Chap#5 nwa

8/7/2019 Chap#5 nwa

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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/dt

2at t = 0+, for element values as

follows:

V 100 VR 1000

C 1 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 1000Element and initial condition Equivalent circuit at t = 0+

C

Sc

Switch

Drop

Rise i(0+) Short circuit

Drop

Applying KVL for t 0

Sum of voltage rise = sum of voltage drop

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1

V = iR + 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 F)(1 k )

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/dt

2at t = 0+ if

R 10

L 1 H

V 100 V

K

+ R

V

- L

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Key closed at t = 0

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 Equivalent circuit at t = 0+

oc



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

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Differentiating [eq. 1]

d2i d V iR

= -

dt2

dt L L

d2i -Rdi

=

dt2

Ldt

d2i(0+) -Rdi(0+)

=

dt2

Ldt


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/dt

2at t = 0+ if

R 1000

L 1 H

C 0.1 F

V 100 V

a K

b R

V C L

Equivalent circuit at t = 0+

b

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sc



1 Ldi

Ri + idt + = 0 [eq. 1]

C dt

Equivalent circuit at t = 0-

a

i(0+)

sc

V 100

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


di(0+)

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= -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


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

V 1

C1 C2 V 2

V1 2 V

V2 1 V

R 1

C1 1 F

C2 ½ F

Af ter switching:

V1 V2

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Applying KCL at node 1:

V1 – V2 dV1

+ C1 = 0

R dt

V1(0+) – V2(0+) dV1(0+)

+ C1 = 0

R dt


2 V – 1 V 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


1 V – 2 V 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/dt

2and

d3v2/dt

3at t = 0+.

NODE

R 1 +

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R 2 V2

V1 C

-

R 1

10 R 2 20

C 1/20 F

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

Applying KCL at NODE for t 0

V2 dV2 V2 – V1

+ C + = 0 … (1)

R 2 dt R 1At t = 0+

V2(0+) dV2(0+) V2(0+) – V1(0+)

+ C + = 0

R 2 dt R 1 V1 = e

-tVolts

V2 = VC(0+) = 0 Volts

0 dV2(0+) 0 – e-0+

+ C + = 0

R 2 dt R 1

Simplifying

dV2(0+)

= 2 Volt/sec

dt

Differentiating eq. 1 with respect to „t‟

1 dV2 d2V2 1 dV2 dV1

+ C + - = 0

R 2 dt dt2

R 1 dt dt

1 dV2 1 d2V2 1 dV2 d(e

-t)

+ + - = 0

20 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)

+ - = 0

20 dt 20 dt2

10 dt

d(e-t) = -e

-t

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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+)

= 0

20 dt 20 dt2

Simplifying

d2V2(0+)

= -8 Volt/sec2

dt2

Differentiating eq. 2

3 d2V2 1 d3V2

+ - 0.1e-t

= 0

20 dt2

20 dt3

At t = 0+

3 d2V2(0+) 1 d

3V2(0+)

+ - 0.1e-t(0+)

= 0

20 dt2

20 dt3

Putti ng corr esponding values and simpli fying

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

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R 1

L 1 H

C ½ F

V 2 V

Equi valent network before switchi ng

i(0+) sc

V 2

iL(0+) = iL(0-) = i(0+) = = = 2 Amp

R 1

1

Also VK = VC = idtC

dVK 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/dt

2if

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I 10 A

R 1000

C 1 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+)

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

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Differentiating (1) with respect to „t‟

1 dV d2V

+ C = 0

R dt dt2

At t = 0+

1 dV(0+) d2V(0+)

+ C = 0

R 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/dt

2at t = 0+ if

I 1 A

R 100

L 1 H

V

Equi valent circuit before switchi ng:

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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


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

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1 d2V 1 dV

+ = 0

R dt2

L dt

At t = 0+

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

+ = 0

R dt2

L dt

Simplifying

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+).

Circui t diagram:

Vb

Va


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sc

R eq = (10 + 20) 10

(10 + 20)(10)

R eq =

(10 + 20) + 10

300

R eq =

40

R eq = 7.5

After simplification R eq

i(0-)

5 V

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V

i(0-) =

R eq

5

i(0-) =

7.5

i(0-) = 0.667 Amp.

i(0-) = iL(0-)

Va(0-)

20

Va(0-) = (5)

(10 + 20)

Va(0-) = 3.334 Volts

Also equivalent network at t = 0+

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t = 0+

Vb

Va

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+).

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CIRCUIT DI AGRAM:

Vb

Va

K

Equi valent circuit before switchi ng

R eq

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VC(0-) = Va(0-) = 0 V

Equivalent network at t = 0+

Vb

Va

Applying KCL at node Va

Va – 5 Va - Vb Va

+ + = 0

10 20 10


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.


+

-

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sc

R eq = (10 + 20) 10

(10 + 20)(10)

R eq =

(10 + 20) + 10

300

R eq =

40

R eq = 7.5

After simplification R eq

i(0-)

5 V

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V

i(0-) =

R eq

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

Applying KCL at node Va iL()

Va – 5 Va Va

+ + = 0

10 20 10


Va() = 2 Volts

V Va

iL() = +

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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

VC(0-) = VC(0+) = 5 Volts

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Also equivalent network at t = is

Vb

Va

According to KCL at Va

Va – 5 Va - Vb Va

+ + = 0

10 20 10


Va = 0.2Vb + 2 … (i)

According to KCL at Vb

Vb – 5 Vb – Va

+ = 0

10 20


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/dt

2at t = 0+.

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CIRCUIT DIAGRAM

R 1

V1

L

C

V2

R 2

Equi valent network af ter switchi ng

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According to statement under Q#5.13

At t = 0-

VC(0-) 0 VVC(0+) 0 V

iL(0-) 0 A

iL(0+) 0 A


V2(0+) = iL(0+)(R 2)

V2(0+) = (0)(R 2)

V2(0+) = 0 V

V1(0+) + V2(0+) = VC(0+)


V1(0+) + 0 = 0

V1(0+) = 0

Equivalent circuit at t =

V1() = iR 2R 2V

iR2 =

R 1 + R 2

VR 2

V1() =

R 1 + R 2VC() = V – iR 1R 1

VR 1

VC() = V –

R 1 + R 2After simplification we get

VR 2

VC() =

R 1 + R 2

SinceVC() = V1() + V2()

VR 2

V2() =

R 1 + R 2V1() = VC() - V2()

VR 2 VR 2

V1() = -

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R 1 + R 2 R 1 + R 2

V1() = 0 Volts Self justified

(c)

V1

V2

According to KCL at node ‘V 1 ’

V – V1 dV1 1

+ C + V1 – V2)dt = 0 … (i)

R 1 dt L


1 dV dV1 d2V1 1

- + C + (V1 – V2) = 0 … (iii)

R 1 dt dt dt2

L

According to KCL at node ‘V 2 ’

V2 1

+ V2 – V1)dt = 0 … (ii)

R 2 L


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1 dV2 1

+ (V2 – V1) = 0 … (iV)

R 2 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

R 1 dt dt dt2

L

1 dV dV1 dV2 d2V1 d

2V2 1

- - + C + + (V1) = 0 … (V)

R 1 dt dt dt dt2

dt2

L

1 dV2 1

+ (V2 – V1) = 0 … (iV)

R 2 dt L

1 dV2 1

+ (V2 – (V1 + V2)) = 0 … (iV)

R 2 dt L

1 dV2 1

+ (V2 – V1 - V2) = 0

R 2 dt L

1 dV2 1

+ ( – V1) = 0 … (Vi)

R 2 dt L

From (V) & (Vi) we can find the values of dV 1/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)


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At t = 0-

V

iL(0-) =

R 1 + R 2VC(0-) = V – iR1(0-)R 1Here

iR1(0-) = iL(0-)

VC(0-) = V – iL(0-)R 1

VR 1 VC(0-) = V -

R 1 + R 2

VR 2

VC(0-) =

R 1 + R 2

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V2 = iR2R 2V

iR2(0-) = iR1(0-) =

R 1 + R 2

VR 2

V2 =

R 1 + R 2

VR 2

V2(0-) =

R 1 + R 2

V1(0-) + V2(0-) = VC(0-)

V1(0-) = VC(0-) - V2(0-)

VR 2 VR 2 V1(0-) = -

R 1 + R 2 R 1 + R 2

V1(0-) = 0 Volts

Equivalent network after switching

V1(0+)

VC(0+)

V2(0+)

VC(0-) = VC(0+)

VR 2

+-

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VC(0+) =

R 1 + R 2

V

iL(0-) = iL(0+) =

R 1 + R 2

V2(0+) = iL(0+)R 2

VR 2

V2(0+) =

R 1 + R 2

V1(0+) + V2(0+) = VC(0+)

V1(0+) = VC(0+) - V2(0+)

VR 2 VR 2

V1(0+) = -

R 1 + R 2 R 1 + R 2

V1(0+) = 0 Volts

(b)


At t = capacitor will be fully discharged and acts as an open circuit.

Hence

VC() = 0 V

iL() = 0 A

V2() = iL()R 2V2() = (0)R 2

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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 R 2

As

V1 = V1 + V2

1 d(V1 + V2)

(V1 + V2 – V2)dt + C = 0

L dt

1 d(V1 + V2)

V1dt + C = 0

L dt

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Differentiating (i) with respect to „t‟

1 d2V1 d

2V2

V1 + C + C = 0 … (iii)

L dt2

dt2

Differentiating (ii) with respect to „t‟

d2

-1 dV2

C (V1 + V2) = … (iV)

dt2

R 2 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/dt

2at t = 0+. (b)

Determine V1, dV1/dt, and d2V1/dt

2at 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 + iR 2 = V0 … (i)

dt

At t = 0+

di(0+)

L + i(0+)R 2 = V0 … (i)

dt

By putting corresponding values we get

di(0+) V0

=

dt L


d2i di

L + R 2 = 0

dt2

dt

At t = 0+

d2i(0+) di(0+)

L + R 2 = 0

dt2

dt

Simplifying we get

d2i(0+) -RV0

=

dt2

L2

Referring to the network at t = 0+

V1(0+) = VR1(0+) = iR1(0+)(R 1)

V0

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iR1(0+) =

R 1

V0R 1

V1(0+) =

R 1

V1(0+) = V0

(b)

Also

V1 = 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+.


V0

R 2

V0

iL(0-) =

R 2

VC(0-) = VR2(0-) = iL(0-)(R 2)

V0

VC(0-) = VR2(0-) = (R 2)

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+

-

R 2

VC(0-) = VR2(0-) = V0

As

VC(0-) = VC(0+) = V0

Equi valent network at t = 0+

i(0+)

V0

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

R 2V1(0+) = V0 – VR1(0+)

V1(0+) = V0 – iL(0+)R 1

V0

V1(0+) = V0 – R 1R 2

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R 2 – R 1

V1(0+) = V0 –

R 2

Equivalent cir cuit for t 0

R 1

L

i

C

R 2

Applying KVL around the loop

1 di

i(R 1 + R 2) + idt + L = 0 … (i)

C dt

Also

VR1 + VR2 + VL + VC = 0

i(0+)R 1 + i(0+)R 2 + VL(0+) + VC(0+) = 0

VL(0+) = -i(0+)R 1 - i(0+)R 2 - VC(0+)

Here

VC(0+) = -V0

V0

i(0+) =

R 2V0 V0

VL(0+) = - R 1 - R 2 – (-V0)

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R 2 R 2

V0

VL(0+) = - R 1 – V0 + V0

R 2

V0

VL(0+) = - R 1

R 2

And

di

VL = L

dt

di(0+) VL(0+)

=

dt L

Putting corresponding value

di(0+) -V0R 1

=

dt R 2L

Differentiating eq. (i) with respect to „t‟

1 di

i(R 1 + R 2) + idt + L = 0 … (i)

C dt

di i d2i

(R 1 + R 2) + + L = 0

dt C dt2

At t = 0+

di(0+) i(0+) d2i(0+)

(R 1 + R 2) + + L = 0

dt C dt2

Here

di(0+) -V0R 1

=

dt R 2L

V0

i(0+) =

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R 2Putting corresponding values and simplifying

d2i(0+) V0 R 1(R 1 + R 2) 1

= -

dt2 R 2 L C

Also

di

V1 = L + iR 2

dt


dV1 d2i di

= L + R 2

dt dt2 dt

At t = 0+

dV1(0+) d2i(0+) di(0+)

= L + R 2

dt dt2

dt

By putting corresponding values and simplifying

dV1(0+) V0 R 12

1

= -

dt R 2 L C

We know

-1

V1 = idt - iR 1

C


dV1 -i di

= - R 1

dt C dt


d2V1 -di 1 d

2i

= - R 1 dt

2dt C dt

2

At t = 0+

d2V1(0+) -di(0+) 1 d

2i(0+)

= - R 1

dt2

dt C dt2

Putting corresponding values and simplifying

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d2V1(0+) V0R 1 2 R 1(R 1 + R 2)

= -

dt2

R 2L 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 = -

R 1 + R 2 + R 3

i3 = 0

a C3

b

V R 2

+ R 3

i1 i2

- i3

R 1 L1

C1 C2


L2

+ -

8/7/2019 Chap#5 nwa




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



i2

i1

+ -

+ -

8/7/2019 Chap#5 nwa




Hence

-V

i1 = i2 =

R 1 + R 2 + R 3C1 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

R 1 2 M

V0 1000 V

R 2 1 M

C1 10 F

C2 20 F

solve for d2i2/dt

2at t = 0+.

+

C1 i1

V0 C2

- i2

R 2

R 1


i2

8/7/2019 Chap#5 nwa




VC1(0-) = V0

VC2(0-) = 0 VEquivalent circuit after switching

K

+

V0 i1 i2

-

VC1(0+) = V0

VC2(0+) = 0 V

For t 0

For loop 1:

1

R 2(i1 – i2) + i1dt = 0 … (i)

C1

For loop 2:

1

R 2(i2 – i1) + i2dt + R 1i2 = 0 … (ii)

C2

1

R 2(i1 – i2) = - i1dt

8/7/2019 Chap#5 nwa




C1

1

R 2(i2 – i1) = i1dt … (iii)

C1

Taking loop around outside

i2R 1 = V0

V0

i2 = … (a)

R 1

In loop 1

According to KVL :

R 2(i1 – i2) = V0

R 2i1 – R 2i2 = V0

Putting the value of i2 and simplifying

V0(R 1 + R 2)

i1 = … (b)

R 1R 2

Putting corresponding values we get

i1(0+) 1.5(10-

) Amp.

i2(0+) 5(10-4

) Amp.

Substituting value of R 2(i2 – i1) in eq. (ii)

1 1

i1dt + i2dt + R 1i2 = 0

C1 C2

Differentiating with respect to „t‟ i1 di2 i2

+ R 1 + = 0 … (iv)

C1 dt C2

At t = 0+

i1(0+) di2(0+) i2(0+)

+ R 1 + = 0

C1 dt C2


di2(0+)

= -8.75(10-5

) Amp/sec.dt

Differentiating eq. (iii) with respect to „t‟

di2 di1 i1

R 2 - R 2 =

dt dt C1

At t = 0+

8/7/2019 Chap#5 nwa




di2(0+) di1(0+) i1(0+)

R 2 - R 2 =

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

+ R 1 + = 0

C1 dt dt2

C2 dt

At t = 0+

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

+ R 1 + = 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 t, to the parallel RL-RC circuit. Find (a) di1/dt and (b) di2/dt at t =

0+.

i1 i2


+

-

+

8/7/2019 Chap#5 nwa




At t 0


di2

Ri2 + L = V0sin t … (i)

dt

Applying KVL around inside loop

1

Ri1 + i1dt = V0sin t … (ii)

C


iL(0+) = iL(0-) = 0 A

VC(0+) = 0 V

V0sin t

i1 =

R

At t = 0+

V0sin (0+)

i1(0+) =

R

V0sin 0

+

-

8/7/2019 Chap#5 nwa




i1(0+) =

R

i1(0+) = 0 A

From (i)

At t = 0+

di2(0+)

Ri2(0+) + L = V0sin (0+) … (i)

dt


di2(0+)

= 0 Amp/sec

dt

Differentiating eq. (ii) with respect to „t‟

1

Ri1 + i1dt = V0sin t … (ii)

C

di1 i1

R + = V0cos t

dt C

At t = 0+

di1(0+) i1(0+)

R + = V0cos (0+)

dt C

By putting corresponding values & simplifying we get

di1(0+) V0

=

dt R

Q#5.20: In the network shown, a steady state is reached with the switch K open with

V 100 V

R 1 10

R 2 20

R 3 20

L 1 HC 1 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+).

8/7/2019 Chap#5 nwa




(d) Solve for the values of di1/dt and di2/dt at t = 0+.

(e) What is the value of di1/dt at ?

Circui t diagram:

i

i2 i1


R 2

R 1

R 3

8/7/2019 Chap#5 nwa




Simplifying

i

VC(0-)

VC(0-) = iR2(R 2) = VR2

Here

V

iR2 =

R 1 + R 2

VR 2

VR2 =

R 1 + R 2


VC(0-) = VR2 = 66.667 V

V

iL(0-) =

R 1 + R 2

iL(0-) = 3.334 A

For t 0

i2 i1

8/7/2019 Chap#5 nwa





di1

R 2i1 + L = V … (i)

dt

Applying KVL around inside loop

1

R 3i2 + i2dt = V … (ii)

C

Since

10

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

3

V – VC(0+)

i2(0+) =

R 3

Here

VC(0+) = 6.667 Volts

Putting corresponding values & simplifying

i2(0+) = 1.667 Amp.

From eq. (i)

At t = 0+

di1(0+)

R 2i1(0+) + L = V … (i)

dt


di1(0+)

= 33.334 Amp/sec.

dt

Differentiating eq. 2:

di2 i2

R 3 + = 0

dt C

At t = 0+

di2(0+) i2(0+)

R 3 + = 0

dt C


di2(0+)

= 83.334(104) Amp/sec.

dt

From eq. (i)

8/7/2019 Chap#5 nwa




At t =

di1()

R 2i1() + L = V … (i)

dt

Heredi1()

= 0 Amp/sec

dt

i1() = 5 Amp.

Equivalent circui t 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

Circui t diagram:

V1 V2

L

i(t) K R 1 C

R 2

+

-

8/7/2019 Chap#5 nwa




I nitial conditions:

iL(0-) = iL(0+) = 0 A

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


1 V1

(V1 – V2)dt + = i(t) … (i)

L R 1


(V1 – V2) 1 dV1 di(t)

+ =

L R 1 dt dt

At t = 0+

(V1(0+) – V2(0+)) 1 dV1(0+) di(0+)

+ =

L R 1 dt dt


dV1(0+) di(t)(0+) V1(0+) R 1

= -

dt dt L


1 V2 dV2

(V2 – V1)dt + + C = 0 … (ii)

L R 2 dt

At t = 0+

(V2(0+) – V1(0+)) 1 dV2(0+) dV2(0+)

+ + C = 0

L R 2 dt dt


dV2(0+)

= 0 V/sec.

dt

Equivalent circuit at t = 0+ V1(0+) V2(0+)

8/7/2019 Chap#5 nwa




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

R 1

L3

R 3

i1 i2

R 2

iL1(0-) = iL1(0+) = 0 A

iL2(0-) = iL2(0+) = 0 A

For t 0

According to KVL

Loop 1:

di1 d(i1 – i2) d(i1 – i2) di1 di2

R 1i1 + L1 + L3 + R 2(i1 – i2) + M13 + M31 + (-M32) = V(t)

dt dt dt dt dt

Af ter simpli fi cation

di1 di2

i(R 1 + R 2) – i2R 2 + (L1 + L3 + 2M13) - (L3 + M13 + M23) = V(t) … (i)

+

-

8/7/2019 Chap#5 nwa




dt dt

i1(0+) = i2(0+) = 0

At t = 0+

di1(0+) di2(0+)

i(0+)(R 1 + R 2) – i2(0+)R 2 + (L1 + L3 + 2M13) - (L3 + M13 + M23) = V(t)

dt dt


di1(0+) di2(0+)

(L1 + L3 + 2M13) - (L3 + M13 + M23) = V … (ii)

dt dt

According to KVL

Loop 2:

di2 d(i2 – i1) d(i2 – i1) di1 di2

R 3i2 + L2 + L3 + R 2(i2 – i1) + M23 - M31 + M32 = 0

dt dt dt dt dt

After simplifying

di2 di1

i2(R 3 + R 2) – i1R 2 + (L2 + L3 + 2M23) - (L3 + M13 + M23) = 0 … (iii)

dt dt

At t = 0+

di2(0+) di1(0+)

i2(0+)(R 3 + R 2) – i1(0+)R 2 + (L2 + L3 + 2M23) - (L3 + M13 + M23) = 0

dt dt


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 dotconvention. 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+)/dt

2?

8/7/2019 Chap#5 nwa




CIRCUIT DI AGRAM:

L

R 1

i1 C i2 R 2

V(t)

I nitial conditions:

iL(0-) = iL(0+) = 0 A = i2(0+)

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


V(t)

i1(t) =

R 1

At t = 0+

V(0+)

i1(0+) =

R 1

For t 0

Loop 1:

1

R 1i1 + (i1 – i2)dt = V(t) … (i)

+

-

+

-

8/7/2019 Chap#5 nwa




C


di1 (i1 – i2) dV(t)

R 1 + = … (ii)

dt C dt

At t = 0+

di1(0+) (i1(0+) – i2(0+)) dV(0+)

R 1 + =

dt C dt


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

= -

dt dt R 1C R 1

Differentiating (ii) with respect to „t‟

d2i1 1 di1 di2 d

2V(t)

R 1 + - =

dt2

C dt dt dt2

At t = 0+

d2i1(0+) 1 di1(0+) di2(0+) d

2V(0+)

R 1 + - =

dt2

C dt dt dt2

From here we can determine the value of d2i1(0+)/dt

2.

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+)/dt

2= 0.

CIRCUIT DI AGRAM:

M

K

a

+

-

8/7/2019 Chap#5 nwa




Va

V(t)


After simplification

diL(0+)

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

dt

+

-

8/7/2019 Chap#5 nwa




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


di(0+)

= 0 Amp/sec.

dt

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

Now

di(0+)

Va(0+) = VC(0+) + M … (i)

dt


Va(0+) = 0 Volt

Now


dVa dVC d2i

= + M … (b)

dt dt dt2

Differentiating (a) with respect to „t‟

d2i i dV(t)

L + = … (a)

dt2

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)

dt2

C dt


d2i(0+) V

=

8/7/2019 Chap#5 nwa




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

dVa dVC d2i

= + M … (b)

dt dt dt2

At t = 0+

dVa(0+) dVC(0+) d2i(0+)

= + M … (b)

dt dt dt2


dVa(0+) V M=

dt L C

Differentiating (b) with respect to „t‟

d2Va d

2VC d

3i

= + M

dt2

dt2

dt3

Differentiating (c) with respect to „t‟

d3i(0+) 1 di(0+) d

2V(0+)

L + = … (c)

dt3 C dt dt2 d

2V(0+)

=?

dt2

V

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

)

(MC)1/2

8/7/2019 Chap#5 nwa




-V

d2(V(t)) = sin (t/(MC)

1/2)

(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

dt2

d3i(0+)

= 0 Amp/sec3

dt3

Differentiating (c) with respect to „t‟

(VC – V(t)) d2VC

+ C = 0 … (c) L dt

2

At t = 0+

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

L dt2


d2VC(0+)

= 0 V/sec2

dt2

d2Va d

2VC d

3i

= + M

dt2

dt2

dt3

At t = 0+

d2Va(0+) d

2VC(0+) d

3i(0+)

= + M

dt2

dt2

dt3

d2Va(0+)

= 0 Volt/sec2

8/7/2019 Chap#5 nwa




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 DI AGRAM:

+ V K -

R 2

R 1 C

V

i

L

Initial conditions:

i(0+) = 1

di(0+)

= -1

dt

8/7/2019 Chap#5 nwa




Equi valent network af ter switchi ng:

VK

sc

V

iL =

R 2

V

iL(0-) = iL(0+) =

R 2

At t = 0+

VK (0+) = VR1(0+)

VR1(0+) = iL(0+)(R 1)

Putting corresponding value we get

V

VR1(0+) = R 1

R 2

For t 0

1

VK = iR 1 + idt

C


dVK di i

= R 1 +

dt dt C

8/7/2019 Chap#5 nwa




At t = 0+

dVK (0+) di(0+) i(0+)

= R 1 +

dt dt C

Putting 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 DI AGRAM:

R 1

Va

K

C1 C2

R 2

V L


8/7/2019 Chap#5 nwa




After simplification:

R 2

Va(0+) = V


VC1() = V.

8/7/2019 Chap#5 nwa




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 DI AGRAM:

V1 V3

V2

Using KCL at node ‘V 1 ’

For t 0

(V1 – V(t)) dV1 (V1 – V2) 1

+ C1 + + (V1 – V3)dt = 0 … (i)

R 1 dt R 2 L


(V2 – V1) dV2 1

+ C2 + V2dt = 0 … (ii)

R 2 dt L2


1 dV3

(V3 – V1)dt + C3 = 0 … (iii)

L1 dt

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

+

-

8/7/2019 Chap#5 nwa




V1(0+) 0 V

V2(0+) 0 V

V3(0+) 0 V

At t = 0+

1 dV3

(V3 – V1)dt + C3 = 0 … (iii)

L1 dt

dV3(0+)

iL1(0+) + C3 = 0

dt


dV3(0+)

= 0 Volt/secdt

Here

iL1(0-) = iL1(0+) = 0 A

At t = 0+

(V2(0+) – V1(0+)) dV2(0+)

+ C2 + iL2(0+) = 0 … (ii)

R 2 dt

iL2(0-) = iL2(0+) = 0 A


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

R 1 dt R 2 L


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+.

iL3(0+)

8/7/2019 Chap#5 nwa




(b) Find dVK /dt at t = 0+.

CIRCUIT DI AGRAM:

V K

+ -

Equi valent network before switchi ng

R 1 R 2

i(0-)

8/7/2019 Chap#5 nwa




R 3

At t = 0-

VC1(0-) = i(o-) R 2

V

i(0-) =

R 1 + R 2 + R 3

VR 2

VC1(0-) =

R 1 + R 2 + R 3

VC1(0-) = i(o-) R 2

VR 3

VC1(0-) =

R 1 + R 2 + R 3VC2(0-) = V – VR1(0-)

Equi valent network af ter switchi ng

V1 K

8/7/2019 Chap#5 nwa




At t 0

At node K, according to KCL

d dVk VK – V1

C1 (VK – V1) + C3 + = 0

dt dt R 2


dV1 1 dVK (VK – V1)

= (C3 + C1) + … (i)

dt C1 dt R 2

At node ‘V 1 ’ according to KCL

(V1 - V) dV1 d(V1 - VK ) (V1 - VK )

+ C2 + C1 + = 0

R 1 dt dt R 2After simplification we get

dV1 1 dVK (VK – V1) (V – V1)

= C1 + + … (ii)

dt (C1 + C2) dt R 2 R 1

Equating (i) & (ii) we get dVK /dt at t = 0+.

Hint:

V1(0+) = VC2(0-) = V – VR1(0-)

Here

VR 1

VR1(0-) =

R 1 + R 2 + R 3

VR 2 + VR 3

V – VR1(0-) =

R 1 + R 2 + R 3

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 DI AGRAM:

1 2

R 2 +

V2

C

I0

8/7/2019 Chap#5 nwa




R 1 R 3 L

-

Equi valent network at t = 0-


R 1 R 2

I0

I0

8/7/2019 Chap#5 nwa




Now

According to current divider ru le:

R 1I0

iR2 =

R 1 + R 2We know V2(0-) = VL(0-) = 0

Equi valent network at t = 0+

R 1

I0R 1 VC(0+)

iL(0+)

I0

+ -

+

-

+ -

8/7/2019 Chap#5 nwa




After simplification

V2(0+)

VC(0+)

I0R 1

At node „V1‟

V1 d(V1 – V2)

+ C = I0 … (i)

R 1 dt

At node „V2‟

V2 d(V2 – V1) 1

+ C + V2dt … (ii)

R 3 dt L

From eq. (ii)

d(V1 – V2) V2 1

C = + V2dt

+

-

+ -

8/7/2019 Chap#5 nwa



dt R 3 L

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

V2 1 V1

+

V2dt + = I

0

R 3 L R 1

At t = 0+

V2(0+) 1 V1(0+)

+ V2(0+)dt + = I0 … (iii)

R 3 L R 1


I0R 1R 2

V1(0+) =

R 1 + R 2

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+) I0R 1R 2

=

R 1 (R 1 + R 2)R 1

dV2(0+) -I0R 1R 3

=

dt C(R 1 + R 2)(R 1 + R 3)

dV1(0+) -R 1 dV2(0+)

=

dt R 3 dt

THE END.

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Chap#5 nwa

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