Solution:

← Back to Fundamentals of Electric Circuits Overview Equivalent circuits at dc and high frequencies: (a) inductor, (b) capacitor. Comparing Eqs. (9.41) and (9.42), we infer that

$$Z = R \pm jX = |Z|/\theta \tag{9.43}$$

where

$$|Z| = \sqrt{R^2 + X^2}, \qquad \theta = \tan^{-1} \frac{\pm X}{R}$$

(9.44)

and

$$R = |\mathbf{Z}| \cos \theta, \qquad X = |\mathbf{Z}| \sin \theta \tag{9.45}$$

It is sometimes con venient to w ork with the reciprocal of imped ance, known as admittance.

The admittance Y is the reciprocal of impedance, measured in siemens (S).

The admittance Y of an element (or a circuit) is the ratio of the phasor current through it to the phasor voltage across it, or

$$Y = \frac{1}{Z} = \frac{I}{V}$$

(9.46)

The admittances of resistors, inductors, and capacitors can be obtained from Eq. (9.39). They are also summarized in Table 9.3.

As a complex quantity, we may write Y as

$$Y = G + jB \tag{9.47}$$

where G = Re Y is called the conductance and B = Im Y is called the susceptance. Admittance, conductance, and susceptance are all expressed in the unit of siemens (or mhos). From Eqs. (9.41) and (9.47),

$$G + jB = \frac{1}{R + jX} \tag{9.48}$$

By rationalization,

$$G + jB = \frac{1}{R + jX} \cdot \frac{R - jX}{R - jX} = \frac{R - jX}{R^2 + X^2}$$

(9.49)

Equating the real and imaginary parts gives

$$G = \frac{R}{R^2 + X^2}, \qquad B = -\frac{X}{R^2 + X^2}$$

(9.50)

showing that G ≠ 1∕R as it is in resisti ve circuits. Of course, if X = 0, then G = 1∕R.

Find v(t) and i(t) in the circuit shown in Fig. 9.16. Example 9.9

Solution:

From the voltage source 10 cos 4t, ω = 4,

$$V_s = 10/0^{\circ} V$$

The impedance is

$$\mathbf{Z} = 5 + \frac{1}{j\omega C} = 5 + \frac{1}{j4 \times 0.1} = 5 - j2.5 \ \Omega$$

Hence the current

$$\mathbf{I} = \frac{\mathbf{V}_s}{\mathbf{Z}} = \frac{10/0^{\circ}}{5 - j2.5} = \frac{10(5 + j2.5)}{5^2 + 2.5^2}$$

(9.9.1)
= 1.6 + j0.8 = 1.789/26.57° A

The voltage across the capacitor is

The voltage across the capacitor is
\n

$$\mathbf{V} = IZ_C = \frac{I}{j\omega C} = \frac{1.789/26.57^\circ}{j4 \times 0.1} = \frac{1.789/26.57^\circ}{0.4/90^\circ} = 4.47/-63.43^\circ \text{ V}$$

\n(9.9.2)

Converting I and V in Eqs. (9.9.1) and (9.9.2) to the time domain, we get

$$i(t) = 1.789 \cos(4t + 26.57^{\circ}) \text{ A}$$ $$v(t) = 4.47 \cos(4t - 63.43^{\circ}) \text{ V}$$

Notice that i(t) leads v(t) by 90° as expected.

Refer to Fig. 9.17. Determine v(t) and i(t). Practice Problem 9.9

Answer: 8.944 sin (10t + 93.43°) V, 4.472 sin(10t + 3.43°) A.

9.6 Kirchhoff’s Laws in the Frequency Domain

We cannot do circuit analysis in the frequenc y domain without Kirch hoff’s current and v oltage laws. Therefore, we need to e xpress them in the frequency domain.

For KVL, let v1,v2, … , vn be the voltages around a closed loop. Then

$$v_1 + v_2 + \dots + v_n = 0 \tag{9.51}$$

In the sinusoidal steady state, each v oltage may be written in cosine form, so that Eq. (9.51) becomes

$$V_{m1}\cos(\omega t + \theta_1) + V_{m2}\cos(\omega t + \theta_2)$$

• … + $V_{mn}\cos(\omega t + \theta_n) = 0$ (9.52)