9.7 Impedance Combinations

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9.7 Impedance Combinations

Consider the N series-connected impedances sho wn in Fig. 9.18. The same current I flows through the impedances. Applying KVL around the loop gives

$$V = V_1 + V_2 + \dots + V_N = I(Z_1 + Z_2 + \dots + Z_N)$$

(9.58)

N impedances in series.

The equivalent impedance at the input terminals is

$$\mathbf{Z}_{\text{eq}} = \frac{\mathbf{V}}{\mathbf{I}} = \mathbf{Z}_1 + \mathbf{Z}_2 + \dots + \mathbf{Z}_N$$ $$Z_{eq} = Z_1 + Z_2 + \dots + Z_N \tag{9.59}$$

or

showing that the total or equi valent impedance of series-connected impedances is the sum of the indi vidual impedances. This is similar to the series connection of resistances.

If N = 2, as shown in Fig. 9.19, the current through the imped ances is

$$\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}_1 + \mathbf{Z}_2} \tag{9.60}$$

Because V1 = Z1I and V2 = Z2I, then

$$V_1 = \frac{Z_1}{Z_1 + Z_2} V, \qquad V_2 = \frac{Z_2}{Z_1 + Z_2} V$$

(9.61)

which is the voltage-division relationship.

In the same manner , we can obtain the equi valent impedance or admittance of the N parallel-connected impedances shown in Fig. 9.20. The voltage across each impedance is the same. Applying KCL at the top node,

$$\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 + \dots + \mathbf{I}_N = \mathbf{V} \left( \frac{1}{\mathbf{Z}_1} + \frac{1}{\mathbf{Z}_2} + \dots + \frac{1}{\mathbf{Z}_N} \right) \tag{9.62}$$

N impedances in parallel.

The equivalent impedance is

$$\frac{1}{Z_{\text{eq}}} = \frac{I}{V} = \frac{1}{Z_1} + \frac{1}{Z_2} + \dots + \frac{1}{Z_N}$$

(9.63)

and the equivalent admittance is

$$\mathbf{Y}_{\text{eq}} = \mathbf{Y}_1 + \mathbf{Y}_2 + \dots + \mathbf{Y}_N \tag{9.64}$$

This indicates that the equivalent admittance of a parallel connection of admittances is the sum of the individual admittances.

When N = 2, as sho wn in Fig. 9.21, the equi valent impedance becomes

$$Z_{\text{eq}} = \frac{1}{Y_{\text{eq}}} = \frac{1}{Y_1 + Y_2} = \frac{1}{1/Z_1 + 1/Z_2} = \frac{Z_1 Z_2}{Z_1 + Z_2}$$

(9.65)

Figure 9.21 Current division.

Voltage division.

Also, since

$$\mathbf{V} = \mathbf{I}\mathbf{Z}_{\text{eq}} = \mathbf{I}_1\mathbf{Z}_1 = \mathbf{I}_2\mathbf{Z}_2$$

the currents in the impedances are

$$\mathbf{I}_1 = \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{I}, \qquad \mathbf{I}_2 = \frac{\mathbf{Z}_1}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{I}$$

(9.66)

which is the current-division principle.

The delta-to-wye and wye-to-delta transformations that we applied to resisti ve circuits are also v alid for impedances. With reference to Fig. 9.22, the conversion formulas are as follows.

Figure 9.22 Superimposed Y and ∆ networks.

Y-∆ Conversion:

$$Z_a = \frac{Z_1 Z_2 + Z_2 Z_3 + Z_3 Z_1}{Z_1}$$

\n

$$Z_b = \frac{Z_1 Z_2 + Z_2 Z_3 + Z_3 Z_1}{Z_2}$$

\n

$$Z_c = \frac{Z_1 Z_2 + Z_2 Z_3 + Z_3 Z_1}{Z_3}$$

\n(9.67)

∆-Y Conversion:

$$\mathbf{Z}_{1} = \frac{\mathbf{Z}_{b}\mathbf{Z}_{c}}{\mathbf{Z}_{a} + \mathbf{Z}_{b} + \mathbf{Z}_{c}}$$

\n

$$\mathbf{Z}_{2} = \frac{\mathbf{Z}_{c}\mathbf{Z}_{a}}{\mathbf{Z}_{a} + \mathbf{Z}_{b} + \mathbf{Z}_{c}}$$

\n

$$\mathbf{Z}_{3} = \frac{\mathbf{Z}_{a}\mathbf{Z}_{b}}{\mathbf{Z}_{a} + \mathbf{Z}_{b} + \mathbf{Z}_{c}}$$

\n(9.68)

A delta or wye circuit is said to be balanced if it has equal impedances in all three branches.

When a ∆-Y circuit is balanced, Eqs. (9.67) and (9.68) become

$$\mathbf{Z}_{\Delta} = 3\mathbf{Z}_{Y} \qquad \text{or} \qquad \mathbf{Z}_{Y} = \frac{1}{3}\mathbf{Z}_{\Delta} \qquad (9.69)$$

where ZY = Z1 = Z2 = Z3 and Z∆ = Za = Zb = Zc.

As you see in this section, the principles of voltage division, current division, circuit reduction, impedance equi valence, and Y-∆ transformation all apply to ac circuits. Chapter 10 will sho w that other c ircuit techniques—such as superposition, nodal analysis, mesh analysis, source transformation, the Thevenin theorem, and the Norton theorem are all applied to ac circuits in a manner similar to their application in dc circuits.

Find the input impedance of the circuit in Fig. 9.23. Assume that the Example 9.10 circuit operates at ω = 50 rad/s.

Solution:

Let

• Z1 = Impedance of the 2-mF capacitor
• Z2 = Impedance of the 3-Ω resistor in series with the10-mF capacitor
• Z3 = Impedance of the 0.2-H inductor in series with the 8-Ω resistor

Then

$$\mathbf{Z}_1 = \frac{1}{j\omega C} = \frac{1}{j50 \times 2 \times 10^{-3}} = -j10 \text{ }\Omega$$ $$\mathbf{Z}_2 = 3 + \frac{1}{j\omega C} = 3 + \frac{1}{j50 \times 10 \times 10^{-3}} = (3 - j2) \text{ }\Omega$$ $$\mathbf{Z}_3 = 8 + j\omega L = 8 + j50 \times 0.2 = (8 + j10) \text{ }\Omega$$

The input impedance is

at impedance is
\n

$$\mathbf{Z}_{in} = \mathbf{Z}_1 + \mathbf{Z}_2 \parallel \mathbf{Z}_3 = -j10 + \frac{(3 - j2)(8 + j10)}{11 + j8}$$

\n

$$= -j10 + \frac{(44 + j14)(11 - j8)}{11^2 + 8^2} = -j10 + 3.22 - j1.07 \Omega$$

Thus,

$$\mathbf{Z}_{\text{in}} = 3.22 - j11.07 \ \Omega$$

For Example 9.10.

Figure 9.25

Figure 9.26

circuit in Fig. 9.25.

For Example 9.11.

The frequency domain equivalent of the

vs = 20 cos(4t − 15°) ⇒ Vs = 20⧸−15° V, ω = 4 10 mF ^⇒ ____ ^{1 j}ωC = ____________ 1 j4 × 10 × 10⁻³ = −j25 Ω 5 H ⇒ jωL = j4 × 5 = j20 Ω

Z1 = Impedance of the 60-Ω resistor

Z2 = Impedance of the parallel combination of the 10-mF capacitor and the 5-H inductor

Then Z1 = 60 Ω and

$$\mathbf{Z}_2 = -j25 \parallel j20 = \frac{-j25 \times j20}{-j25 + j20} = j100 \text{ }\Omega$$

By the voltage-division principle,

$$\mathbf{V}_o = \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{V}_s = \frac{j100}{60 + j100} (20/15^\circ)$$

= (0.8575/30.96°)(20/15°) = 17.15/15.96° V

We convert this to the time domain and obtain

vo(t) = 17.15 cos(4t + 15.96°) V

For Practice Prob. 9.11.

Find current I in the circuit of Fig. 9.28. Example 9.12

Solution:

The delta netw ork connected to nodes a, b, and c can be con verted to the Y network of Fig. 9.29. We obtain the Y impedances as follows using Eq. (9.68):

68):
\n

$$\mathbf{Z}_{an} = \frac{j4(2-j4)}{j4+2-j4+8} = \frac{4(4+j2)}{10} = (1.6+j0.8) \ \Omega$$

\n

$$\mathbf{Z}_{bn} = \frac{j4(8)}{10} = j3.2 \ \Omega, \qquad \mathbf{Z}_{cn} = \frac{8(2-j4)}{10} = (1.6-j3.2) \ \Omega$$

The total impedance at the source terminals is

$$\mathbf{Z} = 12 + \mathbf{Z}_{an} + (\mathbf{Z}_{bn} - j3) \parallel (\mathbf{Z}_{cn} + j6 + 8)$$

= 12 + 1.6 + j0.8 + (j0.2) || (9.6 + j2.8)
= 13.6 + j0.8 + $\frac{j0.2(9.6 + j2.8)}{9.6 + j3}$
= 13.6 + j1 = 13.64/4.204° Ω

The desired current is

$$I = \frac{V}{Z} = \frac{50/0^{\circ}}{13.64/4.204^{\circ}} = 3.666/-4.204^{\circ} A$$

Practice Problem 9.12 Find I in the circuit of Fig. 9.30.

Figure 9.30 For Practice Prob. 9.12.

Figure 9.31

Series RC shift circuits: (a) leading output, (b) lagging output.

Answer: 12.728 ⧸ 63.8° A.