Example 10.5

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Example 10.5

(a)

From Eqs. (10.5.4) and (10.5.5),

$$(4 - j4)\mathbf{I}_2 + j2\mathbf{I}_1 + j10 = 0$$

Expressing I1 in terms of I2 gives

$$\mathbf{I}_1 = (2 + j2)\mathbf{I}_2 - 5\tag{10.5.6}$$

Substituting Eqs. (10.5.5) and (10.5.6) into Eq. (10.5.3), we get

$$(8+j8)[(2+j2)I2 - 5] - j50 + j2I2 = 0$$

or

$$I_2 = \frac{90 - j40}{34} = 2.647 - j1.176$$

Current I″ o is obtained as

$$\mathbf{I}_{o}'' = -\mathbf{I}_{2} = -2.647 + j1.176$$

(10.5.7)

From Eqs. (10.5.2) and (10.5.7), we write

$$I_o = I'_o + I''_o = -5 + j3.529 = 6.12 \underline{144.78^\circ} A$$

which agrees with what we got in Example 10.3. It should be noted that applying the superposition theorem is not the best way to solve this prob‑ lem. It seems that we have made the problem twice as hard as the origi ‑ nal one by using superposition. However, in Example 10.6, superposi ‑ tion is clearly the easiest approach.

Practice Problem 10.5	Find current	Io in the circuit of Fig. 10.8 using the superposition
	theorem.

Answer: 5.97∕65.45° A.

Because the circuit operates at three different frequencies (ω = 0 for the dc voltage source), one way to obtain a solution is to use superposition, which breaks the problem into single‑frequency problems. So we let

$$v_o = v_1 + v_2 + v_3 \tag{10.6.1}$$

where v1 is due to the 5‑V dc voltage source, v2 is due to the 10 cos 2t V voltage source, and v3 is due to the 2 sin 5t A current source.

To find v1, we set to zero all sources except the 5‑V dc source. We recall that at steady state, a capacitor is an open circuit to dc while an inductor is a short circuit to dc. There is an alternative way of looking at this. Because ω = 0, jωL = 0, 1∕jωC = ∞. Either way, the equivalent circuit is as shown in Fig. 10.14(a). By voltage division,

$$-v_1 = \frac{1}{1+4} (5) = 1 \text{ V}$$

(10.6.2)

To find v2, we set to zero both the 5‑V source and the 2 sin 5t current source and transform the circuit to the frequency domain.

10 cos 2t

$$\Rightarrow$$

$10\underline{/0}^{\circ}$ , $\omega = 2 \text{ rad/s}$
2 H $\Rightarrow$ $j\omega L = j4 \Omega$
0.1 F $\Rightarrow$ $\frac{1}{j\omega C} = -j5 \Omega$

The equivalent circuit is now as shown in Fig. 10.14(b). Let

$$\mathbf{Z} = -j5 \| 4 = \frac{-j5 \times 4}{4 - j5} = 2.439 - j1.951$$

Figure 10.14

Solution of Example 10.6: (a) setting all sources to zero except the 5‑V dc source, (b) setting all sources to zero except the ac voltage source, (c) setting all sources to zero except the ac current source.

By voltage division,

oltage division,
\n

$$\mathbf{V}_2 = \frac{1}{1 + j4 + \mathbf{Z}} (10/0^\circ) = \frac{10}{3.439 + j2.049} = 2.498 \underline{\text{ } 20.79^\circ}$$

In the time domain,

$$v_2 = 2.498 \cos(2t - 30.79^\circ) \tag{10.6.3}$$

To obtain v3, we set the voltage sources to zero and transform what is left to the frequency domain.

$$2 \sin 5t \Rightarrow 2(-90^\circ, \omega = 5 \text{ rad/s})$$ $$2 \text{ H} \Rightarrow j\omega L = j10 \Omega$$ $$0.1 \text{ F} \Rightarrow \frac{1}{j\omega C} = -j2 \Omega$$

The equivalent circuit is in Fig. 10.14(c). Let

$$\mathbf{Z}_1 = -j2 \parallel 4 = \frac{-j2 \times 4}{4 - j2} = 0.8 - j1.6 \ \Omega$$

By current division,

$$\mathbf{I}_1 = \frac{j10}{j10 + 1 + \mathbf{Z}_1} (2 \angle -90^\circ) \text{ A}$$ $$\mathbf{V}_3 = \mathbf{I}_1 \times 1 = \frac{j10}{1.8 + j8.4} (-j2) = 2.328 \angle -80^\circ \text{ V}$$

In the time domain,

$$v_3 = 2.33 \cos(5t - 80^\circ) = 2.33 \sin(5t + 10^\circ) \text{ V} \qquad (10.6.4)$$

Substituting Eqs. (10.6.2) to (10.6.4) into Eq. (10.6.1), we have

$$v_o(t) = -1 + 2.498 \cos(2t - 30.79^\circ) + 2.33 \sin(5t + 10^\circ) \text{ V}$$

Practice Problem 10.6 Calculate vo in the circuit of Fig. 10.15 using the superposition theorem.

Answer: 11.577 sin(5t − 81.12°) + 3.154 cos(10t − 86.24°) V.