ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<
ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<
9th Edition
ISBN: 9781260540666
Author: Hayt
Publisher: MCG CUSTOM
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Textbook Question
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Chapter 8, Problem 38E

The switch in Fig. 8.70 is moved from A to B at t = 0 after being at A for a long time. This places the two capacitors in series, thus allowing equal and opposite dc voltages to be trapped on the capacitors. (a) Determine v1(0), v2(0), and vR(0). (b) Find v1(0+), v2(0+), and vR(0+). (c) Determine the time constant of vR(t). (d) Find vR(t), t > 0. (e) Find i(t). (f) Find v1(t) and v2(t) from i(t) and the initial values. (g) Show that the stored energy at t = ∞ plus the total energy dissipated in the 20 kΩ resistor is equal to the energy stored in the capacitors at t = 0.

Chapter 8, Problem 38E, The switch in Fig. 8.70 is moved from A to B at t = 0 after being at A for a long time. This places

■ FIGURE 8.70

(a)

Expert Solution
Check Mark
To determine

Find the value of v1(0), v2(0) and vR(0).

Answer to Problem 38E

The value of v1(0) is 100V, v2(0) is 0V and vR(0) is 0V.

Explanation of Solution

Given data:

Refer to Figure 8.70 in the textbook.

The switch is moved from A to B at time t=0 and allows equal and opposite dc voltage to be trapped on the capacitors.

Calculation:

The given circuit is redrawn as shown in Figure 1.

ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<, Chapter 8, Problem 38E , additional homework tip  1

Refer to Figure 1, the voltage across the capacitor (C1) is v1, the voltage across the resistor (R2) is vR and the voltage across the capacitor (C2) is v2.

For a DC circuit at steady state condition, when the switch is in position A for time t<0, the capacitor acts like open circuit.

Now, the Figure 1 is reduced as shown in Figure 2.

ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<, Chapter 8, Problem 38E , additional homework tip  2

Refer to Figure 2, the voltage source (v) is connected in parallel with the capacitor (C1). For the parallel connection, the voltage is same.

The voltage across the capacitor (C1) is calculated as follows:

v1(0)=100V

Refer to Figure 2, the resistor (R2) and capacitor (C2) is not placed in a circuit. Therefore, the voltage across the resistor (R2) and capacitor (C2) is equal to zero.

vR(0)=0V

v2(0)=0V

The voltage across the capacitor is always continuous so that,

v1(0)=v1(0)=v1(0+)=100V

v2(0)=v2(0)=v2(0+)=0V

Conclusion:

Thus, the value of v1(0) is 100V, v2(0) is 0V and vR(0) is 0V.

(b)

Expert Solution
Check Mark
To determine

Find the value of v1(0+), v2(0+) and vR(0+).

Answer to Problem 38E

The value of v1(0+) is 100V, v2(0+) is 0V and vR(0+) is 100V.

Explanation of Solution

Calculation:

For time t>0, the switch is moved to position B and the Figure 1 is reduced as shown in Figure 3.

ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<, Chapter 8, Problem 38E , additional homework tip  3

Refer to part (a),

v1(0+)=100V

v2(0+)=0V

Apply Kirchhoff’s voltage for the circuit shown in Figure 3 for time t=0+.

v1(0+)+vR(0+)+v2(0+)=0

Substitute 100V for v1(0+), and 0V for v2(0+) in above equation to find vR(0+).

100V+vR(0+)+0V=0

vR(0+)=100V

Conclusion:

Thus, the value of v1(0+) is 100V, v2(0+) is 0V and vR(0+) is 100V.

(c)

Expert Solution
Check Mark
To determine

Find the value of the time constant.

Answer to Problem 38E

The value of time constant (τ) is 80ms.

Explanation of Solution

Formula used:

Write a general expression to calculate the time constant.

τ=RC        (1)

Here,

R is the value of resistance, and

C is the value of capacitance.

Calculation:

Refer to Figure 3, the capacitors C1 and C2 are connected in series.

The equivalent capacitance (C) is calculated as follows:

C=(20μF)(5μF)20μF+5μF=(20μF)(5μF)25μF=4μF

Now, the Figure 3 is reduced as shown in Figure 4.

ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<, Chapter 8, Problem 38E , additional homework tip  4

Refer to Figure 4, it shows the RC circuit.

Use equation (1) to find τ with a resistance of R2.

τ=R2C

Substitute 20kΩ for R2, and 4μF for C in above equation to find τ.

τ=(20kΩ)(4μF)=(20×103Ω)(4×106sΩ){1F=1s1Ω,1k=103,1μ=106}=80×103s=80ms{1m=103}

Conclusion:

Thus, the value of time constant (τ) is 80ms.

(d)

Expert Solution
Check Mark
To determine

Find the expression of voltage vR(t) for time t>0.

Answer to Problem 38E

The expression of voltage vR(t) for time t>0 is 100et80msV.

Explanation of Solution

Formula used:

Write a general expression to calculate the voltage response of an RC circuit.

vC(t)=vC(0+)etτ        (2)

Here,

vR(0) is the initial voltage across the resistor R2.

Refer to Figure 3, the resistor (R2) and capacitance (C) is connected in parallel. For the parallel connection, the voltage is same.

vR(t)=vC(t)

Substitute vR(t) for vC(t) in equation (2) to find vR(t).

vR(t)=vR(0+)etτ

Substitute 100V for vR(0+), and 80ms for τ in above equation to find vR(t).

vR(t)=100et80msV

Conclusion:

Thus, the expression of voltage vR(t) for time t>0 is 100et80msV.

(e)

Expert Solution
Check Mark
To determine

Find the expression of current i(t).

Answer to Problem 38E

The expression of current i(t) is 5et80msmA.

Explanation of Solution

Calculation:

Refer to Figure 3, the capacitor (C1), resistor (R2) and the capacitor (C2) are connected in series. For the series connection, the current is same.

The current i(t) is calculated as follows:

i(t)=vR(t)R2

Substitute 100et80msV for vR(t), and 20kΩ for R2 in above equation to find i(t).

i(t)=100et80msV20kΩ=100et80msV20×103Ω=5×103et80msA=5et80msmA

Conclusion:

Thus, the expression of current i(t) is 5et80msmA.

(f)

Expert Solution
Check Mark
To determine

Find the value of voltage v1(t) and v2(t) from i(t) and the initial values.

Answer to Problem 38E

The value of voltage v1(t) is [10020(1et80ms)]V and v2(t) is 80(1et80ms)V.

Explanation of Solution

Formula used:

Write a general expression to calculate the voltage across the capacitor.

v(t)=1Cidt+v(0)        (3)

Calculation:

From the given data, the capacitor (C1) and (C2) has equal and opposite dc voltages.

Use equation (1) to find the voltage v1(t) across the capacitor (C1).

v1(t)=[1C1idt]+v1(0)

Substitute 20μF for C1, 5et80msmA for i(t), and 100V for v1(0) in above equation to find v1(t).

v1(t)=[120μF0t5et80msmAdt]+100V=[120×106F0t5×103et80×103sAdt]+100V{1m=1031μ=106}=[5×10320×106F([et80×103s(180×103)]0tAs)]+100V=[5×103(80×103)20×106AsV([et80×103s]0tAs)]+100V{1F=1A1s1V}

Simplify the above equation to find v1(t).

v1(t)=[5×103(80×103)[et80×103s+e080×103s]20×106V]+100V=20(et80×103s+1)V+100V{e0=1}=[10020(1et80ms)]V{1m=103}

Use equation (1) to find the voltage v2(t) across the capacitor (C2).

v2(t)=[1C2idt]+v2(0)

Substitute 5μF for C2, 5et80msmA for i(t), and 0V for v2(0) in above equation to find v2(t).

v2(t)=[15μF0t5et80msmAdt]+0V=[15×106F0t5×103et80×103sAdt]{1m=1031μ=106}=[5×1035×106F([et80×103s(180×103)]0tAs)]=[5×103(80×103)5×106AsV([et80×103s]0tAs)]{1F=1A1s1V}

Simplify the above equation to find v1(t).

v2(t)=[5×103(80×103)[et80×103s+e080×103s]5×106V]=80(et80×103s+1)V{e0=1}=80(1et80ms)V{1m=103}

Conclusion:

Thus, the value of voltage v1(t) is [10020(1et80ms)]V and v2(t) is 80(1et80ms)V.

(g)

Expert Solution
Check Mark
To determine

Show that the stored energy at t= plus the total energy dissipated in 20kΩ resistor is equal to the energy stored in the capacitors at t=0.

Explanation of Solution

Formula used:

Write a general expression to calculate the energy stored in a capacitor.

w=12Cv2        (4)

Write a general expression to calculate the energy stored in a resistor.

wR=i2(t)Rdt        (5)

Here,

R is the value of resistance.

Calculation:

Refer to Figure 1, there are two capacitors placed in a circuit. Therefore, the total energy stored in a capacitor is,

w=w1+w2        (6)

Use equation (1) to find w1 for capacitor C1.

w1=12C1v12(t)        (7)

Use equation (1) to find w2 for capacitor C2.

w2=12C2v22(t)        (8)

Substitute equation (7) and (8) in (6).

w=12C1v12(t)+12C2v22(t)        (9)

Substitute 0 for t in equation (9) to find w(0).

w(0)=12C1(v1(0))2+12C2(v2(0))2

Substitute 20μF for C1, 5μF for C2, 100V for v1(0), and 0V for v2(0) in above equation to find w(0).

w(0)=12(20μF)(100V)2+12(5μF)(0V)2=12(20×106F)(10000V2)+0{1μ=106}=12(20×106CV)(10000V2){1F=1C1V}=0.1CV

Simplify the above equation to find w(0).

w(0)=0.1×103×103J{1J=1C1V}=100mJ{1m=103}

Refer to part (f),

v1(t)=[10020(1et80ms)]V

Substitute for t in above equation to find v1().

v1()=[10020(1e80ms)]V=[10020(1e)]V=[10020(10)]V{e=0}=80V

Refer to part (f),

v2(t)=80(1et80ms)V

Substitute for t in above equation to find v2().

v2()=80(1e80ms)V=80(1e)V=80(10)V{e=0}=80V

Substitute for t in equation (9) to find w().

w()=12C1(v1())2+12C2(v2())2

Substitute 20μF for C1, 5μF for C2, 80V for v1(), and 80V for v2() in above equation to find w().

w()=12(20μF)(80V)2+12(5μF)(80V)2=12(80V)2[20μF+5μF]=12(80V)2[25μF]=12(6400V2)(25×106F){1μ=106}

Simplify the above equation to find w(0).

w()=12(25×106CV)(6400V2){1F=1C1V}=0.08CV=0.08×103×103J{1J=1C1V}=80mJ{1m=103}

Substitute 20kΩ for R, and 5et80msmA for i(t) in equation (5) to find wR.

wR=0t(5et80msmA)2(20kΩ)dt=(20kΩ)0t(5et80msmA)2dt=(20×103Ω)0t(5×103et80×103A)2dt{1k=103}=(20×103Ω)(5×103)20te2t80×103A2dt

Simplify the above equation to find wR.

wR=(20×103Ω)(5×103)20tet40×103A2dt=(20×103Ω)(5×103)2[et40×103]0t140×103A2s=(20×103)(25×106)(40×103)[et40×103+e040×103]A2H{1H=1Ω1s}=0.02[1et40×103]A2JA2{1H=1J1A2}

Simplify the above equation to find wR.

wR=0.02[1et40×103]J=0.02×103×103[1et40×103]J=20[1et40×103]mJ{1m=103}

Substitute for t in above equation to find wR().

wR()=20[1e40×103]mJ=20[1e]mJ=20[10]mJ{e=0}=20mJ

Add wR() and w().

wR()+w()=100mJ

Therefore,

wR()+w()=w(0)=100mJ

Therefore, the stored energy at t= plus the total energy dissipated in 20kΩ resistor is equal to the energy stored in the capacitors at t=0.

Conclusion:

Thus, the stored energy at t= plus the total energy dissipated in 20kΩ resistor is equal to the energy stored in the capacitors at t=0 is showed.

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