R2L

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

di i d2i(R1 + R2) + + L = 0 dt C dt2

At t = 0+ di(0+) i(0+) d2i(0+)(R1 + R2) + + L = 0 dt C dt2

Here

di(0+) -V0R1

= dt R2L

V0

i(0+) = R2


d2i(0+) V0 R1(R1 + R2) 1 = - dt2 R2 L C


Also diV1 = L + iR2

dtDifferentiating with respect to ‘t’

dV1 d2i di = L + R2 dt dt2 dtAt t = 0+dV1(0+) d2i(0+) di(0+) = L + R2 dt dt2 dtBy putting corresponding values and simplifying

dV1(0+) V0 R12 1

= - dt R2 L C

We know -1V1 = ∫ idt - iR1

CDifferentiating with respect to ‘t’

dV1 -i di = - R1 dt C dtDifferentiating 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


d2V1(0+) V0R1 2 R1(R1 + R2) = - dt2 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


i1 = i2 = - R1 + R2 + R3

i3 = 0

a C3

bV 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 AiL2(0-) = iL2(0+) = 0 AVC1(0-) = VC1(0+) = 0 VVC2(0-) = VC2(0+) = 0 VEquivalent circuit after switching

L2

+ -


After simplification we get

i2

i1

Hence -Vi1 = i2 = R1 + R2 + R3

C1 behaves short circuit being uncharged at t = 0- & L1 behaving open circuit sinceiL(0-) = iL(0+) = 0 Aand 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

+ -

+ -


V0 1000 VR2 1 MΩC1 10 µ FC2 20 µ F

solve for d2i2/dt2 at t = 0+.

+C1 i1

V0 C2

- i2

R2

R1


i2

VC1(0-) = V0

VC2(0-) = 0 VEquivalent circuit after switching


K

+

V0 i1 i2

-

VC1(0+) = V0

VC2(0+) = 0 VFor t ≥ 0 For loop 1: 1R2(i1 – i2) + ∫ i1dt = 0 … (i) C1

For loop 2: 1R2(i2 – i1) + ∫ i2dt + R1i2 = 0 … (ii) C2

1R2(i1 – i2) = - ∫ i1dt C1

1R2(i2 – i1) = ∫ i1dt … (iii) C1

Taking loop around outsidei2R1 = 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


V0(R1 + R2)i1 = … (b) R1R2

Putting corresponding values we geti1(0+) 1.5(10-3) Amp.i2(0+) 5(10-4) Amp.

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 + = 0C1 dt dt2 C2 dtAt t = 0+

1 di1(0+) d2i2(0+) 1 di2(0+)


+ R1 + = 0C1 dt dt2 C2 dtPutting 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+.

i1 i2

Equivalent circuit after switching

At t ≥ 0 Applying KVL around outside loop di2

Ri2 + L = V0sin ω t … (i) dtApplying KVL around inside loop

+ -

+ -


1Ri1 + ∫ i1dt = V0sin ω t … (ii) CEquivalent circuit at t = 0+

iL(0+) = iL(0-) = 0 AVC(0+) = 0 V V0sin ω ti1 = RAt t = 0+ V0sin ω (0+)i1(0+) = R V0sin 0i1(0+) = Ri1(0+) = 0 AFrom (i)At t = 0+ di2(0+)Ri2(0+) + L = V0sin ω (0+) … (i) dtBy putting corresponding values we get

di2(0+) = 0 Amp/sec dt

Differentiating eq. (ii) with respect to ‘t’ 1Ri1 + ∫ i1dt = V0sin ω t … (ii) C

+ -


di1 i1

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

di1(0+) V0ω =

dt R

Q#5.20: In the network shown, a steady state is reached with the switch K open with V 100 VR1 10 ΩR2 20 ΩR3 20 ΩL 1 HC 1 µ 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


i2 i1


R2

R1

R3

Simplifying

i

VC(0-)

VC(0-) = iR2(R2) = VR2

Here ViR2 = R1 + R2

VR2

VR2 = R1 + R2


By putting corresponding values we get VC(0-) = VR2 = 66.667 V

ViL(0-) = R1 + R2

iL(0-) = 3.334 A

For t ≥ 0

i2 i1

Applying KVL around outside loop di1

R2i1 + L = V … (i) dtApplying KVL around inside loop 1R3i2 + ∫ i2dt = V … (ii) CSince

10iL(0+) = iL(0-) = iR2(0-) = i1(0-) = i1(0+) = Amp.

3 V – VC(0+)i2(0+) = R3

HereVC(0+) = 6.667 VoltsPutting corresponding values & simplifyingi2(0+) = 1.667 Amp.From eq. (i)At t = 0+ di1(0+)R2i1(0+) + L = V … (i) dt


Putting corresponding values we get

di1(0+) = 33.334 Amp/sec. dt

Differentiating eq. 2:

di2 i2

R3 + = 0 dt CAt t = 0+ di2(0+) i2(0+)R3 + = 0 dt CPutting corresponding values

di2(0+) = 83.334(104) Amp/sec. dt

From eq. (i)At t = ∞ di1(∞ )R2i1(∞ ) + L = V … (i) dtHeredi1(∞ ) = 0 Amp/sec dt

i1(∞ ) = 5 Amp.

Equivalent circuit after switching:

+ -


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 CR2

Initial conditions:iL(0-) = iL(0+) = 0 AVC(0-) = VC(0+) = 0 VApplying 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 dtAt t = 0+(V1(0+) – V2(0+)) 1 dV1(0+) di(0+) + = L R1 dt dtPutting corresponding values we get

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


= - dt dt L

Applying KCL at node ‘V2’1 V2 dV2

∫ (V2 – V1)dt + + C = 0 … (ii) L R2 dtAt t = 0+(V2(0+) – V1(0+)) 1 dV2(0+) dV2(0+) + + C = 0 L R2 dt dtPutting 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

+ -


R2

iL1(0-) = iL1(0+) = 0 AiL2(0-) = iL2(0+) = 0 AFor t ≥ 0According to KVLLoop 1:

di1 d(i1 – i2) d(i1 – i2) di1 di2

R1i1 + L1 + L3 + R2(i1 – i2) + M13 + M31 + (-M32) = V(t) dt dt dt dt dtAfter simplification di1 di2

i(R1 + R2) – i2R2 + (L1 + L3 + 2M13) - (L3 + M13 + M23) = V(t) … (i) dt dt

i1(0+) = i2(0+) = 0At t = 0+ di1(0+) di2(0+)i(0+)(R1 + R2) – i2(0+)R2 + (L1 + L3 + 2M13) - (L3 + M13 + M23) = V(t) dt dtPutting corresponding values we get

di1(0+) di2(0+) (L1 + L3 + 2M13) - (L3 + M13 + M23) = V … (ii) dt dt

According to KVLLoop 2: di2 d(i2 – i1) d(i2 – i1) di1 di2

R3i2 + L2 + L3 + R2(i2 – i1) + M23 - M31 + M32 = 0 dt dt dt dt dtAfter simplifying

di2 di1

i2(R3 + R2) – i1R2 + (L2 + L3 + 2M23) - (L3 + M13 + M23) = 0 … (iii) dt dt

At t = 0+ di2(0+) di1(0+)i2(0+)(R3 + R2) – i1(0+)R2 + (L2 + L3 + 2M23) - (L3 + M13 + M23) = 0


dt dtPutting corresponding values we get

di2(0+) di1(0+) (L2 + L3 + 2M23) - (L3 + M13 + M23) = 0 … (iv) dt 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 C i2 R2

V(t)

Initial conditions:iL(0-) = iL(0+) = 0 A = i2(0+)VC(0-) = VC(0+) = 0 VEquivalent circuit after switching

+ -

+ -


V(t)i1(t) = R1

At t = 0+

V(0+)i1(0+) = R1

For t ≥ 0Loop 1: 1R1i1 + ∫ (i1 – i2)dt = V(t) … (i) CDifferentiating (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 dtPutting corresponding values we get

di1(0+) dV(0) V(0) 1= -

dt dt R1C R1

Differentiating (ii) with respect to ‘t’ d2i1 1 di1 di2 d2V(t)R1 + - =

dt2 C dt dt dt2

At t = 0+

d2i1(0+) 1 di1(0+) di2(0+) d2V(0+)R1 + - =


dt2 C dt dt 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 dtWe know for t ≥ 0, according to KVL di 1L + ∫ idt = V(t) … (a) dt CAt t = 0+ di(0+) L + VC(0+) = V(0+) dt HereVC(0+) = 0 VV(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 VPutting corresponding values we getdi(0+) = 0 Amp/sec. dt iL(0-) = iL(0+) = i(0+) = 0 ANow di(0+)Va(0+) = VC(0+) + M … (i) dtPutting corresponding values we getVa(0+) = 0 VoltNowDifferentiating (i) with respect to ‘t’dVa dVC d2i


= + M … (b) dt dt dt2

Differentiating (a) with respect to ‘t’ d2i i dV(t)L + = … (a) dt2 C dtHere Vd(V(t)) = cos (t/(MC)1/2) (MC)1/2

At t = 0+ d2i(0+) i(0+) dV(0+)L + = … (c) dt2 C dtPutting corresponding values we getd2i(0+) V

= dt2 L(MC)1/2

At node a, apply KCL1 dVC

∫ (VC – V(t)) + C = 0 … (c)L dtRearranging dVC(0+)iL(0+) + C = 0 dt dVC(0+)0 + C = 0 dt

dVC(0+) = 0 dt

dVa dVC 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(0+) V M =


dt L C

Differentiating (b) with respect to ‘t’d2Va d2VC d3i = + M dt2 dt2 dt3

Differentiating (c) with respect to ‘t’ d3i(0+) 1 di(0+) d2V(0+)L + = … (c) dt3 C dt dt2

d2V(0+) =? dt2

Vd(V(t)) = cos (t/(MC)1/2) (MC)1/2

-Vd2(V(t)) = sin (t/(MC)1/2) (MC)

-Vd2(V(0+)) = sin (0+/(MC)1/2) (MC) -Vd2(V(0+)) = sin (0) (MC) -Vd2(V(0+)) = (0) (MC)

d2V(0+) = 0 dt2

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

+ C = 0 … (c) L dt2

At t = 0+(VC(0+) – V(0+)) d2VC(0+) + C = 0 … (c)


L dt2

Putting corresponding values

d2VC(0+)= 0 V/sec2

dt2

d2Va d2VC d3i = + M dt2 dt2 dt3

At t = 0+d2Va(0+) d2VC(0+) d3i(0+) = + M dt2 dt2 dt3

d2Va(0+)= 0 Volt/sec2

dt2

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+) = 1di(0+) = -1 dt


Equivalent network after switching:

VK

sc

V iL = R2

ViL(0-) = iL(0+) = R2

At t = 0+VK(0+) = VR1(0+)VR1(0+) = iL(0+)(R1)Putting corresponding value we get


VVR1(0+) = R1

R2

For t ≥ 0 1VK = iR1 + ∫ idt CDifferentiating with respect to ‘t’dVK di i = R1 +

dt dt CAt t = 0+dVK(0+) di(0+) i(0+) = R1 +

dt dt CPutting corresponding value we get

dVK(0+) 1 = - R dt 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


Equivalent network at t = 0+

After simplification:

R2

Va(0+) = VEquivalent network at t = ∞


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)) dV1 (V1 – V2) 1 + C1 + + ∫ (V1 – V3)dt = 0 … (i) R1 dt R2 L

Using KCL at node ‘V2’

(V2 – V1) dV2 1+ C2 + ∫ V2dt = 0 … (ii)

+ -


R2 dt L2

Using KCL at node ‘V3’

1 dV3

∫ (V3 – V1)dt + C3 = 0 … (iii)L1 dt

At t = 0+, capacitor C1 becomes short circuit as a result of whichV1(0+) 0 VV2(0+) 0 VV3(0+) 0 V

At t = 0+1 dV3

∫ (V3 – V1)dt + C3 = 0 … (iii)L1 dt

dV3(0+)iL1(0+) + C3 = 0 dtAfter simplification we get

dV3(0+) = 0 Volt/sec dt

HereiL1(0-) = iL1(0+) = 0 AAt t = 0+

(V2(0+) – V1(0+)) dV2(0+)

+ C2 + iL2(0+) = 0 … (ii) R2 dt iL2(0-) = iL2(0+) = 0 APutting corresponding values we get

dV2(0+) = 0 dt

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 LPutting corresponding values we get

iL3

(0+)


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


R1 R2

i(0-)

R3

At t = 0-VC1(0-) = i(o-) R2

Vi(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


At t ≥ 0 At node K, according to KCL d dVk VK – V1

C1 (VK – V1) + C3 + = 0 dt dt R2


dV1 1 dVK (VK – V1) = (C3 + C1) + … (i) dt C1 dt R2

At node ‘V1’ according to KCL(V1 - V) dV1 d(V1 - VK) (V1 - VK) + C2 + C1 + = 0 R1 dt dt R2


dV1 1 dVK (VK – V1) (V – V1) = C1 + + … (ii) dt (C1 + C2) dt R2 R1

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 +


V2

C

R1 R3 L

-

Equivalent network at t = 0-


R1 R2

I0

I0

I0


NowAccording to current divider rule: R1I0

iR2 = R1 + R2

We know V2(0-) = VL(0-) = 0Equivalent network at t = 0+

R1

I0R1 VC(0+)

iL(0+)

I0

+ -

+ -

+ -


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) 1 + C + ∫ V2dt … (ii)R3 dt L

From eq. (ii)

d(V1 – V2) V2 1C = + ∫ V2dt dt R3 L

+ -

+ -


Substituting the value of Cd(V1 – V2)/dt in (i) we getV2 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.

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Network Analysis by Van Valkenburg chap 5 solution manual