Power is associated by many people with electricity. Knowing that power is the rate of energy use or energy conversion, what is the expression for electric power? Power transmission lines might come to mind. We also think of lightbulbs in terms of their power ratings in watts. Let us compare a 25-W bulb with a 60-W bulb. (See Figure 1(a).) Since both operate on the same voltage, the 60-W bulb must draw more current to have a greater power rating. Thus the 60-W bulb’s resistance must be lower than that of a 25-W bulb. If we increase voltage, we also increase power. For example, when a 25-W bulb that is designed to operate on 120 V is connected to 240 V, it briefly glows very brightly and then burns out. Precisely how are voltage, current, and resistance related to electric power?
Electric energy depends on both the voltage involved and the charge moved. This is expressed most simply as PE=qVPE=qV size 12{"PE"= ital "qV"} {}, where qq size 12{q} {} is the charge moved and VV size 12{V} {} is the voltage (or more precisely, the potential difference the charge moves through). Power is the rate at which energy is moved, and so electric power is
P=PEt=qVt.P=PEt=qVt. size 12{P = { { ital "PE"} over {t} } = { { ital "qV"} over {t} } "."} {}
(1)Recognizing that current is I=q/tI=q/t size 12{I = q/t} {} (note that Δt=tΔt=t size 12{Δt=t} {} here), the expression for power becomes
P=IV.P=IV. size 12{P = ital "IV."} {}
(2)Electric power (PP size 12{P} {} ) is simply the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Thus, 1 A ⋅V= 1 W1 A ⋅V= 1 W size 12{"1 A " cdot V=" 1 W"} {}. For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. These outlets may be rated at 20 A, so that the circuit can deliver a maximum power
P=IV=(20 A)(12 V)=240 WP=IV=(20 A)(12 V)=240 W. In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes (
1 kA ⋅V= 1 kW1 kA ⋅V= 1 kW size 12{"1 kA " cdot V=" 1 kW"} {}).
To see the relationship of power to resistance, we combine Ohm’s law with P=IVP=IV size 12{P = ital "IV"} {}. Substituting I=V/RI=V/R size 12{I = ital "V/R"} {} gives P=(V/R)V=V2/RP=(V/R)V=V2/R size 12{P = \( V/R \) V=V rSup { size 8{2} } R} {}. Similarly, substituting V=IRV=IR size 12{V= ital "IR"} {} gives P=I(IR)=I2RP=I(IR)=I2R size 12{P =I \( ital "IR" \) = I rSup { size 8{2} } R} {}. Three expressions for electric power are listed together here for convenience:
P
=
IV
P
=
IV
size 12{P = ital "IV"} {}
(3)
P
=
V
2
R
P
=
V
2
R
size 12{P = { {V rSup { size 8{2} } } over {R} } } {}
(4)P=I2R.P=I2R. size 12{P = I rSup { size 8{2} } R"."} {}
(5)Note that the first equation is always valid, whereas the other two can be used only for resistors. In a simple circuit, with one voltage source and a single resistor, the power supplied by the voltage source and that dissipated by the resistor are identical. (In more complicated circuits, PP size 12{P} {} can be the power dissipated by a single device and not the total power in the circuit.)
Different insights can be gained from the three different expressions for electric power. For example, P=V2/RP=V2/R size 12{P = V rSup { size 8{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=V2/RP=V2/R size 12{P = V rSup { size 8{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.
(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=IVP=IV size 12{P = ital "IV"} {} to find the power. For the cold headlight, we know the voltage and resistance, so we can use P=V2/RP=V2/R size 12{P = V rSup { size 8{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.P=IV=(2.50 A)(12.0 V)= 30.0 W. size 12{P = ital "IV" = \( 2 "." "50 A" \) \( "12" "." "0 V" \) =" 30" "." "0 W."} {}
(6)The cold resistance was 0.350Ω0.350Ω size 12{0 "." "350" %OMEGA } {}, and so the power it uses when first switched on is
P=V2R=(12.0 V)20.350Ω= 411 W.P=V2R=(12.0 V)20.350Ω= 411 W. size 12{P = { {V rSup { size 8{2} } } over {R} } = { { \( "12" "." "0 V" \) rSup { size 8{2} } } over {0 "." "350" %OMEGA } } =" 411 W."} {}
(7)
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=I2RP=I2R size 12{P = I rSup { size 8{2} } R} {}, and enter known values, obtaining
I=PR=411 W0.350 Ω= 34.3 A.I=PR=411 W0.350 Ω= 34.3 A. size 12{I = sqrt { { {P} over {R} } } = sqrt { { {"411 W"} over {0 "." "350 " %OMEGA } } } =" 34" "." "3 A."} {}
(8)
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.
