Chapter09

Chapter 9, Problem 46.

If i s = 5 cos(10t + 40 o ) A in the circuit of Fig. 9.53, find i o .

Figure 9.53

For Prob. 9.46.

Chapter 9, Solution 46.

Thus, io (t) = 2.325 cos(10t + 94.46°) A

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Chapter 9, Problem 47.

In the circuit of Fig. 9.54, determine the value of i s (t).

Figure 9.54

For Prob. 9.47.

Chapter 9, Solution 47.

First, we convert the circuit into the frequency domain.

is(t) = 460.7cos(2000t +52.63˚) mA

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Chapter 9, Problem 48.

Given that v s (t) = 20 sin(100t - 40 o ) in Fig. 9.55, determine i x (t).

Figure 9.55

For Prob. 9.48.

Chapter 9, Solution 48.

Converting the circuit to the frequency domain, we get:

10 Ω V1 30 Ω

We can solve this using nodal analysis.

V1(0.1− j0.05 +0.02307 + j0.01538) = 2 −40°

ix = 0.4338sin(100t +9.4°) A

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Chapter 9, Problem 49.

Find v s (t) in the circuit of Fig. 9.56 if the current i x through the 1- Ω resistor is 0.5 sin 200t A.

Figure 9.56

For Prob. 9.49.

Chapter 9, Solution 49.

vs (t) = 1.414 sin(200t – 45°) V

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Chapter 9, Problem 50.

Determine v x in the circuit of Fig. 9.57. Let i s (t) = 5 cos(100t + 40 o )A.

Figure 9.57

For Prob. 9.50.

Chapter 9, Solution 50.

Since ω = 100, the inductor = j100x0.1 = j10 Ω and the capacitor = 1/(j100x10-3) = -j10Ω.

Vx = 20Ix = 50 −50° vx = 50cos(100t −50°) V

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Chapter 9, Problem 51.

If the voltage v o across the 2- Ω resistor in the circuit of Fig. 9.58 is 10 cos2t V, obtain i s .

Figure 9.58

For Prob. 9.51.

Chapter 9, Solution 51.

Is = (5)(3 − j4) = 25 - 53.13°

Therefore,

is (t) = 25 cos(2t – 53.13°) A

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Chapter 9, Problem 52.

If V o = 8 30 o V in the circuit of Fig. 9.59, find I s. .

Figure 9.59

For Prob. 9.52.

Chapter 9, Solution 52.

Vo = I2 (2.5 + j2.5)

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Chapter 9, Problem 53.

Find I o in the circuit of Fig. 9.60.

Figure 9.60

For Prob. 9.53.

Chapter 9, Solution 53.

Convert the delta to wye subnetwork as shown below.

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Chapter 9, Problem 54.

In the circuit of Fig. 9.61, find V s if I o = 2 0 o A.

Figure 9.61

For Prob. 9.54.

Chapter 9, Solution 54.

Since the left portion of the circuit is twice as large as the right portion, the equivalent circuit is shown below.

+− Vs

+−

−+

V1 = Io (1− j) = 2(1− j)

V2 = 2V1 = 4(1− j)

Vs = −V1 −V2 = −6(1− j)

Vs = 8.485 –135° V

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Chapter 9, Problem 55.

* Find Z in the network of Fig. 9.62, given that V o = 4 0 o V.

Figure 9.62

For Prob. 9.55.

* An asterisk indicates a challenging problem.

Chapter 9, Solution 55.

Z = 2.798 – j16.403 Ω

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