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?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy for (a)
For the hot headlight, we know voltage and current, so we can use $P = IV$ 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
$P = IV = (2 . 50 A) (12 . 0 V) = 30 . 0 W.$
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The cold resistance was $0.350 Ω$ , and so the power it uses when first switched on is
$P = \frac{V^{2}}{R} = \frac{(12 . 0 V)^{2}}{0.350 Ω} = 411 W.$
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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

$I = \sqrt{\frac{P}{R}} = \sqrt{\frac{411 W}{0.350 Ω}} = 34 . 3 A.$
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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.

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