In the circuit of Fig. 7.140, find v o and i o , given that v s = 10[1 – e –t ] u ( t ) V.

Question

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Answer 1

Textbook Question

Chapter 7, Problem 75P

In the circuit of Fig. 7.140, find v_o and i_o, given that v_s = 10[1 – e^–t]u(t) V.

Chapter 7, Problem 75P, In the circuit of Fig. 7.140, find vo and io, given that vs = 10[1 et]u(t) V.

Expert Solution & Answer

To determine

Find the output voltage vo and the current io of the op amp circuit in the given Figure 7.140.

Answer to Problem 75P

The output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

Explanation of Solution

Given data:

Refer to Figure 7.140 in the textbook.

The value of capacitance (C) is 10 μF.

The source voltage vs is 10[1−e−t]u(t) V.

Formula used:

Write the general expression for the unit step function.

u(t)={0, t<01, t>0 (1)

Calculation:

The given Figure 7.140 is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 7, Problem 75P

The given source voltage is,

vs=10[1−e−t]u(t) V (2)

Apply the unit step function in equation (1) to equation (2).

vs={0 V, t<010[1−e−t] V, t>0

Consider the voltage at the non-inverting terminal as va and the voltage at the inverting terminal as vb in Figure 1.

For t<0:

Since the source voltage vs=0 V for all t<0, all the initial voltages are equal to zero.

For t>0:

The voltage,

va=vb=vs=10[1−e−t] V (3)

On differentiating the equation (3).

dvbdt=10[0−(−1)e−t] Vs=10e−t Vs

In Figure 1, apply Kirchhoff’s current law at node vb.

(10 μF)dvbdt+vb100 kΩ+vb−vo100 kΩ=0 (4)

Substitute 10e−t Vs for dvbdt and 10[1−e−t] V for vb in equation (4) to find the output voltage vo of the op amp in volts.

(10 μF)(10e−t Vs)+10[1−e−t] V100 kΩ+10[1−e−t] V−vo100 kΩ=0(10×10−6 F)(10e−t Vs)+10[1−e−t] V100×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 k=103, 1 μ=10−6}(10×10−6 A⋅sV)(10e−t Vs)+[1−e−t] V10×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 F=A⋅sV}(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] A−vo105 Ω=0 {∵1 A=1 V1 Ω}

Rearrange the equation as follows,

vo105 Ω=(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] Avo105 Ω=(10−4e−t A)+2(10−4[1−e−t] A)vo=(105 Ω)(10−4e−t A)+(105 Ω)2(10−4[1−e−t] A)vo=10e−t V+20[1−e−t] V

Simplify the equation as follows,

vo=10e−t V+20 V−20e−t V=20 V−10e−t V

vo=[20−10e−t] V (5)

Apply the unit step function in equation (1) to equation (5).

vo=([20−10e−t] V)u(t)=[20−10e−t]u(t) V

The output current io of the op amp circuit is calculated as follows,

io=vo−vs100 kΩ

Substitute [20−10e−t] V for vo and 10[1−e−t] V for vs to find the current io in amperes.

io=[20−10e−t] V−10[1−e−t] V100 kΩ=20−10e−t−10+10e−t100×103 VΩ {∵1 k=103}=10100×103 A {∵1 A=1 V1 Ω}=10−4 A

Convert the unit A to μA.

io=10−4 A×102×10−2 A=100×10−6 A=100 μA

Conclusion:

Thus, the output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

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Answer 2

Textbook Question

Chapter 7, Problem 75P

In the circuit of Fig. 7.140, find v_o and i_o, given that v_s = 10[1 – e^–t]u(t) V.

Chapter 7, Problem 75P, In the circuit of Fig. 7.140, find vo and io, given that vs = 10[1 et]u(t) V.

Expert Solution & Answer

To determine

Find the output voltage vo and the current io of the op amp circuit in the given Figure 7.140.

Answer to Problem 75P

The output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

Explanation of Solution

Given data:

Refer to Figure 7.140 in the textbook.

The value of capacitance (C) is 10 μF.

The source voltage vs is 10[1−e−t]u(t) V.

Formula used:

Write the general expression for the unit step function.

u(t)={0, t<01, t>0 (1)

Calculation:

The given Figure 7.140 is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 7, Problem 75P

The given source voltage is,

vs=10[1−e−t]u(t) V (2)

Apply the unit step function in equation (1) to equation (2).

vs={0 V, t<010[1−e−t] V, t>0

Consider the voltage at the non-inverting terminal as va and the voltage at the inverting terminal as vb in Figure 1.

For t<0:

Since the source voltage vs=0 V for all t<0, all the initial voltages are equal to zero.

For t>0:

The voltage,

va=vb=vs=10[1−e−t] V (3)

On differentiating the equation (3).

dvbdt=10[0−(−1)e−t] Vs=10e−t Vs

In Figure 1, apply Kirchhoff’s current law at node vb.

(10 μF)dvbdt+vb100 kΩ+vb−vo100 kΩ=0 (4)

Substitute 10e−t Vs for dvbdt and 10[1−e−t] V for vb in equation (4) to find the output voltage vo of the op amp in volts.

(10 μF)(10e−t Vs)+10[1−e−t] V100 kΩ+10[1−e−t] V−vo100 kΩ=0(10×10−6 F)(10e−t Vs)+10[1−e−t] V100×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 k=103, 1 μ=10−6}(10×10−6 A⋅sV)(10e−t Vs)+[1−e−t] V10×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 F=A⋅sV}(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] A−vo105 Ω=0 {∵1 A=1 V1 Ω}

Rearrange the equation as follows,

vo105 Ω=(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] Avo105 Ω=(10−4e−t A)+2(10−4[1−e−t] A)vo=(105 Ω)(10−4e−t A)+(105 Ω)2(10−4[1−e−t] A)vo=10e−t V+20[1−e−t] V

Simplify the equation as follows,

vo=10e−t V+20 V−20e−t V=20 V−10e−t V

vo=[20−10e−t] V (5)

Apply the unit step function in equation (1) to equation (5).

vo=([20−10e−t] V)u(t)=[20−10e−t]u(t) V

The output current io of the op amp circuit is calculated as follows,

io=vo−vs100 kΩ

Substitute [20−10e−t] V for vo and 10[1−e−t] V for vs to find the current io in amperes.

io=[20−10e−t] V−10[1−e−t] V100 kΩ=20−10e−t−10+10e−t100×103 VΩ {∵1 k=103}=10100×103 A {∵1 A=1 V1 Ω}=10−4 A

Convert the unit A to μA.

io=10−4 A×102×10−2 A=100×10−6 A=100 μA

Conclusion:

Thus, the output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

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Answer 3

Textbook Question

Chapter 7, Problem 75P

In the circuit of Fig. 7.140, find v_o and i_o, given that v_s = 10[1 – e^–t]u(t) V.

Chapter 7, Problem 75P, In the circuit of Fig. 7.140, find vo and io, given that vs = 10[1 et]u(t) V.

Expert Solution & Answer

To determine

Find the output voltage vo and the current io of the op amp circuit in the given Figure 7.140.

Answer to Problem 75P

The output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

Explanation of Solution

Given data:

Refer to Figure 7.140 in the textbook.

The value of capacitance (C) is 10 μF.

The source voltage vs is 10[1−e−t]u(t) V.

Formula used:

Write the general expression for the unit step function.

u(t)={0, t<01, t>0 (1)

Calculation:

The given Figure 7.140 is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 7, Problem 75P

The given source voltage is,

vs=10[1−e−t]u(t) V (2)

Apply the unit step function in equation (1) to equation (2).

vs={0 V, t<010[1−e−t] V, t>0

Consider the voltage at the non-inverting terminal as va and the voltage at the inverting terminal as vb in Figure 1.

For t<0:

Since the source voltage vs=0 V for all t<0, all the initial voltages are equal to zero.

For t>0:

The voltage,

va=vb=vs=10[1−e−t] V (3)

On differentiating the equation (3).

dvbdt=10[0−(−1)e−t] Vs=10e−t Vs

In Figure 1, apply Kirchhoff’s current law at node vb.

(10 μF)dvbdt+vb100 kΩ+vb−vo100 kΩ=0 (4)

Substitute 10e−t Vs for dvbdt and 10[1−e−t] V for vb in equation (4) to find the output voltage vo of the op amp in volts.

(10 μF)(10e−t Vs)+10[1−e−t] V100 kΩ+10[1−e−t] V−vo100 kΩ=0(10×10−6 F)(10e−t Vs)+10[1−e−t] V100×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 k=103, 1 μ=10−6}(10×10−6 A⋅sV)(10e−t Vs)+[1−e−t] V10×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 F=A⋅sV}(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] A−vo105 Ω=0 {∵1 A=1 V1 Ω}

Rearrange the equation as follows,

vo105 Ω=(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] Avo105 Ω=(10−4e−t A)+2(10−4[1−e−t] A)vo=(105 Ω)(10−4e−t A)+(105 Ω)2(10−4[1−e−t] A)vo=10e−t V+20[1−e−t] V

Simplify the equation as follows,

vo=10e−t V+20 V−20e−t V=20 V−10e−t V

vo=[20−10e−t] V (5)

Apply the unit step function in equation (1) to equation (5).

vo=([20−10e−t] V)u(t)=[20−10e−t]u(t) V

The output current io of the op amp circuit is calculated as follows,

io=vo−vs100 kΩ

Substitute [20−10e−t] V for vo and 10[1−e−t] V for vs to find the current io in amperes.

io=[20−10e−t] V−10[1−e−t] V100 kΩ=20−10e−t−10+10e−t100×103 VΩ {∵1 k=103}=10100×103 A {∵1 A=1 V1 Ω}=10−4 A

Convert the unit A to μA.

io=10−4 A×102×10−2 A=100×10−6 A=100 μA

Conclusion:

Thus, the output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

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Answer 4

Textbook Question

Chapter 7, Problem 75P

In the circuit of Fig. 7.140, find v_o and i_o, given that v_s = 10[1 – e^–t]u(t) V.

Chapter 7, Problem 75P, In the circuit of Fig. 7.140, find vo and io, given that vs = 10[1 et]u(t) V.

Expert Solution & Answer

To determine

Find the output voltage vo and the current io of the op amp circuit in the given Figure 7.140.

Answer to Problem 75P

The output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

Explanation of Solution

Given data:

Refer to Figure 7.140 in the textbook.

The value of capacitance (C) is 10 μF.

The source voltage vs is 10[1−e−t]u(t) V.

Formula used:

Write the general expression for the unit step function.

u(t)={0, t<01, t>0 (1)

Calculation:

The given Figure 7.140 is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 7, Problem 75P

The given source voltage is,

vs=10[1−e−t]u(t) V (2)

Apply the unit step function in equation (1) to equation (2).

vs={0 V, t<010[1−e−t] V, t>0

Consider the voltage at the non-inverting terminal as va and the voltage at the inverting terminal as vb in Figure 1.

For t<0:

Since the source voltage vs=0 V for all t<0, all the initial voltages are equal to zero.

For t>0:

The voltage,

va=vb=vs=10[1−e−t] V (3)

On differentiating the equation (3).

dvbdt=10[0−(−1)e−t] Vs=10e−t Vs

In Figure 1, apply Kirchhoff’s current law at node vb.

(10 μF)dvbdt+vb100 kΩ+vb−vo100 kΩ=0 (4)

Substitute 10e−t Vs for dvbdt and 10[1−e−t] V for vb in equation (4) to find the output voltage vo of the op amp in volts.

(10 μF)(10e−t Vs)+10[1−e−t] V100 kΩ+10[1−e−t] V−vo100 kΩ=0(10×10−6 F)(10e−t Vs)+10[1−e−t] V100×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 k=103, 1 μ=10−6}(10×10−6 A⋅sV)(10e−t Vs)+[1−e−t] V10×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 F=A⋅sV}(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] A−vo105 Ω=0 {∵1 A=1 V1 Ω}

Rearrange the equation as follows,

vo105 Ω=(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] Avo105 Ω=(10−4e−t A)+2(10−4[1−e−t] A)vo=(105 Ω)(10−4e−t A)+(105 Ω)2(10−4[1−e−t] A)vo=10e−t V+20[1−e−t] V

Simplify the equation as follows,

vo=10e−t V+20 V−20e−t V=20 V−10e−t V

vo=[20−10e−t] V (5)

Apply the unit step function in equation (1) to equation (5).

vo=([20−10e−t] V)u(t)=[20−10e−t]u(t) V

The output current io of the op amp circuit is calculated as follows,

io=vo−vs100 kΩ

Substitute [20−10e−t] V for vo and 10[1−e−t] V for vs to find the current io in amperes.

io=[20−10e−t] V−10[1−e−t] V100 kΩ=20−10e−t−10+10e−t100×103 VΩ {∵1 k=103}=10100×103 A {∵1 A=1 V1 Ω}=10−4 A

Convert the unit A to μA.

io=10−4 A×102×10−2 A=100×10−6 A=100 μA

Conclusion:

Thus, the output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

Find the output voltage vo and the current io of the op amp circuit in the given Figure 7.140.

Answer to Problem 75P

The output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

Explanation of Solution

Given data:

Refer to Figure 7.140 in the textbook.

The value of capacitance (C) is 10 μF.

The source voltage vs is 10[1−e−t]u(t) V.

Formula used:

Write the general expression for the unit step function.

u(t)={0, t<01, t>0 (1)

Calculation:

The given Figure 7.140 is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 7, Problem 75P

The given source voltage is,

vs=10[1−e−t]u(t) V (2)

Apply the unit step function in equation (1) to equation (2).

vs={0 V, t<010[1−e−t] V, t>0

Consider the voltage at the non-inverting terminal as va and the voltage at the inverting terminal as vb in Figure 1.

For t<0:

Since the source voltage vs=0 V for all t<0, all the initial voltages are equal to zero.

For t>0:

The voltage,

va=vb=vs=10[1−e−t] V (3)

On differentiating the equation (3).

dvbdt=10[0−(−1)e−t] Vs=10e−t Vs

In Figure 1, apply Kirchhoff’s current law at node vb.

(10 μF)dvbdt+vb100 kΩ+vb−vo100 kΩ=0 (4)

Substitute 10e−t Vs for dvbdt and 10[1−e−t] V for vb in equation (4) to find the output voltage vo of the op amp in volts.

(10 μF)(10e−t Vs)+10[1−e−t] V100 kΩ+10[1−e−t] V−vo100 kΩ=0(10×10−6 F)(10e−t Vs)+10[1−e−t] V100×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 k=103, 1 μ=10−6}(10×10−6 A⋅sV)(10e−t Vs)+[1−e−t] V10×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 F=A⋅sV}(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] A−vo105 Ω=0 {∵1 A=1 V1 Ω}

Rearrange the equation as follows,

vo105 Ω=(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] Avo105 Ω=(10−4e−t A)+2(10−4[1−e−t] A)vo=(105 Ω)(10−4e−t A)+(105 Ω)2(10−4[1−e−t] A)vo=10e−t V+20[1−e−t] V

Simplify the equation as follows,

vo=10e−t V+20 V−20e−t V=20 V−10e−t V

vo=[20−10e−t] V (5)

Apply the unit step function in equation (1) to equation (5).

vo=([20−10e−t] V)u(t)=[20−10e−t]u(t) V

The output current io of the op amp circuit is calculated as follows,

io=vo−vs100 kΩ

Substitute [20−10e−t] V for vo and 10[1−e−t] V for vs to find the current io in amperes.

io=[20−10e−t] V−10[1−e−t] V100 kΩ=20−10e−t−10+10e−t100×103 VΩ {∵1 k=103}=10100×103 A {∵1 A=1 V1 Ω}=10−4 A

Convert the unit A to μA.

io=10−4 A×102×10−2 A=100×10−6 A=100 μA

Conclusion:

Thus, the output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

Answer 8

Expert Solution & Answer

To determine

Find the output voltage vo and the current io of the op amp circuit in the given Figure 7.140.

Answer to Problem 75P

The output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

Explanation of Solution

Given data:

Refer to Figure 7.140 in the textbook.

The value of capacitance (C) is 10 μF.

The source voltage vs is 10[1−e−t]u(t) V.

Formula used:

Write the general expression for the unit step function.

u(t)={0, t<01, t>0 (1)

Calculation:

The given Figure 7.140 is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 7, Problem 75P

The given source voltage is,

vs=10[1−e−t]u(t) V (2)

Apply the unit step function in equation (1) to equation (2).

vs={0 V, t<010[1−e−t] V, t>0

Consider the voltage at the non-inverting terminal as va and the voltage at the inverting terminal as vb in Figure 1.

For t<0:

Since the source voltage vs=0 V for all t<0, all the initial voltages are equal to zero.

For t>0:

The voltage,

va=vb=vs=10[1−e−t] V (3)

On differentiating the equation (3).

dvbdt=10[0−(−1)e−t] Vs=10e−t Vs

In Figure 1, apply Kirchhoff’s current law at node vb.

(10 μF)dvbdt+vb100 kΩ+vb−vo100 kΩ=0 (4)

Substitute 10e−t Vs for dvbdt and 10[1−e−t] V for vb in equation (4) to find the output voltage vo of the op amp in volts.

(10 μF)(10e−t Vs)+10[1−e−t] V100 kΩ+10[1−e−t] V−vo100 kΩ=0(10×10−6 F)(10e−t Vs)+10[1−e−t] V100×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 k=103, 1 μ=10−6}(10×10−6 A⋅sV)(10e−t Vs)+[1−e−t] V10×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 F=A⋅sV}(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] A−vo105 Ω=0 {∵1 A=1 V1 Ω}

Rearrange the equation as follows,

vo105 Ω=(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] Avo105 Ω=(10−4e−t A)+2(10−4[1−e−t] A)vo=(105 Ω)(10−4e−t A)+(105 Ω)2(10−4[1−e−t] A)vo=10e−t V+20[1−e−t] V

Simplify the equation as follows,

vo=10e−t V+20 V−20e−t V=20 V−10e−t V

vo=[20−10e−t] V (5)

Apply the unit step function in equation (1) to equation (5).

vo=([20−10e−t] V)u(t)=[20−10e−t]u(t) V

The output current io of the op amp circuit is calculated as follows,

io=vo−vs100 kΩ

Substitute [20−10e−t] V for vo and 10[1−e−t] V for vs to find the current io in amperes.

io=[20−10e−t] V−10[1−e−t] V100 kΩ=20−10e−t−10+10e−t100×103 VΩ {∵1 k=103}=10100×103 A {∵1 A=1 V1 Ω}=10−4 A

Convert the unit A to μA.

io=10−4 A×102×10−2 A=100×10−6 A=100 μA

Conclusion:

Thus, the output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Find the output voltage vo and the current io of the op amp circuit in the given Figure 7.140.

Answer 12

Answer to Problem 75P

The output voltage vo is [20−10e−t]u(t) V and the current io is 100 μA of the op amp circuit.

Answer 13

Explanation of Solution

Given data:

Refer to Figure 7.140 in the textbook.

The value of capacitance (C) is 10 μF.

The source voltage vs is 10[1−e−t]u(t) V.

Formula used:

Write the general expression for the unit step function.

u(t)={0, t<01, t>0 (1)

Calculation:

The given Figure 7.140 is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 7, Problem 75P

The given source voltage is,

vs=10[1−e−t]u(t) V (2)

Apply the unit step function in equation (1) to equation (2).

vs={0 V, t<010[1−e−t] V, t>0

Consider the voltage at the non-inverting terminal as va and the voltage at the inverting terminal as vb in Figure 1.

For t<0:

Since the source voltage vs=0 V for all t<0, all the initial voltages are equal to zero.

For t>0:

The voltage,

va=vb=vs=10[1−e−t] V (3)

On differentiating the equation (3).

dvbdt=10[0−(−1)e−t] Vs=10e−t Vs

In Figure 1, apply Kirchhoff’s current law at node vb.

(10 μF)dvbdt+vb100 kΩ+vb−vo100 kΩ=0 (4)

Substitute 10e−t Vs for dvbdt and 10[1−e−t] V for vb in equation (4) to find the output voltage vo of the op amp in volts.

(10 μF)(10e−t Vs)+10[1−e−t] V100 kΩ+10[1−e−t] V−vo100 kΩ=0(10×10−6 F)(10e−t Vs)+10[1−e−t] V100×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 k=103, 1 μ=10−6}(10×10−6 A⋅sV)(10e−t Vs)+[1−e−t] V10×103 Ω+10[1−e−t] V−vo100×103 Ω=0 {∵1 F=A⋅sV}(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] A−vo105 Ω=0 {∵1 A=1 V1 Ω}

Rearrange the equation as follows,

vo105 Ω=(10−4e−t A)+10−4[1−e−t] A+10−4[1−e−t] Avo105 Ω=(10−4e−t A)+2(10−4[1−e−t] A)vo=(105 Ω)(10−4e−t A)+(105 Ω)2(10−4[1−e−t] A)vo=10e−t V+20[1−e−t] V