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Figure 10.2

← Back to Fundamentals of Electric Circuits Overview 0.5 \text{ H} \Rightarrow j \Omega L = j2

\n \n

0.1 \text{ F} \Rightarrow \frac{1}{j \Omega C} = -j2.5

Thus, the frequency domain equivalent circuit is as shown in Fig. 10.2. # **Figure 10.2** Frequency domain equivalent of the circuit in Fig. 10.1. Applying KCL at node 1,

\frac{20 - V_1}{10} = \frac{V_1}{-j2.5} + \frac{V_1 - V_2}{j4}

or(1+βˆ—jβˆ—1.5)βˆ—βˆ—Vβˆ—βˆ—1+βˆ—jβˆ—2.5βˆ—βˆ—Vβˆ—βˆ—<sup>2</sup>=20βˆ—βˆ—(10.1.1)βˆ—βˆ—Atnode2, or (1 + *j*1.5)**V**1 + *j*2.5**V**<sup>2</sup> = 20 **(10.1.1)** At node 2,

2\mathbf{I}_x + \frac{\mathbf{V}_1 - \mathbf{V}_2}{j4} = \frac{\mathbf{V}_2}{j2}

Butβˆ—βˆ—Iβˆ—βˆ—βˆ—xβˆ—=βˆ—βˆ—Vβˆ—βˆ—1βˆ•βˆ’βˆ—jβˆ—2.5.Substitutingthisgives But **I***x* = **V**1βˆ•βˆ’*j*2.5. Substituting this gives

\frac{2V_1}{-j2.5} + \frac{V_1 - V_2}{j4} = \frac{V_2}{j2}

Bysimplifying,weget By simplifying, we get

11V_1 + 15V_2 = 0 \tag{10.1.2}

Equations(10.1.1)and(10.1.2)canbeputinmatrixformas Equations (10.1.1) and (10.1.2) can be put in matrix form as

\begin{bmatrix} 1+j1.5 & j2.5 \ 11 & 15 \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} 20 \ 0 \end{bmatrix}

Weobtainthedeterminantsas We obtain the determinants as

\Delta = \begin{vmatrix} 1+j1.5 & j2.5 \ 11 & 15 \end{vmatrix} = 15 - j5

\Delta_1 = \begin{vmatrix} 20 & j2.5 \ 0 & 15 \end{vmatrix} = 300, \qquad \Delta_2 = \begin{vmatrix} 1+j1.5 & 20 \ 11 & 0 \end{vmatrix} = -220

\mathbf{V}_1 = \frac{\Delta_1}{\Delta} = \frac{300}{15 - j5} = 18.97 \underline{/18.43^\circ} \text{ V}

\mathbf{V}_2 = \frac{\Delta_2}{\Delta} = \frac{-220}{15 - j5} = 13.91 \underline{/198.3^\circ} \text{ V}

Thecurrentβˆ—βˆ—Iβˆ—βˆ—βˆ—xβˆ—isgivenby The current **I***x* is given by

\mathbf{I}_x \text{ is given by}

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\mathbf{I}_x = \frac{\mathbf{V}_1}{-j2.5} = \frac{18.97 ; / 18.43^\circ}{2.5 ; / -90^\circ} = 7.59 ; / 108.4^\circ ; \text{A}

Transformingthistothetimedomain, Transforming this to the time domain,

i_x = 7.59 \cos(4t + 108.4^\circ)

A Using nodal analysis, find *v*1 and *v*<sup>2</sup> Practice Problem 10.1 in the circuit of Fig. 10.3. **Figure 10.3** For Practice Prob. 10.1. **Answer:** *v*1(*t*) = 28.31 cos(2*t* + 60.01Β°) V, *v*2(*t*) = 82.56 cos(2*t* + 57.12Β°) V. Compute **V**1 and **V**<sup>2</sup> Example 10.2 in the circuit of Fig. 10.4. For Example 10.2. # **Solution:** Nodes 1 and 2 form a supernode as shown in Fig. 10.5. Applying KCL at the supernode gives

3 = \frac{\mathbf{V}_1}{-j3} + \frac{\mathbf{V}_2}{j6} + \frac{\mathbf{V}_2}{12}

or or

36 = j4V_1 + (1 - j2)V_2 \tag{10.2.1}

<span id="page-437-0"></span> # **Figure 10.5** A supernode in the circuit of Fig. 10.4. But a voltage source is connected between nodes 1 and 2, so that

V_1 = V_2 + 10 \frac{\angle 45^{\circ}}{\angle 0.2.2}

SubstitutingEq.(10.2.2)inEq.(10.2.1)resultsin Substituting Eq. (10.2.2) in Eq. (10.2.1) results in

36 - 40 \underline{1135^\circ} = (1 + i2) \mathbf{V}_2 \implies \mathbf{V}_2 = 31.41 \underline{18^\circ} \text{ V}

FromEq.(10.2.2), From Eq. (10.2.2),

V_1 = V_2 + 10 \underline{745^\circ} = 25.78 \underline{770.48^\circ} \text{ V}

Calculate Calculate

V_1

and $V_2$ in the circuit shown in Fig. 10.6. **Practice Problem 10.2** For Practice Prob. 10.2. **Answer: V**1 = 96.8 βˆ•69.66Β° V, **V**2 = 16.88βˆ•165.72Β° V.