Figure 9.9

← Back to Fundamentals of Electric Circuits Overview By following the same steps as we took for the inductor or by applying Eq. (9.27) on Eq. (9.36), we obtain

$$\mathbf{I} = j\omega C \mathbf{V} \qquad \Rightarrow \qquad \mathbf{V} = \frac{\mathbf{I}}{j\omega C} \tag{9.37}$$

Figure 9.9

Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain.

Phasor diagram for the resistor.

Figure 9.11

Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain.

Figure 9.12 Phasor diagram for the inductor; I lags V.

Although it is equally correct to say that the inductor voltage leads the current by 90°, convention gives the current phase relative to the voltage.

384 Chapter 9 Sinusoids and Phasors

showing that the current and voltage are 90° out of phase. To be specific, the current leads the voltage by 90°. Figure 9.13 shows the voltage-current relations for the capacitor; Fig. 9.14 gives the phasor diagram. Table 9.2 summarizes the time domain and phasor domain representations of the circuit elements.

TABLE 9.2

Summary of voltage-current relationships.

Element	Time domain	Frequency domain
R	v = Ri	V = RI
L	v = L __di dt	V = jωLI
C	i = C ___ dv dt	V = ____ I jωC

Example 9.8 The voltage v = 12cos(60t + 45°) is applied to a 0.1-H inductor. Find the steady-state current through the inductor.

Solution:

For the inductor , V = jωLI, where ω = 60 rad/s and V = 12 ⧸45° V. Hence,

$$I = \frac{V}{j\omega L} = \frac{12/45^{\circ}}{j60 \times 0.1} = \frac{12/45^{\circ}}{6/90^{\circ}} = 2/45^{\circ}$$

A

Converting this to the time domain,

$$i(t) = 2\cos(60t - 45^\circ) \,\mathrm{A}$$

Practice Problem 9.8 If voltage v = 25 sin(100t − 15°) V is applied to a 50 μF capacitor, calculate the current through the capacitor.

Answer: 125 sin(100t + 75°) mA.

$$0.125 \sim$$

9.5 Impedance and Admittance

In the preceding section, we obtained the v oltage-current relations for the three passive elements as

$$V = RI, \t V = j\omega LI, \t V = \frac{I}{j\omega C}$$

(9.38)

These equations may be written in terms of the ratio of the phasor v oltage to the phasor current as

$$\frac{V}{I} = R, \qquad \frac{V}{I} = j\omega L, \qquad \frac{V}{I} = \frac{1}{j\omega C}$$

(9.39)

From these three e xpressions, we obtain Ohm’s law in phasor form for any type of element as

$$Z = \frac{V}{I} \qquad \text{or} \qquad V = ZI \tag{9.40}$$

where Z is a frequenc y-dependent quantity known as impedance, measured in ohms.

The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms (Ω).

The impedance represents the opposition that the circuit e xhibits to the flow of sinusoidal current. Although the impedance is the ratio of two phasors, it is not a phasor, because it does not correspond to a sinusoidally varying quantity.

The impedances of resistors, inductors, and capacitors can be readily obtained from Eq. (9.39). Table 9.3 summarizes their impedances. From the table we notice that ZL = jωL and ZC = −j∕ωC. Consider two extreme cases of angular frequenc y. When ω = 0 (i.e., for dc sources), ZL = 0 and ZC → ∞, confirming what we already know—that the inductor acts like a short circuit, while the capacitor acts like an open circuit. When ω → ∞ (i.e., for high frequencies), ZL → ∞ and ZC = 0, indicating that the inductor is an open circuit to high frequencies, while the capacitor is a short circuit. Figure 9.15 illustrates this.

As a complex quantity, the impedence may be e xpressed in rectangular form as

$$Z = R \pm jX \tag{9.41}$$

where R = Re Z is the resistance and X = Im Z is the reactance. The reactance, X, is just a magnitude, a positi ve value, but when used as a vector, a j is associated with inductance and a −j is associated with capacitance. Thus, impedance Z = R + jX is said to be inductive o r lagging since current lags v oltage, while impedance Z = R − jX i s capacitive or leading because current leads v oltage. The impedance, resistance, and reactance are all measured in ohms. The impedance may also be expressed in polar form as

$$\mathbf{Z} = |\mathbf{Z}| \; / \theta \tag{9.42}$$

TABLE 9.3

Impedances and admittances of passive elements.

Impedance	Admittance
Z = R	Y = __1 R
Z = jωL	Y = ____ 1 jωL
Z = ____ 1 jω C	Y = jωC
	Short circuit at dc Open circuit at high frequencies
(a)
	Open circuit at dc
(b)	Short circuit at high frequencies

Figure 9.15