The Analysis and Design of Linear Circuits Eighth Edition

8 Sinusoidal Steady-State Response

8.1 Exercise Solutions
Exercise 8–1. Convert the following sinusoids to phasors in polar and rectangular form:
(a). 𝑣(𝑡) = 20 cos(150𝑡 − 60◦ ) V
By inspection, in polar form we have 𝐕 = 20∠−60◦ V. Compute the rectangular form.

𝐕 = 20 cos(−60◦ ) + 𝑗20 sin(−60◦ ) = 10 − 𝑗17.32 V

(b). 𝑣(𝑡) = 10 cos(1000𝑡 + 180◦ ) V

By inspection, in polar form we have 𝐕 = 10∠180◦ V. Compute the rectangular form.

𝐕 = 10 cos(180◦ ) + 𝑗10 sin(180◦ ) = −10 + 𝑗0 V

(c). 𝑖(𝑡) = −4 cos(3𝑡) + 3 cos(3𝑡 − 90◦ ) A

In polar form we have 𝐈 = −4∠0◦ + 3∠−90◦ , which we will simplify after we find the rectangular form.

𝐈 = −4 cos(0◦ ) − 𝑗4 sin(0◦ ) + 3 cos(−90◦ ) + 𝑗3 sin(−90◦ ) A

𝐈 = −4 − 𝑗0 + 0 − 𝑗3 = −4 − 𝑗3 A
√
𝐈= (−4)2 + (−3)2 ∠ tan−1 (−3∕ − 4) = 5∠−143◦ A

Exercise 8–2. Convert the following phasors to sinusoids:

(a). 𝐕 = 169∠−45◦ V at 𝑓 = 60 Hz
By inspection, we have 𝑣(𝑡) = 169 cos(120𝜋𝑡 − 45◦ ) V.
(b). 𝐕 = 10∠90◦ + 66 − 𝑗10 V at 𝜔 = 10 krad/s
In rectangular form, we have 𝐕 = 𝑗10+66−𝑗10 = 66 V, so the corresponding sinusoid is 𝑣(𝑡) = 66 cos(10000𝑡) V.

(c). 𝐈 = 15 + 𝑗5 + 10∠180◦ mA at 𝜔 = 1000 rad/s

√
In rectangular form,√we have 𝐈 = 15 + 𝑗5 − 10 = 5 + 𝑗5. In polar form, we have 5 2∠45◦ , so the corresponding
sinusoid is 𝑖(𝑡) = 5 2 cos(1000𝑡 + 45◦ ) mA.
Exercise 8–3. (a). Construct the phasors for the following signals:

𝑖1 (𝑡) = 100 cos(2000𝑡) mA

𝑖2 (𝑡) = 50 cos(2000𝑡 − 60◦ ) mA

By inspection, the phasors are

𝐈1 = 100∠0◦ = 100 + 𝑗0 mA

𝐈2 = 50∠−60◦ = 25 − 𝑗43.3 mA

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The Analysis and Design of Linear Circuits Eighth Edition

(b). Use the additive property to find 𝑖(𝑡) = 𝑖1 (𝑡) + 𝑖2 (𝑡) and check the results using MATLAB.
Compute 𝑖(𝑡).

𝐈 = 𝐈1 + 𝐈2 = 100 + 𝑗0 + 25 − 𝑗43.3 = 125 − 𝑗43.3 = 132.3∠−19.1◦ mA

𝑖(𝑡) = 132.3 cos(2000𝑡 − 19.1◦ ) mA

The following MATLAB code verifies the solution:

I1 = 100* exp ( j *0)

I2 = 50* exp ( - j * pi /3)
I = I1 + I2
IMag = abs ( I )
I P h a s e = 180* a n g l e ( I )/ pi

The corresponding MATLAB output is shown below.

I1 = 100 . 0 0 0 0 e +000
I2 = 25 . 0 0 0 0 e +000 - 43 . 3 0 1 3 e +000 i
I = 125 . 0 0 0 0 e +000 - 43 . 3 0 1 3 e +000 i
IMag = 132 . 2 8 7 6 e +000
IPhase = - 19 . 1 0 6 6 e +000

Exercise 8–4. Show that the phasors 𝐈A , 𝐈B , and 𝐈C would still sum to zero if they were all rotated 90◦ counterclockwise.
The new phasors and their sum are shown below

𝐈A = 5∠140◦ = −3.83 + 𝑗3.21 A

𝐈B = 5∠260◦ = −0.87 − 𝑗4.92 A

𝐈C = 5∠20◦ = 4.70 + 𝑗1.71 A

𝐈A + 𝐈B + 𝐈C = (−3.83 − 0.87 + 4.70) + 𝑗(3.21 − 4.92 + 1.71) = 0 + 𝑗0 A

Exercise 8–5. Find the phasor corresponding to the time derivative of the waveform:

𝑣(𝑡) = 100 cos(1000𝑡) V

The phasor and its derivative are shown below.

𝐕 = 100∠0◦ V

𝑗𝜔𝐕 = (𝑗1000)100∠0◦ = (1000∠90◦ )(100∠0◦ ) = 105 ∠90◦ V/s

Exercise 8–6. Find the phasor corresponding to the waveform 𝑣(𝑡) = 𝑉A cos(𝜔𝑡) + 2𝑉A sin(𝜔𝑡).
The solution is presented below.

𝑣(𝑡) = 𝑉A cos(𝜔𝑡) + 2𝑉A sin(𝜔𝑡)

𝑣(𝑡) = 𝑉A cos(𝜔𝑡) + 2𝑉A cos(𝜔𝑡 − 90◦ )

𝐕 = 𝐕A ∠0◦ + 2𝐕A ∠−90◦ = 𝐕A − 𝑗2𝐕A

√
𝐕= 5𝐕A ∠−63.4◦

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Exercise 8–7. A series circuit is composed of a 1-kΩ resistor, a 1-𝜇F capacitor, and a 100-mH inductor.

(a). At what frequency will the magnitude of the impedance of the inductor equal that of the resistor?
We have the following relationships:

𝑍R = 𝑅

𝑍L = 𝑗𝜔𝐿

|𝑍 | = 𝜔𝐿
| L|
1
𝑍C =
𝑗𝜔𝐶

|𝑍 | = 1
| C | 𝜔𝐶

𝜔𝐿 = 𝑅

𝑅 1 kΩ
𝜔= = = 10 krad/s
𝐿 100 mH

(b). At what frequency will the magnitude of the impedance of the capacitor equal that of the resistor?
We have the following relationship:

1
=𝑅
𝜔𝐶
1 1
𝜔= = = 1 krad/s
𝑅𝐶 (1 kΩ)(1 𝜇F)

(c). At what frequency will the magnitude of the impedance of the inductor equal the magnitude of the impedance
of the capacitor? What is this frequency called?
We have the following relationship:

1
𝜔𝐿 =
𝜔𝐶
√ √
1 1
𝜔= = = 3.162 krad/s
𝐿𝐶 (100 mH)(1 𝜇F)

This frequency is known as the resonant frequency.

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Exercise 8–8. An element in a circuit operating in the sinusoidal steady state has a voltage across it and a current
through it as follows:
𝑣(𝑡) = 50 cos(500𝑡) V and 𝑖(𝑡) = 4 cos(500𝑡 − 60◦ ) A
Find the impedance of the element.
The impedance of the element is the ratio of the phasor voltage to the phasor current and we can compute it as
follows:

𝐕 = 50∠0◦ = 50 + 𝑗0 V

𝐈 = 4∠−60◦ = 2 − 𝑗3.4641 A

𝐕 = 𝑍𝐈

𝐕
𝑍=
𝐈
50∠0◦ V
𝑍= = 12.5∠60◦ Ω = 6.25 + 𝑗10.83 Ω
4∠−60◦ A

Exercise 8–9. A series connection consists of a 12-mH inductor and a 20-pF capacitor. The current flowing through
the circuit is 𝑖L (𝑡) = 20 cos(106 𝑡) mA.

(a). Find the impedance of each element.

We have the following impedances:

𝑍L = 𝑗𝜔𝐿 = 𝑗(106 rad/s)(12 mH) = 𝑗12 kΩ

1 1
𝑍C = = = −𝑗50 kΩ
𝑗𝜔𝐶 6
𝑗(10 rad/s)(20 pF)

(b). Find the phasor voltage across each element.

We have the following voltages:

𝐕 = 𝑍𝐈

𝐈 = 20∠0◦ mA = 20 + 𝑗0 mA

𝐕L = 𝑍L 𝐈 = (𝑗12 kΩ)(20 mA) = 𝑗240 V = 240∠90◦ V

𝐕C = 𝑍C 𝐈 = (−𝑗50 kΩ)(20 mA) = −𝑗1000 V = 1000∠−90◦ V

(c). Using both hand calculations and a Multisim simulation, find the voltages across each element.
Convert each phasor to a waveform.

𝑣L (𝑡) = 240 cos(106 𝑡 + 90◦ ) V

𝑣C (𝑡) = 1 cos(106 𝑡 − 90◦ ) kV

The Multisim simulation and results in the figure below verify the solution.

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The Analysis and Design of Linear Circuits Eighth Edition

(d). Does the current in the inductor lead or lag the voltage across it?
In the inductor, the current’s angle is 90◦ less than that of the voltage, so the current lags the voltage in the
inductor by 90◦ .
(e). Does the current in the capacitor lead or lag the voltage across it?
In the capacitor, the current’s angle is 90◦ greater than that of the voltage, so the current leads the voltage in the
capacitor by 90◦ .
Exercise 8–10. The circuit in Figure 8–15(a) is operating in the sinusoidal steady state with 𝑣S (𝑡) = 100 cos(2000𝑡 −
45◦ ) V.
(a). Transform the circuit into the phasor domain.
In the phasor domain, we have the following values:
𝐕S = 100∠−45◦ V

𝑍R = 50 Ω

𝑍L = 𝑗𝜔𝐿 = 𝑗(2000 rad/s)(25 mH) = 𝑗50 Ω

1 1
𝑍C = = = −𝑗50 Ω
𝑗𝜔𝐶 𝑗(2000 rad/s)(10 𝜇F)
The phasor-domain circuit is shown below.

(b). Solve for the phasor current 𝐈.

Sum the impedances in series and solve for the current.
𝐕S 𝐕S
𝐈= =
𝑍EQ 𝑍R + 𝑍C + 𝑍L

100∠−45◦ V 100∠−45◦ V
𝐈= = = 2∠−45◦ A
50 − 𝑗50 + 𝑗50 Ω 50 Ω

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The Analysis and Design of Linear Circuits Eighth Edition

(c). Solve for the phasor voltage across each element.

We have the following voltages:

𝐕 = 𝑍𝐈

𝐕R = 𝑍R 𝐈 = (50 Ω)(2∠−45◦ A) = 100∠−45◦ V

𝐕L = 𝑍L 𝐈 = (𝑗50 Ω)(2∠−45◦ A) = 100∠45◦ V

𝐕C = 𝑍C 𝐈 = (−𝑗50 Ω)(2∠−45◦ A) = 100∠−135◦ V

(d). Find the waveforms corresponding to the phasors found in (b) and (c).
By inspection, we have the following waveforms:

𝑖(𝑡) = 2 cos(2000𝑡 − 45◦ ) A

𝑣R (𝑡) = 100 cos(2000𝑡 − 45◦ ) V

𝑣L (𝑡) = 100 cos(2000𝑡 + 45◦ ) V

𝑣C (𝑡) = 100 cos(2000𝑡 − 135◦ ) V

(e). Draw a phasor diagram of all three voltages and the current.
The phasor diagram is shown below.

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The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–11. Design the voltage divider in Figure 8–16(a) so that an input 𝑣S (𝑡) = 50 cos(2000𝑡) V produces an
output 𝑣O (𝑡) = 25 cos(2000𝑡 − 30◦ ) V.
There are many correct solutions. First, find the relationship between 𝑍1 and 𝑍2 as follows:

𝑍2
𝐕O = 𝐕
𝑍1 + 𝑍2 S

𝑍2
25∠−30◦ = 50∠0◦
𝑍1 + 𝑍2

(21.65 − 𝑗12.5)(𝑍1 + 𝑍2 ) = 50𝑍2

50 − (21.65 − 𝑗12.5)
𝑍1 = 𝑍2
21.65 − 𝑗12.5

28.35 + 𝑗12.5
𝑍1 = 𝑍
21.65 − 𝑗12.5 2

𝑍1 = (0.7321 + 𝑗)𝑍2 = (1.239∠53.79◦ )𝑍2

Select a reasonable value for 𝑍2 , such as 1 kΩ. We then have:

𝑍1 = (0.7321 + 𝑗)(1 kΩ) = 732 + 𝑗1000 Ω

𝑅1 = 732 Ω

1000 Ω
𝐿1 = = 500 mH
2000 rad/s

𝑅2 = 1 kΩ

As an alternative, if we select 𝑍2 = 1000 − 𝑗1000 Ω, then we have the following relationships:

𝑍1 = (0.7321 + 𝑗)(1000 − 𝑗1000 Ω)

𝑍1 = 1732 + 𝑗268 Ω = 1753∠8.8◦ Ω

𝑅1 = 1.732 kΩ

268 Ω
𝐿1 = = 134 mH
2000 rad/s

𝑅2 = 1 kΩ

1
𝐶2 = = 0.5 𝜇F
(1000 Ω)(2000 rad/s)

The two designs are shown in the figure below.

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The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–12. The circuit shown in Figure 8–17(a) is operating in the sinusoidal steady state at a frequency of
100 krad/s. It requires a load of 𝑍L = 1500∠−57.5◦ Ω to operate properly. Design the load using standard parts
to within ±5% of the desired values.
We have the following relationships:

𝜔 = 100000 rad/s

𝑍L = 1500∠−57.5◦ Ω = 805.9 − 𝑗1265 Ω

One way to design the load is with a resistor in series with a capacitor. The resistance should be 805.9 Ω, and the
closest standard value is 820 Ω. The capacitor should be 𝐶 = 1∕(1265𝜔) = 7905 pF, and the closest standard value is
8200 pF. The resulting impedance is 𝑍L = 820 − 𝑗1220 Ω, which is within ±5% of the desired value.

Exercise 8–13. Consider the Maxwell bridge shown in Figure 8–18. Suppose we know that the unknown impedance
is an unknown capacitor 𝐶X in parallel with an unknown resistance 𝑅X . Let 𝑍1 be a resistance 𝑅1 in series with an
inductance 𝐿1 . Let 𝑍2 be a resistance 𝑅2 and 𝑍3 be a resistance 𝑅3 . Find the relationships that will allow the bridge
to be balanced.
The desired Maxwell bridge is shown in the figure below.

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The Analysis and Design of Linear Circuits Eighth Edition

Using the results from Example 8–8, we have the following relationships:

𝑍2 𝑍3 𝑅2 𝑅3 𝑅2 𝑅3 ∕𝑅1
𝑍X = 𝑅X + 𝑗𝑋X = = =
𝑍1 𝑅1 + 𝑗𝜔𝐿1 1 + 𝑗𝜔𝐿1 ∕𝑅1

𝑅X
1 𝑗𝜔𝐶X 𝑅X
𝑍X = 𝑅X ∥ = =
𝑗𝜔𝐶X 1 𝑗𝜔𝑅X 𝐶X + 1
𝑅X +
𝑗𝜔𝐶X

Equating the expressions and then equating like terms, we get the following relationships:

𝑅2 𝑅3 ∕𝑅1 𝑅X
=
1 + 𝑗𝜔𝐿1 ∕𝑅1 𝑗𝜔𝑅X 𝐶X + 1

𝑅2 𝑅3
𝑅X =
𝑅1

𝐿1
𝑅X 𝐶X =
𝑅1

𝐿1
𝐶X =
𝑅2 𝑅3

Exercise 8–14. The circuit in Figure 8–21(a) is operating in the sinusoidal steady state with 𝑖S (𝑡) = 100 cos(1000𝑡 −
45◦ ) mA.

(a). Transform the circuit into the phasor domain.

In the phasor domain, we have the following values:

𝐈S = 0.1∠−45◦ A

1 1
𝑍C = = = −𝑗1000 Ω
𝑗𝜔𝐶 𝑗(1000 rad/s)(1 𝜇F)

𝑍R = 500 Ω

𝑍L = 𝑗𝜔𝐿 = 𝑗(1000 rad/s)(0.5 H) = 𝑗500 Ω

The phasor-domain circuit is shown below.

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The Analysis and Design of Linear Circuits Eighth Edition

(b). Solve for the phasor voltage 𝐕.

Find the equivalent impedance and then the phasor voltage.

(−𝑗1000)(500 + 𝑗500)
𝑍EQ = 𝑍C ∥ (𝑍R + 𝑍L ) = = 1000 Ω
−𝑗1000 + 500 + 𝑗500

𝐕 = 𝑍𝐈S = (1000 Ω)(0.1∠−45◦ A) = 100∠−45◦ V

(c). Solve for the phasor current through each element.

We have the following currents:

𝐕
𝐈=
𝑍
100∠−45◦ 𝑉 𝑉 𝑉
𝐈C = = 100∠45◦ mA
1000∠−90◦ Ω
100∠−45◦ V
𝐈R = 𝐈L = √ = 141.4∠−90◦ mA
◦
500 2∠45 Ω

(d). Find the waveforms corresponding to the phasors found in (b) and (c).
By inspection, we have the following waveforms:

𝑣(𝑡) = 100 cos(1000𝑡 − 45◦ ) V

𝑖C (𝑡) = 100 cos(1000𝑡 + 45◦ ) mA

𝑖R (𝑡) = 𝑖L (𝑡) = 141.4 cos(1000𝑡 − 90◦ ) mA

Exercise 8–15. Using the values in Example 8–10, find the voltage 𝑣L (𝑡) across the inductor in the circuit shown in
Figure 8–22. Validate your answer using Multisim
We have the following relationships:

𝐕L = 𝑍L 𝐈L

𝐈L = 𝐈 = 0.2∠53.1◦ A

𝑍L = 𝑗500 Ω = 500∠90◦ Ω

𝐕L = (500∠90◦ Ω)(0.2∠53.1◦ A) = 100∠143.1◦ V

𝑣L (𝑡) = 100 cos(2000𝑡 + 143.1◦ ) V

The Multisim simulation and results in the figure below verify the solution.

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The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–16. The circuit in Figure 8–25 is operating in the sinusoidal steady state. If 𝑅 = 1 kΩ, 𝐿 = 200 mH, and
𝐶 = 1 𝜇F:

(a). Find the value of 𝜔 that will cause the circuit to be in resonance.
Using the results of Example 8–11, we have the following relationship:
√ √
1 1 1 1
𝜔0 = − = − = 2 krad/s
𝐿𝐶 (𝑅𝐶)2 (200 mH)(1 𝜇F) [(1 kΩ)(1 𝜇F)]2

(b). What will the value of 𝑍EQ be under those conditions?

Again, using the results of Example 8–11, we have the following relationship:

𝑅
𝑍EQ = 𝑗𝜔𝐿 + 𝑍RC = 𝑗𝜔𝐿 + = 200 + 𝑗0 Ω
1 + 𝑗𝜔𝑅𝐶

As expected, at a resonant frequency, the impedance does not have an imaginary component.

Exercise 8–17. The circuit in Figure 8–25 is operating in the sinusoidal steady state at 𝜔 = 1 krad/s. If 𝑅 = 1 kΩ,
𝐿 = 200 mH, and 𝐶 = 1 𝜇F:

(a). Find the value of 𝑍EQ classically under those conditions.

We have the following relationships:

𝑍EQ = 𝑍L + (𝑍C ∥ 𝑍R )

𝑍L = 𝑗𝜔𝐿 = 𝑗(1000 rad/s)(200 mH) = 𝑗200 Ω

𝑍R = 1 kΩ

1 1
𝑍C = = = −𝑗1000 Ω
𝑗𝜔𝐶 𝑗(1000 rad/s)(1 𝜇F)

(1000)(−𝑗1000)
𝑍EQ = 𝑗200 + = 500 − 𝑗300 Ω = 583.1∠−31◦ Ω
1000 − 𝑗1000

(b). Repeat the problem using Multisim. (Hint: Drive the circuit with a 1-V ac signal source at 1 krad/s. The equiv-
alent impedance will be found when doing the simulation by creating your own expression at the output using
“Add expression” and adding the following to the “Selected variables for analysis” column: V(1)/I(L1).)
The Multisim simulation and results in the figure below verify the solution.

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Exercise 8–18. The circuit in Figure 8–27 is operating at 𝜔 = 10 krad/s.

(a). Find the equivalent impedance 𝑍.
Compute the impedance.

𝑍 = (500 − 𝑗125 Ω) ∥ (100 + 𝑗400 Ω)

(500 − 𝑗125 Ω)(100 + 𝑗400 Ω)

𝑍=
500 − 𝑗125 + 100 + 𝑗400 Ω

𝑍 = 256.1 + 𝑗195.1 Ω = 321.96∠37.3◦ Ω

(b). What element should be connected in series with 𝑍 to make the total reactance zero?
To make the total reactance zero, we need to connect 𝑍C = −𝑗195.1 Ω. The capacitance is 𝐶 = 1∕(195.1𝜔) =
1∕[(195.1 Ω)(10000 rad/s)] = 0.513 𝜇F.

Exercise 8–19. In Figure 8–28, 𝑣S (𝑡) = 12.5 cos(1000𝑡) V and 𝑖S (𝑡) = 0.2 cos(1000𝑡−36.9◦ ) A. What is the impedance
seen by the voltage source and what element is in the box?
Compute the impedance.

𝐕S
𝑍=
𝐈S

12.5∠0◦
𝑍=
0.2∠−36.9◦

𝑍 = 62.5∠36.9◦ = 50 + 𝑗37.5 Ω

The component is an inductor with 𝐿 = (37.5 Ω)∕𝜔 = (37.5 Ω)∕(1000 rad/s) = 37.5 mH.
Exercise 8–20. For the circuit of Figure 8–29(a), design a low-pass filter using standard parts so that the cutoff fre-
quency is 1000 rad/s.
There are many correct solutions. One approach is to use a series 𝑅𝐶 circuit with the output taken across the
capacitor. Select the components such that the cutoff frequency is 𝜔C = 1∕𝑅𝐶 = 1000. Use 𝑅 = 1 kΩ and 𝐶 = 1 𝜇F,
as one approach with standard parts.

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The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–21. For the circuit of Figure 8–29(a), replace the capacitor with an inductor.
(a). Find the ratio ||𝐕L ∕𝐕S ||.
Use voltage division to find the requested ratio.
𝑍L
𝐕L = 𝐕
𝑍R + 𝑍L S

𝑗𝜔𝐿
𝐕L = 𝐕
𝑅 + 𝑗𝜔𝐿 S

𝐕L 𝑗𝜔
=
𝐕S 𝑅
+ 𝑗𝜔
𝐿
| 𝐕L | 𝜔
| |= √
|𝐕 | ( )2
| S| 𝑅
𝜔2 +
𝐿

(b). Comment on its behavior as the frequency changes from 0 to ∞.

For small values of 𝜔, the output voltage is small. As 𝜔 increases, the output voltage increases and eventually
approaches the value of the input voltage. This is a high-pass filter because it blocks low-frequency signals and
passes high-frequency signals. The cutoff frequency is 𝜔C = 𝑅∕𝐿 rad/s.
Exercise 8–22. Use the unit output method to find the output current 𝐈O in the circuit of Figure 8–31.
The following calculations provide the solution. First, define the output current to be 1 A.
𝐈O = 1 A
Compute the voltage across the right horizontal resistor.
𝐕C = (𝐈O )(𝑗50) = 𝑗50 V
Compute the current through the right vertical resistor.
𝐕C
𝐈C = = 𝑗1 A
50
Compute the current through the left horizontal resistor.
𝐈B = 𝐈O + 𝐈C = 1 + 𝑗1 A
Compute the voltage across the left vertical resistor
𝐕B = 100𝐈B + 𝐕C = 100 + 𝑗150 V
Compute the current through the left vertical resistor
𝐕B
𝐈A = = 3.4605 + 𝑗1.0126 A
50∠40◦
Compute the current through the source
𝐈S = 𝐈A + 𝐈B = 4.4605 + 𝑗2.0126 A
Find the unit-output gain
1 1A
𝐾= = = 0.18627 − 𝑗0.084046 = 0.20435∠−24.3◦
𝐈S 4.4605 + 𝑗2.0126 A
Compute the actual output current
𝐈O = 𝐾𝐈S = (0.20435∠−24.3◦ )(50∠15◦ mA) = 10.22∠−9.28◦ mA

Solution Manual Chapter 8 Page 8-13

The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–23. The two sources in Figure 8–34 have the same frequency. Use superposition to find the phasor current
𝐈X .
Replace the voltage source with a short circuit and find the contribution from the current source. Find the equivalent
impedance parallel to the 100-Ω resistor and apply two-path current division.

𝑍EQ1 = 𝑗75 + (10 ∥ −𝑗100) = 9.901 + 𝑗74.01 Ω

100
𝐈X1 = (0.1∠−90◦ A) = −42.157 − 𝑗62.601 mA
100 + 9.901 + 𝑗74.01
Replace the current source with an open circuit and find the contribution from the voltage source. Find the equivalent
impedance in series with the 10-Ω resistor and apply voltage division.

𝑍EQ2 = (100 + 𝑗75) ∥ −𝑗100 = 94.118 − 𝑗76.471 Ω

94.118 − 𝑗76.471
𝐕X2 = − (20∠45◦ V) = −13.908 − 12.612 V
10 + 94.118 − 𝑗76.471

−13.908 − 12.612 V
𝐈X2 = = −149.547 − 𝑗13.958 mA
100 + 𝑗75 Ω
Sum the component currents.

𝐈X = 𝐈X1 + 𝐈X2 = −42.157 − 𝑗62.601 − 149.547 − 𝑗13.958 = −191.704 − 𝑗76.559 = 206.426∠−158.23◦ mA

Exercise 8–24. Use superposition to find the output voltage 𝑣O (𝑡) in the circuit of Figure 8–36 if
𝑖S (𝑡) = 100 cos(10000𝑡) mA and 𝑣S (𝑡) = 20 cos(20000𝑡 − 45◦ ) V.
The source frequencies are different, so the impedances will differ and we will not be able to combine the sinusoidal
expressions. Replace the voltage source with a short circuit, find the equivalent impedance, and the output voltage.

1
𝑍C1 = = −𝑗100 Ω
𝑗(10000 rad/s)(1 𝜇F)

𝑍EQ1 = 100 ∥ −𝑗100 = 50 − 𝑗50 Ω

𝐕O1 = 𝑍EQ1 𝐈S = (50 − 𝑗50 Ω)(100∠0◦ mA) = 5 − 𝑗5 V = 7.07∠−45◦ V

Replace the current source with an open circuit and apply voltage division.

1
𝑍C2 = = −𝑗50 Ω
𝑗(20000 rad/s)(1 𝜇F)

−𝑗50
𝐕O2 = (20∠−45◦ V) = −2.828 − 𝑗8.485 V = 8.944∠−108.43◦ V
100 − 𝑗50
Write the final expression.

𝑣O (𝑡) = 7.07 cos(10000𝑡 − 45◦ ) + 8.944 cos(20000𝑡 − 108.43◦ ) V

Solution Manual Chapter 8 Page 8-14

The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–25. Analyze the sinusoidal steady-state behavior of the circuit shown in Figure 8–37 in more detail. To do
so, find the magnitude of the ratio of the output voltage to the input voltage for the range of frequencies from 1 Hz to
1 kHz. For simplicity, assume that the input signal always has a magnitude of 1 V. Examine at least 500 data points
in the frequency range and space them logarithmically with the MATLAB command logspace. Plot the results in
terms of magnitude versus frequency (in hertz) on log-log axes. Use the plot to verify the results in Example 8–18.

Compute the equivalent impedance to the right of the inductor on the left.

( )
1
(100 + 𝑗𝜔)
1 𝑗𝜔𝐶 100 + 𝑗𝜔
𝑍EQ = ∥ (100 + 𝑗𝜔) = =
𝑗𝜔𝐶 1 1 − 𝜔2 𝐶 + 𝑗100𝜔𝐶
+ 𝑗𝜔 + 100
𝑗𝜔𝐶

Apply voltage division twice to determine the output voltage gain.

𝑍EQ
𝐕A = 𝐕S
𝑍EQ + 1 + 𝑗𝜔

100 100𝑍EQ
𝐕L = 𝐕A = 𝐕
100 + 𝑗𝜔 (100 + 𝑗𝜔)(𝑍EQ + 1 + 𝑗𝜔) S

𝐕L 100𝑍EQ
𝐾= =
𝐕S (100 + 𝑗𝜔)(𝑍EQ + 1 + 𝑗𝜔)

The following MATLAB code provides the solution.

% D e f i n e the f i x e d p a r a m e t e r s
Rs = 1;
L1 = 1;
L2 = 1;
C = 14 e - 6;
RL = 100;
% C r e a t e a list of f r e q u e n c i e s to e v a l u a t e
f = l o g s p a c e (0 ,3 ,1000);
% C r e a t e the r a d i a n f r e q u e n c i e s
w = 2* pi * f ;
% C o m p u t e the gain for each f r e q u e n c y
Zeq = ( 1 0 0 + j * w ) . /(1+ j * 1 0 0 * w * C - C * w. ^2);
K = abs ( 1 0 0 * Zeq. / ( ( 1 0 0 + j * w ) . *( Zeq +1+ j * w )));
% Plot the r e s u l t s
figure
l o g l o g ( f , K , 'b ' , ' L i n e W i d t h ' ,2 .5 )
grid on
x l a b e l ( ' Frequency , ( Hz ) ')
y l a b e l ( ' Gain ')
t i t l e ( ' E x e r c i s e 8 - 25 ')

The corresponding MATLAB output is shown below.

Solution Manual Chapter 8 Page 8-15

The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8−25
1
10

0
10

−1
10

Gain
−2
10

−3
10

−4
10

−5
10
0 1 2 3
10 10 10 10
Frequency, (Hz)

The plot is consistent with the results presented in Example 8–18. Specifically, using the cursor function with the plot,
at a frequency of 60 Hz, the gain is approximately one and at a frequency of 180 Hz, the gain is approximately 0.0055.

Exercise 8–26. The Norton circuit in Figure 8–41(b) has a current source 𝐈N = 300−𝑗400 mA and a Norton impedance
𝑍N of 100 + 𝑗100 Ω. Find the equivalent Thévenin circuit.
We have the following relationships:

𝑍T = 𝑍N = 100 + 𝑗100 Ω = 141.4∠45◦ Ω

𝐕T = 𝑍N 𝐈N = (100 + 𝑗100 Ω)(300 − 𝑗400 mA) = 70 − 𝑗10 V = 70.71∠−8.13◦ V

Exercise 8–27. Convert the Thévenin circuit found in Example 8–20 into its Norton equivalent. Then repeat the design
task in that example.
Compute the Norton equivalent circuit.

𝑍N = 𝑍T = 500 + 𝑗500 Ω = 707.1∠45◦ Ω

𝐕T 35.35∠−45◦ V
𝐈N = = = 0 − 𝑗50 mA = 50∠−90◦ mA
𝑍T 707.1∠45◦ 𝑂𝑂

For the load design, first find the current through the Norton impedance:

𝐕L = 10∠−90◦ V

𝐕L 10∠−90◦ V
𝐈1 = = = −10 − 𝑗10 mA
𝑍N 707.1∠45◦ Ω

Apply KCL to find the current through the load impedance and then calculate that impedance as the ratio of the voltage
across it to the current through it.

𝐈L = 𝐈N − 𝐈1 = 0 − 𝑗50 − (−10 − 𝑗10) mA = 10 − 𝑗40 mA

𝐕L 10∠−90◦ V
𝑍L = = = 235.3 − 𝑗58.8 Ω
𝐈L 10 − 𝑗40 mA

As expected, the load matches the design from Example 8–20.

Solution Manual Chapter 8 Page 8-16

The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–28. Repeat Example 8–21 using Multisim if both sources are operating at a frequency of 𝜔 = 20 krad/s.
(Hint: Find 𝐿1 , 𝐿2 , and 𝐶 from the data in Example 8–21 first.)
Compute the component values.

𝜔𝐿1 = 10 Ω

10 Ω
𝐿1 = = 2 mH
5000 rad/s

𝜔𝐿2 = 30 Ω

30 Ω
𝐿2 = = 6 mH
5000 rad/s
1
= 10 Ω
𝜔𝐶
1
𝐶= = 20 𝜇F
(10 Ω)(5000 rad/s)
The Multisim simulation and results are shown in the figure below.

The results are 𝑣R (𝑡) = 9.196 cos(20000𝑡 − 163.8◦ ) V.

Solution Manual Chapter 8 Page 8-17

The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–29.

(a). Find the Thévenin equivalent circuit seen by the inductor in Figure 8–45.
Remove the inductor from the circuit and find the open-circuit voltage. Apply voltage division across the parallel
combination of the resistor and capacitor:

𝑍EQ = 500 Ω ∥ −𝑗500 Ω = 250 − 𝑗250 Ω

𝑍EQ 𝐕S (250 − 𝑗250 Ω)(50∠120◦ V)

𝐕T = = = 0.8925 + 𝑗13.839 V = 13.868∠86.31◦ V
1000 + 𝑍EQ 1250 − 𝑗250 Ω

Replace the voltage source with a short circuit and use the lookback technique to find the Thévenin impedance:

𝑍T = 1000 Ω ∥ 500 Ω ∥ −𝑗500 Ω = 230.8 − 𝑗153.8 Ω = 277.4∠−33.7◦ Ω

(b). Use the Thévenin equivalent to calculate the current 𝐈X .

Compute the current using Ohm’s law

𝐕T 13.868∠86.31◦ V
𝐈X = = = −9.602 + 𝑗57.73 mA = 58.52∠99.44◦ mA
𝑍T + 𝑗100 Ω 230.8 − 𝑗153.8 + 𝑗100 Ω

The following MATLAB code also provides the solution:

% D e f i n e the v o l t a g e s o u r c e
Vs = 50* exp ( j * 1 2 0 * pi / 1 8 0 ) ;
% Find the e q u i v a l e n t i m p e d a n c e to the r i g h t of the i n d u c t o r
Zeq = ( 5 0 0 ) * ( - 500 i ) / ( 5 0 0 - 500 i );
% Find the open - c i r c u i t v o l t a g e u s i n g v o l t a g e d i v i s i o n
VT = Zeq * Vs / ( 1 0 0 0 + Zeq )
% Find the l o o k b a c k r e s i s t a n c e
ZT = 1 / ( 1 / 1 0 0 0 + 1 / 5 0 0 + 1/ - 500 i )
% Find the c u r r e n t t h r o u g h the i n d u c t o r
IX = VT /( ZT +100 i )

The corresponding MATLAB output is shown below.

VT = 892 . 5 0 3 1 e - 003 + 13 . 8 3 8 8 e +000 i

ZT = 230 . 7 6 9 2 e +000 - 153 . 8 4 6 2 e +000 i
IX = - 9 . 6 0 2 2 e - 003 + 57 . 7 2 7 4 e - 003 i

Exercise 8–30. By inspection, determine the Thévenin equivalent circuit seen by the capacitor in Figure 8–30 for
𝐕S = 10∠0◦ V.
The impedance is a parallel combination of two equal branches, so the Thévenin impedance is half of the impedance
of one branch.
50 + 𝑗100 Ω
𝑍T = = 25 + 𝑗50 Ω
2
The open-circuit voltage can be found through voltage division, which occurs between two equal impedances, so it is
half of the source voltage.
𝐕 10∠0◦ V
𝐕T = S = = 5∠0◦ V
2 2

Solution Manual Chapter 8 Page 8-18

The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–31. In the steady state, the short-circuit current at an interface is observed to be

𝑖SC (𝑡) = 0.75 sin(𝜔𝑡) A

When a 150-Ω resistor is connected across the interface, the interface current is observed to be

𝑖(𝑡) = 0.6 cos(𝜔𝑡 − 53.1◦ ) A

Find the Norton equivalent phasor circuit at the interface.

The short-circuit current is the Norton current.

𝑖SC (𝑡) = 0.75 sin(𝜔𝑡) A = 0.75 cos(𝜔𝑡 − 90◦ ) A

𝐈N = 𝐈SC = 0.75∠−90◦ A = 0 − 𝑗0.75 A

Find the interface current phasor, the interface voltage, the current through the Norton impedance, and, finally, the
Norton impedance.

𝐈 = 0.6∠−53.1◦ A = 0.36 − 𝑗0.48 A

𝐕 = 150𝐈 = (150 Ω)(0.36 − 𝑗0.48 A) = 54 − 𝑗72 V

𝐈1 = 𝐈N − 𝐈 = 0 − 𝑗0.75 − (0.36 − 𝑗0.48) A = −0.36 − 𝑗0.27 A = 0.45∠−143.1◦ A

𝐕 54 − 𝑗72 V
𝑍N = = = 0 + 𝑗200 Ω
𝐈1 −0.36 − 𝑗0.27 A
Exercise 8–32. Repeat the problem in Example 8–24, except you cannot use capacitors.
Use an inductor in series with a resistor and take the output across the resistor. We have the following relationships:
𝐕O [ ( )]
𝑅 𝑅 𝑋
= =√ ∠ − tan−1
𝐕S 𝑅 + 𝑗𝑋 𝑅2 + 𝑋 2 𝑅

𝑅
√ = 0.707
𝑅2 + 𝑋 2
( )
𝑋
− tan−1 = −45◦
𝑅
Pick 𝑅 = 1 kΩ and then we solve for 𝑋 = 1 kΩ. Solve for 𝐿 = 𝑋∕𝜔 = (1 kΩ)∕(1000 rad/s) = 1 H. Other solutions
are possible. The Multisim simulation and results in the figure below verify the solution.

Solution Manual Chapter 8 Page 8-19

The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–33. Use Multisim to solve for the node voltages 𝐕A and 𝐕B in the circuit of Figure 8–49(a).

The radian frequency of the circuit is 𝜔 = 10 krad/s or 1591.55 Hz. The Multisim simulation and results are shown
below.

The results are 𝐕A = 12 + 𝑗4 V = 12.6∠18.4◦ V and 𝐕B = 12 − 𝑗6 V = 13.4∠−26.6◦ V and agree with the solution
in Example 8–25.

Exercise 8–34. The circuit in Figure 8–51 is operating in the sinusoidal steady-state at 10000 rad/s. Use Multisim
to find the Thévenin equivalent circuit seen by the 60-Ω resistor. Then use the Thévenin circuit to find the current
waveform through the 60-Ω resistor.

Remove the 60-Ω resistor from the circuit. Create the new circuit twice in Multisim. In one of the new circuits,
measure the open-circuit voltage in place of the 60-Ω resistor. In the other circuit, insert a very small resistor in place of
the 60-Ω resistor and measure the short-circuit current. Compute the ratio of the open-circuit voltage to the short-circuit
current to determine the Thévenin impedance. The simulations and results are shown below.

Solution Manual Chapter 8 Page 8-20

The Analysis and Design of Linear Circuits Eighth Edition

The Thévenin equivalent circuit is 𝐕T = 67.97∠−0.448◦ V and 𝑍T = 45.78∠13.78◦ Ω = 44.46 + 𝑗10.9 Ω. Create
the Thévenin equivalent circuit and connect it to a 60-Ω resistor. The Thévenin equivalent consists of a voltage source
matching 𝐕T in series with a 44.46-Ω resistor and a (10.9 Ω)∕(10000 rad/s) = 1.09-mH inductor. Simulate the circuit
and measure the current through the 60-Ω resistor. The circuit and simulation results are shown below.

The waveform of the current through the 60-Ω resistor is 𝑖R (𝑡) = 647.17 cos(10000𝑡 − 6.4◦ ) mA.

Solution Manual Chapter 8 Page 8-21

The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–35. In the circuit of Figure 8–53, 𝑍1 is a 1-kΩ resistor and 𝑍2 is the parallel combination of a 10-kΩ
resistor and a 1-𝜇F capacitor. Determine the output voltage 𝑣O (𝑡) if the input is 𝑣S (𝑡) = 1 cos(100𝑡) V.
Compute 𝑍2 , the gain of the circuit, and the output voltage.

1
𝑍2 = 10 kΩ ∥ = 10 kΩ ∥ −𝑗10 kΩ = 5 − 𝑗5 kΩ
𝑗𝜔𝐶

𝑍2 5 − 𝑗5 kΩ
𝐾=− =− = −5 + 𝑗5 = 7.07∠135◦
𝑍1 1 kΩ

𝐕O = 𝐾𝐕S = (𝐾)(1∠0◦ V) = 𝐾 = 7.07∠135◦ V

𝑣O (𝑡) = 7.07 cos(100𝑡 + 135◦ ) V

Exercise 8–36. Use the circuit of Figure 8–54 to design a high-pass filter with a pass-band gain of −100 and a cutoff
frequency 𝜔C of 10000 rad/s. Use standard parts. (See the inside rear cover).
Pick 𝑅1 = 1 kΩ and use the cutoff frequency to solve for the capacitance. Then use the gain to solve for the other
resistor.
1
𝜔C =
𝑅1 𝐶

1 1
𝐶= = = 0.1 𝜇F
𝜔C 𝑅1 (10000 rad/s)(1 kΩ)

𝑅2
𝐾=− = −100
𝑅1

𝑅2 = 100𝑅1 = 100 kΩ

Other solutions are possible.

Exercise 8–37. Design a bandpass filter with a lower frequency cutoff of 100 Hz and an upper frequency cutoff of
20 kHz.
Use the circuit in Figure 8-55. The low-pass filter (left side) controls the upper frequency cutoff and the high-pass
filter (right side) controls the lower frequency cutoff. Select both resistors to be equal with 𝑅1 = 𝑅2 = 1 kΩ. Solve
for the capacitor values.

1
= (20000)(2𝜋)
𝑅1 𝐶1

1
𝐶1 = = 7960 pF
40000𝜋𝑅1

1
= (100)(2𝜋)
𝑅2 𝐶2

1
𝐶2 = = 1.59 𝜇F
200𝜋𝑅2

Other solutions are possible.

Solution Manual Chapter 8 Page 8-22

The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–38. Repeat Example 8–32 but use node-voltage analysis to solve for the currents of Figure 8–58. Which
method do you think is easier and why?
Using node-voltage analysis, only the upper right node is not defined by a voltage source. We have the following
results:
𝐕C − 120 𝐕C − 120∠−30◦ 𝐕C
0= + +
𝑗10 𝑗20 50
( )
1 1 1 120 120∠−30◦
+ + 𝐕C = +
𝑗10 𝑗20 50 𝑗10 𝑗20

𝐕C = 110.02 − 𝑗34.67 V

120 − 120∠−30◦ V
𝐈1 = = 12.4∠−15◦ A
𝑗5 Ω

120 − 110.02 + 𝑗34.67 V

𝐈2 = = 3.61∠−16.1◦ A
𝑗10 Ω

120∠−30◦ − 110.02 + 𝑗34.67 V

𝐈3 = = 1.303∠166.5◦ A
𝑗20 Ω
The following MATLAB code also provides a solution.

% D e f i n i e the u n k n o w n node v o l t a g e
syms VC
% D e f i n e the v o l t a g e s o u r c e s
VS1 = 120;
VS2 = 120* exp ( - j * pi /6);
% W r i t e the node - v o l t a g e e q u a t i o n and s o l v e it
Eqn1 = ( VC - VS1 )/10 j + ( VC - VS2 )/20 j + VC /50;
Soln = s o l v e ( Eqn1 , VC );
VC = d o u b l e ( Soln )
% C a l c u l a t e the c u r r e n t s
I1 = ( VS1 - VS2 )/5 j
I2 = ( VS1 - VC )/10 j
I3 = ( VS2 - VC )/20 j
% C o n v e r t to m a g n i t u d e and p h a s e
I 1 M a g = abs ( I1 )
I 1 P h a s e = 180* a n g l e ( I1 )/ pi
I 2 M a g = abs ( I2 )
I 2 P h a s e = 180* a n g l e ( I2 )/ pi
I 3 M a g = abs ( I3 )
I 3 P h a s e = 180* a n g l e ( I3 )/ pi

The corresponding MATLAB output is shown below.

VC = 110 . 0 1 8 5 e +000 - 34 . 6 6 9 1 e +000 i

I1 = 12 . 0 0 0 0 e +000 - 3 . 2 1 5 4 e +000 i
I2 = 3 . 4 6 6 9 e +000 - 998 . 1 5 3 4 e - 003 i
I3 = - 1 . 2 6 6 5 e +000 +304 . 7 7 0 9 e - 003 i
I 1 M a g = 12 . 4 2 3 3 e +000
I 1 P h a s e = - 15 . 0 0 0 0 e +000
I 2 M a g = 3 . 6 0 7 7 e +000
I 2 P h a s e = - 16 . 0 6 1 6 e +000
I 3 M a g = 1 . 3 0 2 7 e +000
I 3 P h a s e = 166 . 4 7 0 0 e +000

The node-voltage approach is easier than the mesh-current approach, because there are fewer simultaneous equations
to solve.

Solution Manual Chapter 8 Page 8-23

The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–39. Use the mesh-current or node-voltage method to find the output voltage 𝐕2 and input impedance 𝑍IN
in Figure 8–60.
Label the undesignated node as 𝐕1 and solve the node-voltage equations.

𝐕1 − 𝐕S 𝐕1 𝐕 − 𝐕2
0= + + 1
50 −𝑗50 𝑗100

𝐕2 − 𝐕1 𝐕2 𝐕
0= + + 2
𝑗100 −𝑗50 50

𝐕1 = 3.75 − 𝑗1.25 V = 3.95∠−18.43◦ V

𝐕2 = −1.25 − 𝑗1.25 V = 1.768∠−135◦ V

5 − 3.75 + 𝑗1.25 V
𝐈S = = 25 + 𝑗25 mA
50 Ω
𝐕S 5V
𝑍IN = = = 100 − 𝑗100 Ω
𝐈S 25 + 𝑗25 mA

Exercise 8–40. Use MATLAB and either mesh-current or node-voltage analysis to find the current 𝐈X in Figure 8–62.
Let the upper phasor mesh current be 𝐈A , the lower left phasor mesh current be 𝐈B , and the lower right phasor mesh
current be 𝐈C . The following MATLAB code provides the mesh-current solution:

% D e f i n e the s y m b o l i c v a r i a b l e s and the v o l t a g e s o u r c e

syms IA IB IC
VS = 500;
% C r e a t e the mesh - c u r r e n t e q u a t i o n s
Eqn1 = IA *60 j + ( IA - IC )*60 + ( IA - IB ) * 6 0 ;
Eqn2 = IB *( - 60 j ) + ( IB - IA )*60 + ( IB - IC )*60 + VS ;
Eqn3 = - VS + ( IC - IB )*60 + ( IC - IA )*60 + IC *( - 200 j );
% S o l v e the e q u a t i o n s and d e t e r m i n e n u m e r i c a l v a l u e s
Soln = s o l v e ( Eqn1 , Eqn2 , Eqn3 , IA , IB , IC );
IAnum = double ( Soln.IA )
IBnum = double ( Soln.IB )
ICnum = double ( Soln.IC )
% Find the d e s i r e d c u r r e n t
IX = I A n u m
I X M a g = abs ( IX )
I X P h a s e = 180* a n g l e ( IX )/ pi

The corresponding MATLAB output is shown below.

I A n u m = - 1 . 4 2 6 2 e +000 +213 . 9 3 6 4 e - 003 i

I B n u m = - 3 . 7 6 9 4 e +000 - 1 . 5 1 7 9 e +000 i
I C n u m = 702 . 9 3 4 0 e - 003 +519 . 5 5 9 9 e - 003 i
IX = - 1 . 4 2 6 2 e +000 +213 . 9 3 6 4 e - 003 i
I X M a g = 1 . 4 4 2 2 e +000
I X P h a s e = 171 . 4 6 9 2 e +000

The result is 𝐈X = −1.4262 + 𝑗0.2139 A = 1.4422∠171.47◦ A.

Solution Manual Chapter 8 Page 8-24

The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–41. The circuit in Figure 8–63 is operating in the sinusoidal steady-state at 10 krad/s. Use Multisim to find
the output waveforms corresponding to 𝐕O and 𝐈O .
With 𝜔 = 10 krad/s, the inductor value is 𝐿 = 200∕10000 = 20 mH and the capacitor value is 𝐶 = 1∕(10000)(5) =
20 𝜇F. The Multisim simulation and results in the figure below provide the solution.

The output waveforms are 𝑣O (𝑡) = 4.04 cos(10000𝑡 − 44.9◦ ) V and 𝑖O (𝑡) = 808.1 cos(10000𝑡 + 45.06◦ ) mA.

Exercise 8–42. The circuit in Figure 8–68 is operating in the sinusoidal steady-state at 60 Hz. Find the average power
delivered to the 25-Ω load. Then use Multisim to validate your answer. (Hint: place a small 0.1 𝜇Ω resistor in series
with either inductor to avoid a singular event that prevents Multisim from running.)
Use node-voltage analysis to solve for the voltage across the 25-Ω resistor and then find the average power delivered
to the 25-Ω load.

𝐕A − 150 𝐕A − 125∠−90◦ 𝐕A
0= + +
𝑗2 𝑗4 25

𝐕A = 97.5 − 𝑗46.87 V

𝐕A
𝐈A = = 3.9 − 𝑗1.875 A = 4.327∠−25.67 A
25 Ω
1
𝑅 |𝐈 | = (0.5)(25 Ω)(4.327 A)2 = 234 W
2
𝑃 =
2 L | A|

With 𝑓 = 60 Hz, the inductor values are 𝐿1 = (2 Ω)∕(120𝜋 rad/s) = 5.305 mH and 𝐿2 = (4 Ω)∕(120𝜋 rad/s) =
10.61 mH. The Multisim simulation and results in the figure below verify the solution.

Solution Manual Chapter 8 Page 8-25

The Analysis and Design of Linear Circuits Eighth Edition

Exercise 8–43. Calculate the maximum average power available at the interface in Figure 8–71.
Find the Thévenin equivalent circuit by finding the lookback impedance and the short-circuit current. Then compute
the maximum average power available.

𝑍T = (50 Ω + 𝑗100 Ω) ∥ −𝑗50 Ω = 25 − 𝑗75 Ω

50 Ω
𝐈SC = (0.1 A) = 0.02 − 𝑗0.04 A
50 Ω + 𝑗100 Ω

𝐕T = 𝑍T 𝐈SC = (25 − 𝑗75 Ω)(0.02 − 𝑗0.04 A) = −2.5 − 𝑗2.5 V = 3.536∠−135◦ V

|𝐕T | 2
(3.536 V)2
𝑃MAX = | | = = 62.5 mW
8𝑅T (8)(25 Ω)

Solution Manual Chapter 8 Page 8-26