# 14.4 Electric power and energy  (Page 2/7)

 Page 2 / 7

Different insights can be gained from the three different expressions for electric power. For example, $P={V}^{2}/R$ implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in $P={V}^{2}/R$ , the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning it out. If the bulb’s resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too.

## Calculating power dissipation and current: hot and cold power

(a) Consider the examples given in Ohm’s Law: Resistance and Simple Circuits and Resistance and Resistivity . Then find the power dissipated by the car headlight in these examples, both when it is hot and when it is cold. (b) What current does it draw when cold?

Strategy for (a)

For the hot headlight, we know voltage and current, so we can use $P=\text{IV}$ to find the power. For the cold headlight, we know the voltage and resistance, so we can use $P={V}^{2}/R$ to find the power.

Solution for (a)

Entering the known values of current and voltage for the hot headlight, we obtain

$P=\text{IV}=\left(2\text{.}\text{50 A}\right)\left(\text{12}\text{.}\text{0 V}\right)=\text{30}\text{.}\text{0 W.}$

The cold resistance was $0\text{.}\text{350}\phantom{\rule{0.25em}{0ex}}\Omega$ , and so the power it uses when first switched on is

$P=\frac{{V}^{2}}{R}=\frac{\left(\text{12}\text{.}\text{0 V}{\right)}^{2}}{0\text{.}\text{350}\phantom{\rule{0.25em}{0ex}}\Omega }=\text{411 W.}$

Discussion for (a)

The 30 W dissipated by the hot headlight is typical. But the 411 W when cold is surprisingly higher. The initial power quickly decreases as the bulb’s temperature increases and its resistance increases.

Strategy and Solution for (b)

The current when the bulb is cold can be found several different ways. We rearrange one of the power equations, $P={I}^{2}R$ , and enter known values, obtaining

$I=\sqrt{\frac{P}{R}}=\sqrt{\frac{\text{411 W}}{0\text{.}\text{350}\phantom{\rule{0.25em}{0ex}}\Omega }}=\text{34}\text{.}\text{3 A.}$

Discussion for (b)

The cold current is remarkably higher than the steady-state value of 2.50 A, but the current will quickly decline to that value as the bulb’s temperature increases. Most fuses and circuit breakers (used to limit the current in a circuit) are designed to tolerate very high currents briefly as a device comes on. In some cases, such as with electric motors, the current remains high for several seconds, necessitating special “slow blow” fuses.

## The cost of electricity

The more electric appliances you use and the longer they are left on, the higher your electric bill. This familiar fact is based on the relationship between energy and power. You pay for the energy used. Since $P=E/t$ , we see that

$E=\text{Pt}$

is the energy used by a device using power $P$ for a time interval $t$ . For example, the more lightbulbs burning, the greater $P$ used; the longer they are on, the greater $t$ is. The energy unit on electric bills is the kilowatt-hour ( $\text{kW}\cdot \text{h}$ ), consistent with the relationship $E=\text{Pt}$ . It is easy to estimate the cost of operating electric appliances if you have some idea of their power consumption rate in watts or kilowatts, the time they are on in hours, and the cost per kilowatt-hour for your electric utility. Kilowatt-hours, like all other specialized energy units such as food calories, can be converted to joules. You can prove to yourself that $\text{1 kW}\cdot \text{h = 3}\text{.}6×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{J}$ .

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