Figure 11.1

← Back to Fundamentals of Electric Circuits Overview We apply the trigonometric identity

$$\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)] \tag{11.4}$$

and express Eq. (11.3) as

$$p(t) = \frac{1}{2}V_m I_m \cos(\theta_v - \theta_i) + \frac{1}{2}V_m I_m \cos(2\omega t + \theta_v + \theta_i)$$

(11.5)

This shows us that the instantaneous power has two parts. The first part is constant or time independent. Its value depends on the phase dif ference between the voltage and the current. The second part is a sinusoidal function whose frequency is 2ω, which is twice the angular frequency of the voltage or current.

A sketch of p(t) in Eq. (11.5) is shown in Fig. 11.2, where T = 2π∕ω is the period of v oltage or current. We observ e that p(t) is periodic, p(t) = p(t + T0), and has a period of T0 = T∕2, since its frequency is twice that of voltage or current. We also observe that p(t) is positive for some part of each c ycle and ne gative for the rest of the c ycle. When p(t) is positive, power is absorbed by the circuit. When p(t) is negative, power is absorbed by the source; that is, power is transferred from the circuit to the source. This is possible because of the storage elements (capacitors and inductors) in the circuit.

Figure 11.2 The instantaneous power p(t) entering a circuit.

The instantaneous power changes with time and is therefore difficult to measure. The average power is more convenient to measure. In f act, the wattmeter, the instrument for measuring power, responds to average power.

The average power, in watts, is the average of the instantaneous power over one period.

Figure 11.1

Sinusoidal source and passive linear circuit.

Thus, the average power is given by

$$P = \frac{1}{T} \int_0^T p(t) \, dt \tag{11.6}$$

Although Eq. (11.6) shows the averaging done over T, we would get the same result if we performed the integration over the actual period of p ( t) which is T0 = T∕2.

Substituting p ( t) in Eq. (11.5) into Eq. (11.6) gives

$$P = \frac{1}{T} \int_0^T \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) dt$$

• $\frac{1}{T} \int_0^T \frac{1}{2} V_m I_m \cos(2\omega t + \theta_v + \theta_i) dt$
  = $\frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) \frac{1}{T} \int_0^T dt$
• $\frac{1}{2} V_m I_m \frac{1}{T} \int_0^T \cos(2\omega t + \theta_v + \theta_i) dt$ (11.7)

The first integrand is constant, and the average of a constant is the same constant. The second integrand is a sinusoid. We know that the average of a sinusoid over its period is zero because the area under the sinusoid during a positi ve half-cycle is canceled by the area under it during the following negative half-cycle. Thus, the second term in Eq. (11.7) v an ishes and the average power becomes

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

(11.8)

Since cos( θv − θ i ) = cos( θi − θv), what is important is the diference in the phases of the voltage and current.

Note that p ( t) is time-varying while P does not depend on time. To find the instantaneous power, we must necessarily have v ( t) and i ( t) in the time domain. But we can find the average power when voltage and cur rent are expressed in the time domain, as in Eq. (11.8), or when they are expressed in the frequency domain. The phasor forms of v ( t) and i ( t) in Eq. (11.2) are V = V^m ⧸ θv and I = Im ⧸ θ ⁱ, respectively. P is calculated using Eq. (11.8) or using phasors V and I. To use phasors, we notice that

$$\frac{1}{2}\mathbf{V}\mathbf{I}^* = \frac{1}{2}V_m I_m / \theta_v - \theta_i$$

=

$$\frac{1}{2}V_m I_m [\cos(\theta_v - \theta_i) + j \sin(\theta_v - \theta_i)]$$

(11.9)

We recognize the real part of this e xpression as the a verage power P according to Eq. (11.8). Thus,

$$P = \frac{1}{2} \text{Re}[\mathbf{V} \mathbf{I}^*] = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i)$$

(11.10)

Consider two special cases of Eq. (11.10). When θ ^v = θ ⁱ, the voltage and current are in phase. This implies a purely resistive circuit or resis tive load R, and

$$P = \frac{1}{2} V_m I_m = \frac{1}{2} I_m^2 R = \frac{1}{2} |\mathbf{I}|^2 R$$

(11.11)

where ∣I∣ 2 = I × I*. Equation (11.11) shows that a purely resistive circuit absorbs power at all times. When θv − θⁱ = ±90°, we have a purely reactive circuit, and

$$P = \frac{1}{2} V_m I_m \cos 90^\circ = 0 \tag{11.12}$$

showing that a purely reacti ve circuit absorbs no a verage po wer. In summary,

A resistive load (R) absorbs power at all times, while a reactive load (L or C ) absorbs zero average power.

$$v(t) = 120 \cos(377t + 45^{\circ})$$

V and $i(t) = 10 \cos(377t - 10^{\circ})$ A

find the instantaneous power and the a verage po wer absorbed by the passive linear network of Fig. 11.1.

Solution:

The instantaneous power is given by

$$p = vi = 1200 \cos(377t + 45^{\circ}) \cos(377t - 10^{\circ})$$

Applying the trigonometric identity

$$\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$$

gives

$$p = 600[\cos(754t + 35^\circ) + \cos 55^\circ]$$

or

$$p(t) = 344.2 + 600 \cos(754t + 35^\circ)$$

W

The average power is

$$P = \frac{1}{2}V_m I_m \cos(\theta_v - \theta_i) = \frac{1}{2}120(10) \cos[45^\circ - (-10^\circ)]$$

= 600 cos 55° = 344.2 W

which is the constant part of p(t) above.

Calculate the instantaneous power and average power absorbed by the passive linear network of Fig. 11.1 if Practice Problem 11.1

v(t) = 330 cos(10t + 20°) V and i(t) = 33 sin(10t + 60°) A

Answer: 3.5 + 5.445 cos(20t − 10°) kW, 3.5 kW.

Calculate the average power absorbed by an impedance Z = 30 − j70 Ω Example 11.2 when a voltage V = 120⧸ 0**°** is applied across it.

Solution:

The current through the impedance is

ent through the impedance is
\n

$$I = \frac{V}{Z} = \frac{120/0^{\circ}}{30 - j70} = \frac{120/0^{\circ}}{76.16/-66.8^{\circ}} = 1.576/66.8^{\circ} A$$

Given that Example 11.1

The average power is

$$P = \frac{1}{2}V_m I_m \cos(\theta_v - \theta_i) = \frac{1}{2}(120)(1.576)\cos(0 - 66.8^\circ) = 37.24 \text{ W}$$

Practice Problem 11.2

A current I = 33⧸ 30**°** A flows through an impedance Z = 40⧸ −22° Ω. Find the average power delivered to the impedance.

Answer: 20.19 kW.

Example 11.3 For the circuit shown in Fig. 11.3, find the average power supplied by the source and the average power absorbed by the resistor.

Solution:

The current I is given by

1 is given by
\n

$$I = \frac{5/30^{\circ}}{4 - j2} = \frac{5/30^{\circ}}{4.472/-26.57^{\circ}} = 1.118/56.57^{\circ} A$$

The average power supplied by the voltage source is

$$P = \frac{1}{2}(5)(1.118)\cos(30^\circ - 56.57^\circ) = 2.5 \text{ W}$$

The current through the resistor is

$$I_R = I = 1.118 / 56.57^{\circ}$$

A

and the voltage across it is

$$V_R = 4I_R = 4.472/56.57^\circ$$

V

The average power absorbed by the resistor is

$$P = \frac{1}{2}(4.472)(1.118) = 2.5$$

W

which is the same as the average power supplied. Zero average power is absorbed by the capacitor.

Practice Problem 11.3

For Practice Prob. 11.3.

In the circuit of Fig. 11.4, calculate the average power absorbed by the resistor and inductor. Find the average power supplied by the voltage source.

Answer: 29.04 kW, 0 W, 29.04 kW.

Determine the average power generated by each source and the average Example 11.4 power absorbed by each passive element in the circuit of Fig. 11.5(a).

For Example 11.4.

Solution:

We apply mesh analysis as shown in Fig. 11.5(b). For mesh 1,

$$\mathbf{I}_1 = 4 \text{ A}$$

For mesh 2,

$$(j10 - j5)$$

I₂ - $j10$ I₁ + 60/ $\cancel{30^{\circ}}$ = 0, I₁ = 4 A

or

$$j5I_2 = -60/30^{\circ} + j40$$

$\Rightarrow$ $I_2 = -12/-60^{\circ} + 8$
= 10.58/79.1° A

For the voltage source, the current flowing from it is I² = 10.58⧸ 79.1**°** A and the voltage across it is 60⧸ 30**°** V, so that the average power is

$$P_5 = \frac{1}{2} (60)(10.58) \cos(30^\circ - 79.1^\circ) = 207.8 \text{ W}$$

Following the passive sign convention (see Fig. 1.8), this average power is absorbed by the source, in view of the direction of I2 and the polarity of the voltage source. That is, the circuit is delivering average power to the voltage source.

For the current source, the current through it is I¹ = 4⧸ 0**°** and the voltage across it is

$$\mathbf{V}_1 = 20\mathbf{I}_1 + j10(\mathbf{I}_1 - \mathbf{I}_2) = 80 + j10(4 - 2 - j10.39)$$

= 183.9 + j20 = 184.984/(6.21° V)

The average power supplied by the current source is

$$P_1 = \frac{1}{2} (184.984)(4) \cos(6.21^\circ - 0) = -367.8 \text{ W}$$

It is negative according to the passive sign convention, meaning that the current source is supplying power to the circuit.

For the resistor, the current through it is I¹ = 4⧸ 0**°** and the voltage across it is 20I¹ = 80⧸ 0**°**, so that the power absorbed by the resistor is

$$P_2 = \frac{1}{2} (80)(4 \pm 160 \text{ W})$$

For the capacitor, the current through it is I² = 10.58⧸ 79.1**°** and the volt age across it is −j5I² = (5⧸ −90**°)(10.58⧸ 79.1°) = 52.9⧸ 79.1°**− 90°. The average power absorbed by the capacitor is

$$P_4 = \frac{1}{2} (52.9)(10.58)\cos(-90^\circ) = 0$$

For the inductor, the current through it is I¹ − I² = 2 − j10.39 = 10.58⧸ −79.1**°. The voltage across it is j10(I¹ − I2) = 105.8⧸ −79.1°** + 90°. Hence, the average power absorbed by the inductor is

$$P_3 = \frac{1}{2} (105.8)(10.58) \text{ as } 90^\circ = 0$$

Notice that the inductor and the capacitor absorb zero average power and that the total power supplied by the current source equals the power absorbed by the resistor and the voltage source, or

P¹ + P2 + P³ + P⁴ + P⁵ = −367.8 + 160 + 0 + 0 + 207.8 = 0

indicating that power is conserved.

Calculate the average power absorbed by each of the five elements in the circuit of Fig. 11.6. Practice Problem 11.4

For Practice Prob. 11.4.

Answer: 40-V Voltage source: −60 W; j20-V Voltage source: −40 W; resistor: 100 W; others: 0 W.