For Practice Prob. 11.8. 11.5 Apparent Power and Power Factor

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For Practice Prob. 11.8. 11.5 Apparent Power and Power Factor

In Section 11.2 we saw that if the voltage and current at the terminals of a circuit are

$$v(t) = V_m \cos(\omega t + \theta_v)$$

and $i(t) = I_m \cos(\omega t + \theta_i)$ (11.32)

or, in phasor form, V = Vm^⧸θv and I = Im⧸θi , the average power is

$$P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i)$$

(11.33)

In Section 11.4, we saw that

$$P = V_{\text{rms}} I_{\text{rms}} \cos(\theta_{\nu} - \theta_{i}) = S \cos(\theta_{\nu} - \theta_{i})$$

(11.34)

We have added a new term to the equation:

$$S = V_{\rm rms} I_{\rm rms}$$

(11.35)

The average power is a product of two terms. The product VrmsIrms is known as the apparent power S . The factor cos( θv − θi) is called the power factor (pf).

The apparent power (in VA) is the product of the rms values of voltage and current.

The apparent power is so called because it seems apparent that the power should be the v oltage-current product, by analogy with dc resisti ve circuits. It is measured in volt-amperes or VA to distinguish it from the average or real power, which is measured in watts. The power factor is dimensionless, since it is the ratio of the average power to the apparent power,

$$pf = \frac{P}{S} = \cos(\theta_v - \theta_i)$$

(11.36)

The angle θv − θi is called the power factor angle, because it is the angle whose cosine is the power factor. The power factor angle is equal to the angle of the load impedance if V is the voltage across the load and I is the current through it. This is evident from the fact that

$$Z = \frac{V}{I} = \frac{V_m/\theta_v}{I_m/\theta_i} = \frac{V_m}{I_m}/\theta_v - \theta_i$$

(11.37)

Alternatively, since

$$\mathbf{V}_{\rm rms} = \frac{\mathbf{V}}{\sqrt{2}} = V_{\rm rms} \underline{\theta_v} \tag{11.38a}$$

and

$$\mathbf{I}_{\rm rms} = \frac{\mathbf{I}}{\sqrt{2}} = I_{\rm rms} / \theta_i \tag{11.38b}$$

the impedance is

$$Z = \frac{V}{I} = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{V_{\text{rms}}}{I_{\text{rms}}} \underbrace{\beta_{\nu} - \theta_{i}} \tag{11.39}$$

The power factor is the cosine of the phase difference between voltage and current. It is also the cosine of the angle of the load impedance.

From Eq. (11.36), the power factor may be seen as that f actor by which the apparent power must be multiplied to obtain the real or average power. The value of pf ranges between zero and unity . For a purely resisti ve load, the voltage and current are in phase, so that θv − θi = 0 and pf = 1. This implies that the apparent po wer is equal to the a verage power. For a purely reactive load, θv − θi = ±90° and pf = 0. In this case the a verage power is zero. In between these tw o extreme cases, pf is said to be leading or lagging. Leading power factor means that current leads v oltage, which implies a capaciti ve load. Lagging po wer factor means that current lags voltage, implying an inductive load. Power factor affects the From Eq. (11.36), the power factor may also be regarded as the ratio of the real power dissipated in the load to the apparent power of the load.

electric bills consumers pay the electric utility companies, as we will see in Section 11.9.2.

Example 11.9 A series-connected load dra ws a current i(t) = 4 cos(100 πt + 10°) A when the applied v oltage is v(t) = 120 cos(100πt − 20°) V. Find the apparent power and the power factor of the load. Determine the element values that form the series-connected load.

Solution:

The apparent power is

$$S = V_{\text{rms}} I_{\text{rms}} = \frac{120}{\sqrt{2}} \frac{4}{\sqrt{2}} = 240 \text{ VA}$$

The power factor is

$$pf = \cos(\theta_v - \theta_i) = \cos(-20^\circ - 10^\circ) = 0.866 \quad \text{(leading)}$$

The pf is leading because the current leads the voltage. The pf may also be obtained from the load impedance.

$$\mathbf{Z} = \frac{\mathbf{V}}{\mathbf{I}} = \frac{120/-20^{\circ}}{4/10^{\circ}} = 30/-30^{\circ} = 25.98 - j15 \ \Omega$$

\n

$$\text{pf} = \cos(-30^{\circ}) = 0.866 \qquad \text{(leading)}$$

The load impedance Z can be modeled by a 25.98-Ω resistor in series with a capacitor with

$$X_C = -15 = -\frac{1}{\omega C}$$

or

$$C = \frac{1}{15\omega} = \frac{1}{15 \times 100\pi} = 212.2 \,\mu\text{F}$$

Practice Problem 11.9	Obtain the power factor and the apparent power of a load whose
	impedance is Z = 60 + j40 Ω when the applied voltage is v(t) = 155.56 cos(377t + 10°) V.

Answer: 0.8321 lagging, 167.69⧸ 33.69° VA.

Solution:

The total impedance is

$$\mathbf{Z} = 6 + 4 \left( (-j2) \right) = 6 + \frac{-j2 \times 4}{4 - j2} = 6.8 - j1.6 = 7 \underline{\text{/} -13.24^{\circ}} \,\Omega$$

The power factor is

$$pf = \cos(-13.24) = 0.9734 \text{ (leading)}$$

since the impedance is capacitive. The rms value of the current is

$$\mathbf{I}_{\rm rms} = \frac{\mathbf{V}_{\rm rms}}{\mathbf{Z}} = \frac{30/0^{\circ}}{7/-13.24^{\circ}} = 4.286/13.24^{\circ} \text{A}$$

The average power supplied by the source is

P = VrmsIrmspf = (30)(4.286)0.9734 = 125 W

or

$$P = I_{\text{rms}}^2 R = (4.286)^2 (6.8) = 125 \text{ W}$$

where R is the resistive part of Z.

Calculate the power factor of the entire circuit of Fig. 11.19 as seen by the source. What is the average power supplied by the source?

Answer: 0.936 lagging, 2.008 kW.