In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to position 2 at t = 0. Find: (a) v (0 + ), dv (0 + )/ dt (b) v ( t ) for t ≥ 0.

Question

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Answer 1

Textbook Question

Chapter 16, Problem 53P

In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to position 2 at t = 0. Find:

(a) v(0⁺), dv(0⁺)/dt
(b) v(t) for t ≥ 0.

Chapter 16, Problem 53P, In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to

a.

Expert Solution

To determine

Find the value of v(0+) and dv(0+)dt.

Answer to Problem 53P

The value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

Explanation of Solution

Given data:

Refer to Figure 16.76 in the textbook.

The switch is in position 1 for a long time and moved to position 2 at t=0.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 1

For a DC circuit, at steady state condition when the switch is in position ‘1’at time t=0− the capacitor acts like open circuit and the inductor acts like short circuit.

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

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 2

Refer to Figure 2, the voltage across the resistor is same as the voltage across the capacitor which is the source voltage.

vC(0−)=10 V.

The current through inductor and voltage across capacitor is always continuous so that,

i(0)=iL(0−)=iL(0+)=0 A

v(0)=vC(0−)=vC(0+)=10 V

When the switch is in position ‘2’, the Figure 1 is reduced as shown in Figure 3.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 3

Refer to Figure 3, the capacitor, resistor and inductor are connected in parallel. For the parallel connection the voltage is same. In Figure 3, the magnitude of voltage is in opposite direction.

vR(0+)=vL(0+)=vC(0+)

Apply Kirchhoff’s current law for Figure 3.

iL(0+)=iC(0+)+iR(0+)

Substitute 0 for iL(0+) in above equation to find iC(0+).

0=iC(0+)+iR(0+)

iC(0+)=−iR(0+) (1)

Write an expression to calculate the current through resistor.

iR(0+)=vR(0+)R

Substitute vC(0+) for vR(0+) in above equation to find iR(0+).

iR(0+)=vC(0+)R

Substitute 10 V for vC(0+) and 0.5 Ω for R in above equation to find iR(0+).

iR(0+)=10 V0.5 Ω=20 A

Substitute 20 A for iR(0+) in equation (1) to find iC(0+).

iC(0+)=−20 A

At time t=0+, the capacitor current is calculated as follows:

iC(0+)=Cdv(0+)dt

Rearrange the above equation to find dv(0+)dt.

dv(0+)dt=iC(0+)C

Substitute −20 A for iC(0+), and 1 F for C in above equation to find dv(0+)dt.

dv(0+)dt=−20 A1 F=−20 A1 (A⋅sV) {∵1 F=1 A⋅1 s1 V}=−20 Vs

Conclusion:

Thus, the value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

b.

Expert Solution

To determine

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

Answer to Problem 53P

The expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

Explanation of Solution

Formula used:

Write a general expression to calculate the impedance of a resistor in s-domain.

ZR=R (2)

Here,

R is the value of resistance.

Write a general expression to calculate the impedance of an inductor in s-domain.

ZL=sL (3)

Here,

L is the value of inductance.

Write a general expression to calculate the impedance of a capacitor in s-domain.

ZC=1sC (4)

Here,

C is the value of capacitance.

Calculation:

Substitute 0.5 Ω for R1 in equation (2) to find ZR1.

ZR1=0.5 Ω

Substitute 0.25 H for L in equation (3) to find ZL.

ZL=s(0.25 H)=0.25s

Substitute 1 F for C in equation (4) to find ZC.

ZC=1s(1 F)=1s

Using element transformation methods with initial conditions convert the Figure 3 into s-domain.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 4

Apply nodal analysis at node V(s) for the circuit shown in Figure 4.

V(s)0.25s+V(s)0.5+V(s)−v(0)s(1s)=0V(s)0.25s+V(s)0.5+sV(s)−(s)v(0)s=0V(s)0.25s+V(s)0.5+sV(s)−v(0)=0V(s)[10.25s+10.5+s]=v(0)

Substitute 10 for v(0) in above equation.

V(s)[10.25s+10.5+s]=10V(s)[2+s+0.5s20.5s]=10V(s)[0.5s2+s+20.5s]=10V(s)[0.5(s2+2s+4)0.5s]=10

Simplify the above equation to find V(s).

V(s)=10ss2+2s+4 (5)

From the above equation , the characteristic equation is

s2+2s+4=0 (6)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (7)

Comparing the equation (6) with the equation (as2+bs+c=0).

a=1b=2c=4

Substitute 1 for a, 2 for b, and 4 for c in equation (7) to find s1,2.

s1,2=−2±(2)2−4(1)(4)2(1)=−2±4−162(1)=−2±−122=−2±j3.46422

Simplify the above equation to find s1,2.

s1,2=−2+j3.46422,−2−j3.46422=−1+j1.7321,−1−j1.7321

Substitute the roots of characteristic equation in equation (5) to find V(s).

V(s)=10ss(s−(−1+j1.7321))((s−(−1−j1.7321)))=10ss(s−(−1+j1.7321))(s−(−1−j1.7321))

Take partial fraction for above equation.

V(s)=10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1+j1.7321))+B(s−(−1−j1.7321))] (8)

The equation (8) can also be written as follows:

10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1−j1.7321))+B(s−(−1+j1.7321))[(s−(−1+j1.7321))(s−(−1−j1.7321))]]

Simplify the above equation as follows:

10s=A(s−(−1−j1.7321))+B(s−(−1+j1.7321)) . (9)

Substitute −1+j1.732 for s in equation (9) to find A.

10(−1+j1.7321)=A((−1+j1.7321)−(−1−j1.7321))+B((−1+j1.7321)−(−1+j1.7321))−10+j17.321=A(j3.4642)+B(0)A(j3.4642)=−10+j17.321

Simplify the above equation to find A.

A=−10+j17.321j3.4642=5+j2.886=5.773∠30°

Substitute −1−j1.7321 for s in equation (9) to find B.

10(−1−j1.7321)=A((−1−j1.7321)−(−1−j1.7321))+B((−1−j1.7321)−(−1+j1.7321))−10−j17.321=A(0)+B(−j3.4642)B(−j3.4642)=−10−j17.321

Simplify the above equation to find B.

B=−10−j17.321−j3.4642=5−j2.886=5.773∠−30°

Substitute 5.773∠30° for A, and 5.773∠−30° for B in equation (8) to find V(s).

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]=[5.773(cos(30°)+jsin(30°))(s−(−1+j1.7321))+20.587(cos(−30°)+jsin(−30°))(s−(−1−j1.7321))]=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))] {∵eiθ=cosθ+isinθ}

Take inverse Laplace transform for above equation to find v(t).

v(t)=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))]=[5.773ej30°e(−1+j1.7321)tu(t)+5.773e−j30°e(−1−j1.7321)tu(t)] {∵L−1(1s−a)=eatu(t)}=[5.773e(−1t+j41.7321t+j30°)+5.773e(−1t−j1.7321t−j30°)]u(t) {∵ea⋅eb=ea+b}=[5.773e−tej1.7321t+j30°+5.773e−te−j1.7321t−j30°]u(t)=[5.773e−tej(1.7321t+30°)+5.773e−te−j(1.7321t+30°)]u(t)

Simplify the above equation to find v(t).

v(t)==[5.773e−t[ej(1.7321t+30°)+e−j(1.7321t+30°)]]u(t)=[5.773e−t[2cos(1.7321t+30°)]]u(t) {∵cosθ=eiθ+e−iθ2}=[11.547e−tcos(1.7321t+30°)]u(t) V

Conclusion:

Thus, the expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

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Answer 2

Textbook Question

Chapter 16, Problem 53P

In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to position 2 at t = 0. Find:

(a) v(0⁺), dv(0⁺)/dt
(b) v(t) for t ≥ 0.

Chapter 16, Problem 53P, In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to

a.

Expert Solution

To determine

Find the value of v(0+) and dv(0+)dt.

Answer to Problem 53P

The value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

Explanation of Solution

Given data:

Refer to Figure 16.76 in the textbook.

The switch is in position 1 for a long time and moved to position 2 at t=0.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 1

For a DC circuit, at steady state condition when the switch is in position ‘1’at time t=0− the capacitor acts like open circuit and the inductor acts like short circuit.

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

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 2

Refer to Figure 2, the voltage across the resistor is same as the voltage across the capacitor which is the source voltage.

vC(0−)=10 V.

The current through inductor and voltage across capacitor is always continuous so that,

i(0)=iL(0−)=iL(0+)=0 A

v(0)=vC(0−)=vC(0+)=10 V

When the switch is in position ‘2’, the Figure 1 is reduced as shown in Figure 3.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 3

Refer to Figure 3, the capacitor, resistor and inductor are connected in parallel. For the parallel connection the voltage is same. In Figure 3, the magnitude of voltage is in opposite direction.

vR(0+)=vL(0+)=vC(0+)

Apply Kirchhoff’s current law for Figure 3.

iL(0+)=iC(0+)+iR(0+)

Substitute 0 for iL(0+) in above equation to find iC(0+).

0=iC(0+)+iR(0+)

iC(0+)=−iR(0+) (1)

Write an expression to calculate the current through resistor.

iR(0+)=vR(0+)R

Substitute vC(0+) for vR(0+) in above equation to find iR(0+).

iR(0+)=vC(0+)R

Substitute 10 V for vC(0+) and 0.5 Ω for R in above equation to find iR(0+).

iR(0+)=10 V0.5 Ω=20 A

Substitute 20 A for iR(0+) in equation (1) to find iC(0+).

iC(0+)=−20 A

At time t=0+, the capacitor current is calculated as follows:

iC(0+)=Cdv(0+)dt

Rearrange the above equation to find dv(0+)dt.

dv(0+)dt=iC(0+)C

Substitute −20 A for iC(0+), and 1 F for C in above equation to find dv(0+)dt.

dv(0+)dt=−20 A1 F=−20 A1 (A⋅sV) {∵1 F=1 A⋅1 s1 V}=−20 Vs

Conclusion:

Thus, the value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

b.

Expert Solution

To determine

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

Answer to Problem 53P

The expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

Explanation of Solution

Formula used:

Write a general expression to calculate the impedance of a resistor in s-domain.

ZR=R (2)

Here,

R is the value of resistance.

Write a general expression to calculate the impedance of an inductor in s-domain.

ZL=sL (3)

Here,

L is the value of inductance.

Write a general expression to calculate the impedance of a capacitor in s-domain.

ZC=1sC (4)

Here,

C is the value of capacitance.

Calculation:

Substitute 0.5 Ω for R1 in equation (2) to find ZR1.

ZR1=0.5 Ω

Substitute 0.25 H for L in equation (3) to find ZL.

ZL=s(0.25 H)=0.25s

Substitute 1 F for C in equation (4) to find ZC.

ZC=1s(1 F)=1s

Using element transformation methods with initial conditions convert the Figure 3 into s-domain.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 4

Apply nodal analysis at node V(s) for the circuit shown in Figure 4.

V(s)0.25s+V(s)0.5+V(s)−v(0)s(1s)=0V(s)0.25s+V(s)0.5+sV(s)−(s)v(0)s=0V(s)0.25s+V(s)0.5+sV(s)−v(0)=0V(s)[10.25s+10.5+s]=v(0)

Substitute 10 for v(0) in above equation.

V(s)[10.25s+10.5+s]=10V(s)[2+s+0.5s20.5s]=10V(s)[0.5s2+s+20.5s]=10V(s)[0.5(s2+2s+4)0.5s]=10

Simplify the above equation to find V(s).

V(s)=10ss2+2s+4 (5)

From the above equation , the characteristic equation is

s2+2s+4=0 (6)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (7)

Comparing the equation (6) with the equation (as2+bs+c=0).

a=1b=2c=4

Substitute 1 for a, 2 for b, and 4 for c in equation (7) to find s1,2.

s1,2=−2±(2)2−4(1)(4)2(1)=−2±4−162(1)=−2±−122=−2±j3.46422

Simplify the above equation to find s1,2.

s1,2=−2+j3.46422,−2−j3.46422=−1+j1.7321,−1−j1.7321

Substitute the roots of characteristic equation in equation (5) to find V(s).

V(s)=10ss(s−(−1+j1.7321))((s−(−1−j1.7321)))=10ss(s−(−1+j1.7321))(s−(−1−j1.7321))

Take partial fraction for above equation.

V(s)=10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1+j1.7321))+B(s−(−1−j1.7321))] (8)

The equation (8) can also be written as follows:

10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1−j1.7321))+B(s−(−1+j1.7321))[(s−(−1+j1.7321))(s−(−1−j1.7321))]]

Simplify the above equation as follows:

10s=A(s−(−1−j1.7321))+B(s−(−1+j1.7321)) . (9)

Substitute −1+j1.732 for s in equation (9) to find A.

10(−1+j1.7321)=A((−1+j1.7321)−(−1−j1.7321))+B((−1+j1.7321)−(−1+j1.7321))−10+j17.321=A(j3.4642)+B(0)A(j3.4642)=−10+j17.321

Simplify the above equation to find A.

A=−10+j17.321j3.4642=5+j2.886=5.773∠30°

Substitute −1−j1.7321 for s in equation (9) to find B.

10(−1−j1.7321)=A((−1−j1.7321)−(−1−j1.7321))+B((−1−j1.7321)−(−1+j1.7321))−10−j17.321=A(0)+B(−j3.4642)B(−j3.4642)=−10−j17.321

Simplify the above equation to find B.

B=−10−j17.321−j3.4642=5−j2.886=5.773∠−30°

Substitute 5.773∠30° for A, and 5.773∠−30° for B in equation (8) to find V(s).

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]=[5.773(cos(30°)+jsin(30°))(s−(−1+j1.7321))+20.587(cos(−30°)+jsin(−30°))(s−(−1−j1.7321))]=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))] {∵eiθ=cosθ+isinθ}

Take inverse Laplace transform for above equation to find v(t).

v(t)=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))]=[5.773ej30°e(−1+j1.7321)tu(t)+5.773e−j30°e(−1−j1.7321)tu(t)] {∵L−1(1s−a)=eatu(t)}=[5.773e(−1t+j41.7321t+j30°)+5.773e(−1t−j1.7321t−j30°)]u(t) {∵ea⋅eb=ea+b}=[5.773e−tej1.7321t+j30°+5.773e−te−j1.7321t−j30°]u(t)=[5.773e−tej(1.7321t+30°)+5.773e−te−j(1.7321t+30°)]u(t)

Simplify the above equation to find v(t).

v(t)==[5.773e−t[ej(1.7321t+30°)+e−j(1.7321t+30°)]]u(t)=[5.773e−t[2cos(1.7321t+30°)]]u(t) {∵cosθ=eiθ+e−iθ2}=[11.547e−tcos(1.7321t+30°)]u(t) V

Conclusion:

Thus, the expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

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Answer 3

Textbook Question

Chapter 16, Problem 53P

In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to position 2 at t = 0. Find:

(a) v(0⁺), dv(0⁺)/dt
(b) v(t) for t ≥ 0.

Chapter 16, Problem 53P, In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to

a.

Expert Solution

To determine

Find the value of v(0+) and dv(0+)dt.

Answer to Problem 53P

The value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

Explanation of Solution

Given data:

Refer to Figure 16.76 in the textbook.

The switch is in position 1 for a long time and moved to position 2 at t=0.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 1

For a DC circuit, at steady state condition when the switch is in position ‘1’at time t=0− the capacitor acts like open circuit and the inductor acts like short circuit.

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

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 2

Refer to Figure 2, the voltage across the resistor is same as the voltage across the capacitor which is the source voltage.

vC(0−)=10 V.

The current through inductor and voltage across capacitor is always continuous so that,

i(0)=iL(0−)=iL(0+)=0 A

v(0)=vC(0−)=vC(0+)=10 V

When the switch is in position ‘2’, the Figure 1 is reduced as shown in Figure 3.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 3

Refer to Figure 3, the capacitor, resistor and inductor are connected in parallel. For the parallel connection the voltage is same. In Figure 3, the magnitude of voltage is in opposite direction.

vR(0+)=vL(0+)=vC(0+)

Apply Kirchhoff’s current law for Figure 3.

iL(0+)=iC(0+)+iR(0+)

Substitute 0 for iL(0+) in above equation to find iC(0+).

0=iC(0+)+iR(0+)

iC(0+)=−iR(0+) (1)

Write an expression to calculate the current through resistor.

iR(0+)=vR(0+)R

Substitute vC(0+) for vR(0+) in above equation to find iR(0+).

iR(0+)=vC(0+)R

Substitute 10 V for vC(0+) and 0.5 Ω for R in above equation to find iR(0+).

iR(0+)=10 V0.5 Ω=20 A

Substitute 20 A for iR(0+) in equation (1) to find iC(0+).

iC(0+)=−20 A

At time t=0+, the capacitor current is calculated as follows:

iC(0+)=Cdv(0+)dt

Rearrange the above equation to find dv(0+)dt.

dv(0+)dt=iC(0+)C

Substitute −20 A for iC(0+), and 1 F for C in above equation to find dv(0+)dt.

dv(0+)dt=−20 A1 F=−20 A1 (A⋅sV) {∵1 F=1 A⋅1 s1 V}=−20 Vs

Conclusion:

Thus, the value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

b.

Expert Solution

To determine

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

Answer to Problem 53P

The expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

Explanation of Solution

Formula used:

Write a general expression to calculate the impedance of a resistor in s-domain.

ZR=R (2)

Here,

R is the value of resistance.

Write a general expression to calculate the impedance of an inductor in s-domain.

ZL=sL (3)

Here,

L is the value of inductance.

Write a general expression to calculate the impedance of a capacitor in s-domain.

ZC=1sC (4)

Here,

C is the value of capacitance.

Calculation:

Substitute 0.5 Ω for R1 in equation (2) to find ZR1.

ZR1=0.5 Ω

Substitute 0.25 H for L in equation (3) to find ZL.

ZL=s(0.25 H)=0.25s

Substitute 1 F for C in equation (4) to find ZC.

ZC=1s(1 F)=1s

Using element transformation methods with initial conditions convert the Figure 3 into s-domain.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 4

Apply nodal analysis at node V(s) for the circuit shown in Figure 4.

V(s)0.25s+V(s)0.5+V(s)−v(0)s(1s)=0V(s)0.25s+V(s)0.5+sV(s)−(s)v(0)s=0V(s)0.25s+V(s)0.5+sV(s)−v(0)=0V(s)[10.25s+10.5+s]=v(0)

Substitute 10 for v(0) in above equation.

V(s)[10.25s+10.5+s]=10V(s)[2+s+0.5s20.5s]=10V(s)[0.5s2+s+20.5s]=10V(s)[0.5(s2+2s+4)0.5s]=10

Simplify the above equation to find V(s).

V(s)=10ss2+2s+4 (5)

From the above equation , the characteristic equation is

s2+2s+4=0 (6)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (7)

Comparing the equation (6) with the equation (as2+bs+c=0).

a=1b=2c=4

Substitute 1 for a, 2 for b, and 4 for c in equation (7) to find s1,2.

s1,2=−2±(2)2−4(1)(4)2(1)=−2±4−162(1)=−2±−122=−2±j3.46422

Simplify the above equation to find s1,2.

s1,2=−2+j3.46422,−2−j3.46422=−1+j1.7321,−1−j1.7321

Substitute the roots of characteristic equation in equation (5) to find V(s).

V(s)=10ss(s−(−1+j1.7321))((s−(−1−j1.7321)))=10ss(s−(−1+j1.7321))(s−(−1−j1.7321))

Take partial fraction for above equation.

V(s)=10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1+j1.7321))+B(s−(−1−j1.7321))] (8)

The equation (8) can also be written as follows:

10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1−j1.7321))+B(s−(−1+j1.7321))[(s−(−1+j1.7321))(s−(−1−j1.7321))]]

Simplify the above equation as follows:

10s=A(s−(−1−j1.7321))+B(s−(−1+j1.7321)) . (9)

Substitute −1+j1.732 for s in equation (9) to find A.

10(−1+j1.7321)=A((−1+j1.7321)−(−1−j1.7321))+B((−1+j1.7321)−(−1+j1.7321))−10+j17.321=A(j3.4642)+B(0)A(j3.4642)=−10+j17.321

Simplify the above equation to find A.

A=−10+j17.321j3.4642=5+j2.886=5.773∠30°

Substitute −1−j1.7321 for s in equation (9) to find B.

10(−1−j1.7321)=A((−1−j1.7321)−(−1−j1.7321))+B((−1−j1.7321)−(−1+j1.7321))−10−j17.321=A(0)+B(−j3.4642)B(−j3.4642)=−10−j17.321

Simplify the above equation to find B.

B=−10−j17.321−j3.4642=5−j2.886=5.773∠−30°

Substitute 5.773∠30° for A, and 5.773∠−30° for B in equation (8) to find V(s).

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]=[5.773(cos(30°)+jsin(30°))(s−(−1+j1.7321))+20.587(cos(−30°)+jsin(−30°))(s−(−1−j1.7321))]=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))] {∵eiθ=cosθ+isinθ}

Take inverse Laplace transform for above equation to find v(t).

v(t)=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))]=[5.773ej30°e(−1+j1.7321)tu(t)+5.773e−j30°e(−1−j1.7321)tu(t)] {∵L−1(1s−a)=eatu(t)}=[5.773e(−1t+j41.7321t+j30°)+5.773e(−1t−j1.7321t−j30°)]u(t) {∵ea⋅eb=ea+b}=[5.773e−tej1.7321t+j30°+5.773e−te−j1.7321t−j30°]u(t)=[5.773e−tej(1.7321t+30°)+5.773e−te−j(1.7321t+30°)]u(t)

Simplify the above equation to find v(t).

v(t)==[5.773e−t[ej(1.7321t+30°)+e−j(1.7321t+30°)]]u(t)=[5.773e−t[2cos(1.7321t+30°)]]u(t) {∵cosθ=eiθ+e−iθ2}=[11.547e−tcos(1.7321t+30°)]u(t) V

Conclusion:

Thus, the expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

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Answer 4

Textbook Question

Chapter 16, Problem 53P

In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to position 2 at t = 0. Find:

(a) v(0⁺), dv(0⁺)/dt
(b) v(t) for t ≥ 0.

Chapter 16, Problem 53P, In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to

a.

Expert Solution

To determine

Find the value of v(0+) and dv(0+)dt.

Answer to Problem 53P

The value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

Explanation of Solution

Given data:

Refer to Figure 16.76 in the textbook.

The switch is in position 1 for a long time and moved to position 2 at t=0.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 1

For a DC circuit, at steady state condition when the switch is in position ‘1’at time t=0− the capacitor acts like open circuit and the inductor acts like short circuit.

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

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 2

Refer to Figure 2, the voltage across the resistor is same as the voltage across the capacitor which is the source voltage.

vC(0−)=10 V.

The current through inductor and voltage across capacitor is always continuous so that,

i(0)=iL(0−)=iL(0+)=0 A

v(0)=vC(0−)=vC(0+)=10 V

When the switch is in position ‘2’, the Figure 1 is reduced as shown in Figure 3.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 3

Refer to Figure 3, the capacitor, resistor and inductor are connected in parallel. For the parallel connection the voltage is same. In Figure 3, the magnitude of voltage is in opposite direction.

vR(0+)=vL(0+)=vC(0+)

Apply Kirchhoff’s current law for Figure 3.

iL(0+)=iC(0+)+iR(0+)

Substitute 0 for iL(0+) in above equation to find iC(0+).

0=iC(0+)+iR(0+)

iC(0+)=−iR(0+) (1)

Write an expression to calculate the current through resistor.

iR(0+)=vR(0+)R

Substitute vC(0+) for vR(0+) in above equation to find iR(0+).

iR(0+)=vC(0+)R

Substitute 10 V for vC(0+) and 0.5 Ω for R in above equation to find iR(0+).

iR(0+)=10 V0.5 Ω=20 A

Substitute 20 A for iR(0+) in equation (1) to find iC(0+).

iC(0+)=−20 A

At time t=0+, the capacitor current is calculated as follows:

iC(0+)=Cdv(0+)dt

Rearrange the above equation to find dv(0+)dt.

dv(0+)dt=iC(0+)C

Substitute −20 A for iC(0+), and 1 F for C in above equation to find dv(0+)dt.

dv(0+)dt=−20 A1 F=−20 A1 (A⋅sV) {∵1 F=1 A⋅1 s1 V}=−20 Vs

Conclusion:

Thus, the value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

b.

Expert Solution

To determine

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

Answer to Problem 53P

The expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

Explanation of Solution

Formula used:

Write a general expression to calculate the impedance of a resistor in s-domain.

ZR=R (2)

Here,

R is the value of resistance.

Write a general expression to calculate the impedance of an inductor in s-domain.

ZL=sL (3)

Here,

L is the value of inductance.

Write a general expression to calculate the impedance of a capacitor in s-domain.

ZC=1sC (4)

Here,

C is the value of capacitance.

Calculation:

Substitute 0.5 Ω for R1 in equation (2) to find ZR1.

ZR1=0.5 Ω

Substitute 0.25 H for L in equation (3) to find ZL.

ZL=s(0.25 H)=0.25s

Substitute 1 F for C in equation (4) to find ZC.

ZC=1s(1 F)=1s

Using element transformation methods with initial conditions convert the Figure 3 into s-domain.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 4

Apply nodal analysis at node V(s) for the circuit shown in Figure 4.

V(s)0.25s+V(s)0.5+V(s)−v(0)s(1s)=0V(s)0.25s+V(s)0.5+sV(s)−(s)v(0)s=0V(s)0.25s+V(s)0.5+sV(s)−v(0)=0V(s)[10.25s+10.5+s]=v(0)

Substitute 10 for v(0) in above equation.

V(s)[10.25s+10.5+s]=10V(s)[2+s+0.5s20.5s]=10V(s)[0.5s2+s+20.5s]=10V(s)[0.5(s2+2s+4)0.5s]=10

Simplify the above equation to find V(s).

V(s)=10ss2+2s+4 (5)

From the above equation , the characteristic equation is

s2+2s+4=0 (6)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (7)

Comparing the equation (6) with the equation (as2+bs+c=0).

a=1b=2c=4

Substitute 1 for a, 2 for b, and 4 for c in equation (7) to find s1,2.

s1,2=−2±(2)2−4(1)(4)2(1)=−2±4−162(1)=−2±−122=−2±j3.46422

Simplify the above equation to find s1,2.

s1,2=−2+j3.46422,−2−j3.46422=−1+j1.7321,−1−j1.7321

Substitute the roots of characteristic equation in equation (5) to find V(s).

V(s)=10ss(s−(−1+j1.7321))((s−(−1−j1.7321)))=10ss(s−(−1+j1.7321))(s−(−1−j1.7321))

Take partial fraction for above equation.

V(s)=10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1+j1.7321))+B(s−(−1−j1.7321))] (8)

The equation (8) can also be written as follows:

10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1−j1.7321))+B(s−(−1+j1.7321))[(s−(−1+j1.7321))(s−(−1−j1.7321))]]

Simplify the above equation as follows:

10s=A(s−(−1−j1.7321))+B(s−(−1+j1.7321)) . (9)

Substitute −1+j1.732 for s in equation (9) to find A.

10(−1+j1.7321)=A((−1+j1.7321)−(−1−j1.7321))+B((−1+j1.7321)−(−1+j1.7321))−10+j17.321=A(j3.4642)+B(0)A(j3.4642)=−10+j17.321

Simplify the above equation to find A.

A=−10+j17.321j3.4642=5+j2.886=5.773∠30°

Substitute −1−j1.7321 for s in equation (9) to find B.

10(−1−j1.7321)=A((−1−j1.7321)−(−1−j1.7321))+B((−1−j1.7321)−(−1+j1.7321))−10−j17.321=A(0)+B(−j3.4642)B(−j3.4642)=−10−j17.321

Simplify the above equation to find B.

B=−10−j17.321−j3.4642=5−j2.886=5.773∠−30°

Substitute 5.773∠30° for A, and 5.773∠−30° for B in equation (8) to find V(s).

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]=[5.773(cos(30°)+jsin(30°))(s−(−1+j1.7321))+20.587(cos(−30°)+jsin(−30°))(s−(−1−j1.7321))]=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))] {∵eiθ=cosθ+isinθ}

Take inverse Laplace transform for above equation to find v(t).

v(t)=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))]=[5.773ej30°e(−1+j1.7321)tu(t)+5.773e−j30°e(−1−j1.7321)tu(t)] {∵L−1(1s−a)=eatu(t)}=[5.773e(−1t+j41.7321t+j30°)+5.773e(−1t−j1.7321t−j30°)]u(t) {∵ea⋅eb=ea+b}=[5.773e−tej1.7321t+j30°+5.773e−te−j1.7321t−j30°]u(t)=[5.773e−tej(1.7321t+30°)+5.773e−te−j(1.7321t+30°)]u(t)

Simplify the above equation to find v(t).

v(t)==[5.773e−t[ej(1.7321t+30°)+e−j(1.7321t+30°)]]u(t)=[5.773e−t[2cos(1.7321t+30°)]]u(t) {∵cosθ=eiθ+e−iθ2}=[11.547e−tcos(1.7321t+30°)]u(t) V

Conclusion:

Thus, the expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Find the value of v(0+) and dv(0+)dt.

Answer to Problem 53P

The value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

Explanation of Solution

Given data:

Refer to Figure 16.76 in the textbook.

The switch is in position 1 for a long time and moved to position 2 at t=0.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 1

For a DC circuit, at steady state condition when the switch is in position ‘1’at time t=0− the capacitor acts like open circuit and the inductor acts like short circuit.

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

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 2

Refer to Figure 2, the voltage across the resistor is same as the voltage across the capacitor which is the source voltage.

vC(0−)=10 V.

The current through inductor and voltage across capacitor is always continuous so that,

i(0)=iL(0−)=iL(0+)=0 A

v(0)=vC(0−)=vC(0+)=10 V

When the switch is in position ‘2’, the Figure 1 is reduced as shown in Figure 3.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 3

Refer to Figure 3, the capacitor, resistor and inductor are connected in parallel. For the parallel connection the voltage is same. In Figure 3, the magnitude of voltage is in opposite direction.

vR(0+)=vL(0+)=vC(0+)

Apply Kirchhoff’s current law for Figure 3.

iL(0+)=iC(0+)+iR(0+)

Substitute 0 for iL(0+) in above equation to find iC(0+).

0=iC(0+)+iR(0+)

iC(0+)=−iR(0+) (1)

Write an expression to calculate the current through resistor.

iR(0+)=vR(0+)R

Substitute vC(0+) for vR(0+) in above equation to find iR(0+).

iR(0+)=vC(0+)R

Substitute 10 V for vC(0+) and 0.5 Ω for R in above equation to find iR(0+).

iR(0+)=10 V0.5 Ω=20 A

Substitute 20 A for iR(0+) in equation (1) to find iC(0+).

iC(0+)=−20 A

At time t=0+, the capacitor current is calculated as follows:

iC(0+)=Cdv(0+)dt

Rearrange the above equation to find dv(0+)dt.

dv(0+)dt=iC(0+)C

Substitute −20 A for iC(0+), and 1 F for C in above equation to find dv(0+)dt.

dv(0+)dt=−20 A1 F=−20 A1 (A⋅sV) {∵1 F=1 A⋅1 s1 V}=−20 Vs

Conclusion:

Thus, the value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

b.

Expert Solution

To determine

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

Answer to Problem 53P

The expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

Explanation of Solution

Formula used:

Write a general expression to calculate the impedance of a resistor in s-domain.

ZR=R (2)

Here,

R is the value of resistance.

Write a general expression to calculate the impedance of an inductor in s-domain.

ZL=sL (3)

Here,

L is the value of inductance.

Write a general expression to calculate the impedance of a capacitor in s-domain.

ZC=1sC (4)

Here,

C is the value of capacitance.

Calculation:

Substitute 0.5 Ω for R1 in equation (2) to find ZR1.

ZR1=0.5 Ω

Substitute 0.25 H for L in equation (3) to find ZL.

ZL=s(0.25 H)=0.25s

Substitute 1 F for C in equation (4) to find ZC.

ZC=1s(1 F)=1s

Using element transformation methods with initial conditions convert the Figure 3 into s-domain.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 4

Apply nodal analysis at node V(s) for the circuit shown in Figure 4.

V(s)0.25s+V(s)0.5+V(s)−v(0)s(1s)=0V(s)0.25s+V(s)0.5+sV(s)−(s)v(0)s=0V(s)0.25s+V(s)0.5+sV(s)−v(0)=0V(s)[10.25s+10.5+s]=v(0)

Substitute 10 for v(0) in above equation.

V(s)[10.25s+10.5+s]=10V(s)[2+s+0.5s20.5s]=10V(s)[0.5s2+s+20.5s]=10V(s)[0.5(s2+2s+4)0.5s]=10

Simplify the above equation to find V(s).

V(s)=10ss2+2s+4 (5)

From the above equation , the characteristic equation is

s2+2s+4=0 (6)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (7)

Comparing the equation (6) with the equation (as2+bs+c=0).

a=1b=2c=4

Substitute 1 for a, 2 for b, and 4 for c in equation (7) to find s1,2.

s1,2=−2±(2)2−4(1)(4)2(1)=−2±4−162(1)=−2±−122=−2±j3.46422

Simplify the above equation to find s1,2.

s1,2=−2+j3.46422,−2−j3.46422=−1+j1.7321,−1−j1.7321

Substitute the roots of characteristic equation in equation (5) to find V(s).

V(s)=10ss(s−(−1+j1.7321))((s−(−1−j1.7321)))=10ss(s−(−1+j1.7321))(s−(−1−j1.7321))

Take partial fraction for above equation.

V(s)=10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1+j1.7321))+B(s−(−1−j1.7321))] (8)

The equation (8) can also be written as follows:

10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1−j1.7321))+B(s−(−1+j1.7321))[(s−(−1+j1.7321))(s−(−1−j1.7321))]]

Simplify the above equation as follows:

10s=A(s−(−1−j1.7321))+B(s−(−1+j1.7321)) . (9)

Substitute −1+j1.732 for s in equation (9) to find A.

10(−1+j1.7321)=A((−1+j1.7321)−(−1−j1.7321))+B((−1+j1.7321)−(−1+j1.7321))−10+j17.321=A(j3.4642)+B(0)A(j3.4642)=−10+j17.321

Simplify the above equation to find A.

A=−10+j17.321j3.4642=5+j2.886=5.773∠30°

Substitute −1−j1.7321 for s in equation (9) to find B.

10(−1−j1.7321)=A((−1−j1.7321)−(−1−j1.7321))+B((−1−j1.7321)−(−1+j1.7321))−10−j17.321=A(0)+B(−j3.4642)B(−j3.4642)=−10−j17.321

Simplify the above equation to find B.

B=−10−j17.321−j3.4642=5−j2.886=5.773∠−30°

Substitute 5.773∠30° for A, and 5.773∠−30° for B in equation (8) to find V(s).

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]=[5.773(cos(30°)+jsin(30°))(s−(−1+j1.7321))+20.587(cos(−30°)+jsin(−30°))(s−(−1−j1.7321))]=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))] {∵eiθ=cosθ+isinθ}

Take inverse Laplace transform for above equation to find v(t).

v(t)=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))]=[5.773ej30°e(−1+j1.7321)tu(t)+5.773e−j30°e(−1−j1.7321)tu(t)] {∵L−1(1s−a)=eatu(t)}=[5.773e(−1t+j41.7321t+j30°)+5.773e(−1t−j1.7321t−j30°)]u(t) {∵ea⋅eb=ea+b}=[5.773e−tej1.7321t+j30°+5.773e−te−j1.7321t−j30°]u(t)=[5.773e−tej(1.7321t+30°)+5.773e−te−j(1.7321t+30°)]u(t)

Simplify the above equation to find v(t).

v(t)==[5.773e−t[ej(1.7321t+30°)+e−j(1.7321t+30°)]]u(t)=[5.773e−t[2cos(1.7321t+30°)]]u(t) {∵cosθ=eiθ+e−iθ2}=[11.547e−tcos(1.7321t+30°)]u(t) V

Conclusion:

Thus, the expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

Answer 8

a.

Expert Solution

To determine

Find the value of v(0+) and dv(0+)dt.

Answer to Problem 53P

The value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

Explanation of Solution

Given data:

Refer to Figure 16.76 in the textbook.

The switch is in position 1 for a long time and moved to position 2 at t=0.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 1

For a DC circuit, at steady state condition when the switch is in position ‘1’at time t=0− the capacitor acts like open circuit and the inductor acts like short circuit.

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

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 2

Refer to Figure 2, the voltage across the resistor is same as the voltage across the capacitor which is the source voltage.

vC(0−)=10 V.

The current through inductor and voltage across capacitor is always continuous so that,

i(0)=iL(0−)=iL(0+)=0 A

v(0)=vC(0−)=vC(0+)=10 V

When the switch is in position ‘2’, the Figure 1 is reduced as shown in Figure 3.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 3

Refer to Figure 3, the capacitor, resistor and inductor are connected in parallel. For the parallel connection the voltage is same. In Figure 3, the magnitude of voltage is in opposite direction.

vR(0+)=vL(0+)=vC(0+)

Apply Kirchhoff’s current law for Figure 3.

iL(0+)=iC(0+)+iR(0+)

Substitute 0 for iL(0+) in above equation to find iC(0+).

0=iC(0+)+iR(0+)

iC(0+)=−iR(0+) (1)

Write an expression to calculate the current through resistor.

iR(0+)=vR(0+)R

Substitute vC(0+) for vR(0+) in above equation to find iR(0+).

iR(0+)=vC(0+)R

Substitute 10 V for vC(0+) and 0.5 Ω for R in above equation to find iR(0+).

iR(0+)=10 V0.5 Ω=20 A

Substitute 20 A for iR(0+) in equation (1) to find iC(0+).

iC(0+)=−20 A

At time t=0+, the capacitor current is calculated as follows:

iC(0+)=Cdv(0+)dt

Rearrange the above equation to find dv(0+)dt.

dv(0+)dt=iC(0+)C

Substitute −20 A for iC(0+), and 1 F for C in above equation to find dv(0+)dt.

dv(0+)dt=−20 A1 F=−20 A1 (A⋅sV) {∵1 F=1 A⋅1 s1 V}=−20 Vs

Conclusion:

Thus, the value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Find the value of v(0+) and dv(0+)dt.

Answer 13

Answer to Problem 53P

The value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

Answer 14

Explanation of Solution

Given data:

Refer to Figure 16.76 in the textbook.

The switch is in position 1 for a long time and moved to position 2 at t=0.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 1

For a DC circuit, at steady state condition when the switch is in position ‘1’at time t=0− the capacitor acts like open circuit and the inductor acts like short circuit.

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

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 2

Refer to Figure 2, the voltage across the resistor is same as the voltage across the capacitor which is the source voltage.

vC(0−)=10 V.

The current through inductor and voltage across capacitor is always continuous so that,

i(0)=iL(0−)=iL(0+)=0 A

v(0)=vC(0−)=vC(0+)=10 V

When the switch is in position ‘2’, the Figure 1 is reduced as shown in Figure 3.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 3

Refer to Figure 3, the capacitor, resistor and inductor are connected in parallel. For the parallel connection the voltage is same. In Figure 3, the magnitude of voltage is in opposite direction.

vR(0+)=vL(0+)=vC(0+)

Apply Kirchhoff’s current law for Figure 3.

iL(0+)=iC(0+)+iR(0+)

Substitute 0 for iL(0+) in above equation to find iC(0+).

0=iC(0+)+iR(0+)

iC(0+)=−iR(0+) (1)

Write an expression to calculate the current through resistor.

iR(0+)=vR(0+)R

Substitute vC(0+) for vR(0+) in above equation to find iR(0+).

iR(0+)=vC(0+)R

Substitute 10 V for vC(0+) and 0.5 Ω for R in above equation to find iR(0+).

iR(0+)=10 V0.5 Ω=20 A

Substitute 20 A for iR(0+) in equation (1) to find iC(0+).

iC(0+)=−20 A

At time t=0+, the capacitor current is calculated as follows:

iC(0+)=Cdv(0+)dt

Rearrange the above equation to find dv(0+)dt.

dv(0+)dt=iC(0+)C

Substitute −20 A for iC(0+), and 1 F for C in above equation to find dv(0+)dt.

dv(0+)dt=−20 A1 F=−20 A1 (A⋅sV) {∵1 F=1 A⋅1 s1 V}=−20 Vs

Conclusion:

Thus, the value of v(0+) is 10 V and dv(0+)dt is −20 Vs.

Answer 15

b.

Expert Solution

To determine

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

Answer to Problem 53P

The expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

Explanation of Solution

Formula used:

Write a general expression to calculate the impedance of a resistor in s-domain.

ZR=R (2)

Here,

R is the value of resistance.

Write a general expression to calculate the impedance of an inductor in s-domain.

ZL=sL (3)

Here,

L is the value of inductance.

Write a general expression to calculate the impedance of a capacitor in s-domain.

ZC=1sC (4)

Here,

C is the value of capacitance.

Calculation:

Substitute 0.5 Ω for R1 in equation (2) to find ZR1.

ZR1=0.5 Ω

Substitute 0.25 H for L in equation (3) to find ZL.

ZL=s(0.25 H)=0.25s

Substitute 1 F for C in equation (4) to find ZC.

ZC=1s(1 F)=1s

Using element transformation methods with initial conditions convert the Figure 3 into s-domain.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 4

Apply nodal analysis at node V(s) for the circuit shown in Figure 4.

V(s)0.25s+V(s)0.5+V(s)−v(0)s(1s)=0V(s)0.25s+V(s)0.5+sV(s)−(s)v(0)s=0V(s)0.25s+V(s)0.5+sV(s)−v(0)=0V(s)[10.25s+10.5+s]=v(0)

Substitute 10 for v(0) in above equation.

V(s)[10.25s+10.5+s]=10V(s)[2+s+0.5s20.5s]=10V(s)[0.5s2+s+20.5s]=10V(s)[0.5(s2+2s+4)0.5s]=10

Simplify the above equation to find V(s).

V(s)=10ss2+2s+4 (5)

From the above equation , the characteristic equation is

s2+2s+4=0 (6)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (7)

Comparing the equation (6) with the equation (as2+bs+c=0).

a=1b=2c=4

Substitute 1 for a, 2 for b, and 4 for c in equation (7) to find s1,2.

s1,2=−2±(2)2−4(1)(4)2(1)=−2±4−162(1)=−2±−122=−2±j3.46422

Simplify the above equation to find s1,2.

s1,2=−2+j3.46422,−2−j3.46422=−1+j1.7321,−1−j1.7321

Substitute the roots of characteristic equation in equation (5) to find V(s).

V(s)=10ss(s−(−1+j1.7321))((s−(−1−j1.7321)))=10ss(s−(−1+j1.7321))(s−(−1−j1.7321))

Take partial fraction for above equation.

V(s)=10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1+j1.7321))+B(s−(−1−j1.7321))] (8)

The equation (8) can also be written as follows:

10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1−j1.7321))+B(s−(−1+j1.7321))[(s−(−1+j1.7321))(s−(−1−j1.7321))]]

Simplify the above equation as follows:

10s=A(s−(−1−j1.7321))+B(s−(−1+j1.7321)) . (9)

Substitute −1+j1.732 for s in equation (9) to find A.

10(−1+j1.7321)=A((−1+j1.7321)−(−1−j1.7321))+B((−1+j1.7321)−(−1+j1.7321))−10+j17.321=A(j3.4642)+B(0)A(j3.4642)=−10+j17.321

Simplify the above equation to find A.

A=−10+j17.321j3.4642=5+j2.886=5.773∠30°

Substitute −1−j1.7321 for s in equation (9) to find B.

10(−1−j1.7321)=A((−1−j1.7321)−(−1−j1.7321))+B((−1−j1.7321)−(−1+j1.7321))−10−j17.321=A(0)+B(−j3.4642)B(−j3.4642)=−10−j17.321

Simplify the above equation to find B.

B=−10−j17.321−j3.4642=5−j2.886=5.773∠−30°

Substitute 5.773∠30° for A, and 5.773∠−30° for B in equation (8) to find V(s).

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]=[5.773(cos(30°)+jsin(30°))(s−(−1+j1.7321))+20.587(cos(−30°)+jsin(−30°))(s−(−1−j1.7321))]=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))] {∵eiθ=cosθ+isinθ}

Take inverse Laplace transform for above equation to find v(t).

v(t)=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))]=[5.773ej30°e(−1+j1.7321)tu(t)+5.773e−j30°e(−1−j1.7321)tu(t)] {∵L−1(1s−a)=eatu(t)}=[5.773e(−1t+j41.7321t+j30°)+5.773e(−1t−j1.7321t−j30°)]u(t) {∵ea⋅eb=ea+b}=[5.773e−tej1.7321t+j30°+5.773e−te−j1.7321t−j30°]u(t)=[5.773e−tej(1.7321t+30°)+5.773e−te−j(1.7321t+30°)]u(t)

Simplify the above equation to find v(t).

v(t)==[5.773e−t[ej(1.7321t+30°)+e−j(1.7321t+30°)]]u(t)=[5.773e−t[2cos(1.7321t+30°)]]u(t) {∵cosθ=eiθ+e−iθ2}=[11.547e−tcos(1.7321t+30°)]u(t) V

Conclusion:

Thus, the expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

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

Answer 20

Answer to Problem 53P

The expression of voltage v(t) for t>0 is [11.547e−tcos(1.7321t+30°)]u(t) V.

Answer 21

Explanation of Solution

Formula used:

Write a general expression to calculate the impedance of a resistor in s-domain.

ZR=R (2)

Here,

R is the value of resistance.

Write a general expression to calculate the impedance of an inductor in s-domain.

ZL=sL (3)

Here,

L is the value of inductance.

Write a general expression to calculate the impedance of a capacitor in s-domain.

ZC=1sC (4)

Here,

C is the value of capacitance.

Calculation:

Substitute 0.5 Ω for R1 in equation (2) to find ZR1.

ZR1=0.5 Ω

Substitute 0.25 H for L in equation (3) to find ZL.

ZL=s(0.25 H)=0.25s

Substitute 1 F for C in equation (4) to find ZC.

ZC=1s(1 F)=1s

Using element transformation methods with initial conditions convert the Figure 3 into s-domain.

Fundamentals of Electric Circuits, Chapter 16, Problem 53P , additional homework tip 4

Apply nodal analysis at node V(s) for the circuit shown in Figure 4.

V(s)0.25s+V(s)0.5+V(s)−v(0)s(1s)=0V(s)0.25s+V(s)0.5+sV(s)−(s)v(0)s=0V(s)0.25s+V(s)0.5+sV(s)−v(0)=0V(s)[10.25s+10.5+s]=v(0)

Substitute 10 for v(0) in above equation.

V(s)[10.25s+10.5+s]=10V(s)[2+s+0.5s20.5s]=10V(s)[0.5s2+s+20.5s]=10V(s)[0.5(s2+2s+4)0.5s]=10

Simplify the above equation to find V(s).

V(s)=10ss2+2s+4 (5)

From the above equation , the characteristic equation is

s2+2s+4=0 (6)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (7)

Comparing the equation (6) with the equation (as2+bs+c=0).

a=1b=2c=4

Substitute 1 for a, 2 for b, and 4 for c in equation (7) to find s1,2.

s1,2=−2±(2)2−4(1)(4)2(1)=−2±4−162(1)=−2±−122=−2±j3.46422

Simplify the above equation to find s1,2.

s1,2=−2+j3.46422,−2−j3.46422=−1+j1.7321,−1−j1.7321

Substitute the roots of characteristic equation in equation (5) to find V(s).

V(s)=10ss(s−(−1+j1.7321))((s−(−1−j1.7321)))=10ss(s−(−1+j1.7321))(s−(−1−j1.7321))

Take partial fraction for above equation.

V(s)=10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1+j1.7321))+B(s−(−1−j1.7321))] (8)

The equation (8) can also be written as follows:

10s(s−(−1+j1.7321))(s−(−1−j1.7321))=[A(s−(−1−j1.7321))+B(s−(−1+j1.7321))[(s−(−1+j1.7321))(s−(−1−j1.7321))]]

Simplify the above equation as follows:

10s=A(s−(−1−j1.7321))+B(s−(−1+j1.7321)) . (9)

Substitute −1+j1.732 for s in equation (9) to find A.

10(−1+j1.7321)=A((−1+j1.7321)−(−1−j1.7321))+B((−1+j1.7321)−(−1+j1.7321))−10+j17.321=A(j3.4642)+B(0)A(j3.4642)=−10+j17.321

Simplify the above equation to find A.

A=−10+j17.321j3.4642=5+j2.886=5.773∠30°

Substitute −1−j1.7321 for s in equation (9) to find B.

10(−1−j1.7321)=A((−1−j1.7321)−(−1−j1.7321))+B((−1−j1.7321)−(−1+j1.7321))−10−j17.321=A(0)+B(−j3.4642)B(−j3.4642)=−10−j17.321

Simplify the above equation to find B.

B=−10−j17.321−j3.4642=5−j2.886=5.773∠−30°

Substitute 5.773∠30° for A, and 5.773∠−30° for B in equation (8) to find V(s).

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]

V(s)=[5.773∠30°(s−(−1+j1.7321))+5.773∠−30°(s−(−1−j1.7321))]=[5.773(cos(30°)+jsin(30°))(s−(−1+j1.7321))+20.587(cos(−30°)+jsin(−30°))(s−(−1−j1.7321))]=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))] {∵eiθ=cosθ+isinθ}

Take inverse Laplace transform for above equation to find v(t).

v(t)=[5.773ej30°(s−(−1+j1.7321))+5.773e−j30°(s−(−1−j1.7321))]=[5.773ej30°e(−1+j1.7321)tu(t)+5.773e−j30°e(−1−j1.7321)tu(t)] {∵L−1(1s−a)=eatu(t)}=[5.773e(−1t+j41.7321t+j30°)+5.773e(−1t−j1.7321t−j30°)]u(t) {∵ea⋅eb=ea+b}=[5.773e−tej1.7321t+j30°+5.773e−te−j1.7321t−j30°]u(t)=[5.773e−tej(1.7321t+30°)+5.773e−te−j(1.7321t+30°)]u(t)