1.5 Power and Energy

← Back to Fundamentals of Electric Circuits Overview

1.5 Power and Energy

Although current and v oltage are the tw o basic v ariables in an electric circuit, the y are not suf ficient by themselves. F or practical purposes, we need to kno w how much power an electric de vice can handle. We all know from e xperience that a 100-w att bulb gives more light than a 60-watt bulb. We also know that when we pay our bills to the electric utility companies, we are paying for the electric energy consumed over a certain period of time. Thus, power and energy calculations are important in circuit analysis.

To relate power and ener gy to v oltage and current, we recall from physics that:

Power is the time rate of expending or absorbing energy, measured in watts (W).

We write this relationship as

$$p \triangleq \frac{dw}{dt} \tag{1.5}$$

where p is power in watts (W), w is energy in joules (J), and t is time in seconds (s). From Eqs. (1.1), (1.3), and (1.5), it follows that

$$p = \frac{dw}{dt} = \frac{dw}{dq} \cdot \frac{dq}{dt} = vi \tag{1.6}$$

or

$$p = vi \tag{1.7}$$

The power p in Eq. (1.7) is a time-varying quantity and is called the instantaneous power. Thus, the power absorbed or supplied by an element is the product of the voltage across the element and the current through it. If the power has a + sign, power is being delivered to or absorbed by the element. If, on the other hand, the power has a − sign, power is being supplied by the element. But how do we know when the power has a negative or a positive sign?

Current direction and voltage polarity play a major role in determining the sign of po wer. It is therefore important that we pay attention to the relationship between current i and voltage v in Fig. 1.8(a). The voltage polarity and current direction must conform with those sho wn in Fig. 1.8(a) in order for the power to have a positive sign. This is known as the passive sign convention. By the passive sign convention, current enters through the positive polarity of the voltage. In this case, p = +vi or vi > 0 implies that the element is absorbing po wer. However, if p = −vi or vi < 0, as in Fig. 1.8(b), the element is releasing or supplying power.

Passive sign convention is satisfied when the current enters through the positive terminal of an element and p = +vi. If the current enters through the negative terminal, p = −vi.

Unless otherwise stated, we will follow the passive sign convention throughout this text. For example, the element in both circuits of Fig. 1.9 has an absorbing power of +12 W because a positive current enters the positive terminal in both cases. In Fig. 1.10, ho wever, the element is supplying power of +12 W because a positive current enters the negative terminal. Of course, an absorbing power of −12 W is equivalent to a supplying power of +12 W. In general,

+Power absorbed = −Power supplied

In fact, the law of conservation of energy must be obeyed in any electric circuit. For this reason, the algebraic sum of po wer in a circuit, at any instant of time, must be zero:

$$\sum p = 0 \tag{1.8}$$

This again confirms the fact that the total po wer supplied to the circuit must balance the total power absorbed.

From Eq. (1.6), the energy absorbed or supplied by an element from time t0 to time t is

$$w = \int_{t_0}^{t} p \, dt = \int_{t_0}^{t} v i \, dt \tag{1.9}$$

Figure 1.8

Reference polarities for power using the passive sign convention: (a) absorbing power, (b) supplying power.

When the voltage and current directions conform to Fig. 1.8(b), we have the active sign convention and ^p = +vi.

Figure 1.9

Two cases of an element with an absorbing power of 12 W: (a) p = 4 × 3 = 12 W, (b) p = 4 × 3 = 12 W.

Figure 1.10 Two cases of an element with a supplying power of 12 W: (a) p = −4 × 3 = −12 W, (b) p = −4 × 3 = −12 W.

Energy is the capacity to do work, measured in joules (J).

The electric po wer utility companies measure ener gy in w att-hours (Wh), where

$$1 \text{ Wh} = 3,600 \text{ J}$$

Example 1.4 An energy source forces a constant current of 2 A for 10 s to flow through a light bulb. If 2.3 kJ is gi ven off in the form of light and heat ener gy, calculate the voltage drop across the bulb.

Solution:

The total charge is

$$\Delta q = i \Delta t = 2 \times 10 = 20 \text{ C}$$

The voltage drop is

$$v = \frac{\Delta w}{\Delta q} = \frac{2.3 \times 10^3}{20} = 115 \text{ V}$$

To move charge q from point b to point a requires 25 J. Find the volt age drop vab (the voltage at a positive with respect to b) if: (a) q = 5 C, (b) q = −10 C. Practice Problem 1.4

Answer: (a) 5 V, (b) −2.5 V.

Example 1.5 Find the power delivered to an element at t = 3 ms if the current entering its positive terminal is

i = 5 cos 60π t A

and the voltage is: (a) v = 3i, (b) v = 3 di∕dt.

Solution:

(a) The voltage is v = 3i = 15 cos 60π t; hence, the power is

$$p = vi = 75 \cos^2 60 \pi t \,\mathrm{W}$$ $$At t = 3 ms,$$

p = 75 cos2 (60π × 3 × 10⁻³ ) = 75 cos2 0.18π = 53.48 W

(b) We find the voltage and the power as

$$v = 3\frac{di}{dt} = 3(-60\pi)5 \sin 60\pi t = -900\pi \sin 60\pi t \text{ V}$$ $$p = vi = -4500\pi \sin 60\pi t \cos 60\pi t \text{ W}$$

At t = 3 ms,

p = −4500π sin 0.18π cos 0.18π W

= −14137.167 sin 32.4° cos 32.4° = −6.396 kW

Historical

1884 Exhibition In the United States, nothing promoted the future of electricity like the 1884 International Electrical Exhibition. Just imagine a world without electricity, a world illuminated by candles and gaslights, a world where the most common transportation was by walking and riding on horseback or by horse-drawn carriage. Into this world an exhibi tion was created that highlighted Thomas Edison and reflected his highly developed ability to promote his inventions and products. His exhibit featured spectacular lighting displays powered by an impressive 100-kW “Jumbo” generator.

Edward Weston’s dynamos and lamps were featured in the United States Electric Lighting Company’s display. Weston’s well known col lection of scientific instruments was also shown.

Other prominent exhibitors included Frank Sprague, Elihu Thompson, and the Brush Electric Company of Cleveland. The American Institute of Electrical Engineers (AIEE) held its first technical meeting on October 7–8 at the Franklin Institute during the exhibit. AIEE merged with the Institute of Radio Engineers (IRE) in 1964 to form the Institute of Electrical and Electronics Engineers (IEEE).

Practice Problem 1.5

Source: IEEE History Center

Find the power delivered to the element in Example 1.5 at t = 5 ms if the current remains the same but the voltage is: (a) v = 2i V,

(b)

$$v = \left(10 + 5 \int_0^t i \, dt \right) V.$$

Answer: (a) 17.27 W, (b) 29.7 W.

Example 1.6 How much energy does a 100-W electric bulb consume in two hours?

Solution:

w = pt = 100 (W) × 2 (h) × 60 (min/h) × 60 (s/min) = 720,000 J = 720 kJ

This is the same as

$$w = pt = 100$$

W $\times$ 2 h = 200 Wh

Practice Problem 1.6

A home electric heater dra ws 10 A when connected to a 115 V outlet. How much energy is consumed by the heater over a period of 6 hours?

Answer: 6.9 k watt-hours