10.5 Source Transformation

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10.5 Source Transformation

As Fig. 10.16 shows, source transformation in the frequency domain involves transforming a voltage source in series with an impedance to a current source in parallel with an impedance, or vice versa. As we go from one source type to another, we must keep the following relationship in mind:

$$\mathbf{V}_s = \mathbf{Z}_s \mathbf{I}_s \qquad \Leftrightarrow \qquad \mathbf{I}_s = \frac{\mathbf{V}_s}{\mathbf{Z}_s} \qquad (10.1)$$

Calculate Vx in the circuit of Fig. 10.17 using the method of source transformation.

For Example 10.7.

Solution:

We transform the voltage source to a current source and obtain the cir ‑ cuit in Fig. 10.18(a), where

$$I_s = \frac{20/-90^{\circ}}{5} = 4/-90^{\circ} = -j4 \text{ A}$$

The parallel combination of 5‑Ω resistance and (3 + j4) impedance gives

$$\mathbf{Z}_1 = \frac{5(3+j4)}{8+j4} = 2.5 + j1.25 \ \Omega$$

Converting the current source to a voltage source yields the circuit in Fig. 10.18(b), where

$$\mathbf{V}_s = \mathbf{I}_s \mathbf{Z}_1 = -j4(2.5 + j1.25) = 5 - j10 \text{ V}$$

Figure 10.18 Solution of the circuit in Fig. 10.17.

By voltage division,

oltage division,

$$\mathbf{V}_x = \frac{10}{10 + 2.5 + j1.25 + 4 - j13} (5 - j10) = 5.519 \underline{/-28}^\circ \text{ V}$$

Example 10.7

Practice Problem 10.7 Find Io in the circuit of Fig. 10.19 using the concept of source transformation.

Answer: 1.9727∕99.46° A.

10.6 Thevenin and Norton Equivalent Circuits

Thevenin’s and Norton’s theorems are applied to ac circuits in the same way as they are to dc circuits. The only additional effort arises from the need to manipulate complex numbers. The frequency domain version of a Thevenin equivalent circuit is depicted in Fig. 10.20, where a linear circuit is replaced by a voltage source in series with an impedance. The Norton equivalent circuit is illustrated in Fig. 10.21, where a linear cir ‑ cuit is replaced by a current source in parallel with an impedance. Keep in mind that the two equivalent circuits are related as

$$\mathbf{V}_{\mathrm{Th}} = \mathbf{Z}_N \mathbf{I}_N, \qquad \mathbf{Z}_{\mathrm{Th}} = \mathbf{Z}_N \tag{10.2}$$

just as in source transformation. VTh is the open‑circuit voltage while I*^N* is the short‑circuit current.

If the circuit has sources operating at dif ferent frequencies (see Example 10.6, for example), the Thevenin or Norton equivalent circuit must be determined at each frequenc y. This leads to entirely dif ferent equivalent circuits, one for each frequenc y, not one equi valent circuit with equivalent sources and equivalent impedances.

Example 10.8 Obtain the Thevenin equivalent at terminals a‑b of the circuit in Fig. 10.22.

Figure 10.22 For Example 10.8.

Figure 10.21 Norton equivalent.

Solution:

We find ZTh by setting the voltage source to zero. As shown in Fig. 10.23(a), the 8 ‑Ω resistance is now in parallel with the −j6 reac‑ tance, so that their combination gives

$$\mathbf{Z}_1 = -j6 \| 8 = \frac{-j6 \times 8}{8 - j6} = 2.88 - j3.84 \ \Omega$$

Similarly, the 4 ‑Ω resistance is in parallel with the j12 reactance, and their combination gives

$$\mathbf{Z}_2 = 4 || j12 = \frac{j12 \times 4}{4 + j12} = 3.6 + j1.2 \ \Omega$$

The Thevenin impedance is the series combination of Z1 and Z2; that is,

$$\mathbf{Z}_{\text{Th}} = \mathbf{Z}_1 + \mathbf{Z}_2 = 6.48 - j2.64 \ \Omega$$