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  • Calculate the power dissipated by a resistor and power supplied by a power supply.
  • Calculate the cost of electricity under various circumstances.

Power in electric circuits

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 [link] (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?

Part a has two images. The image on the left is a photograph of a twenty five watt incandescent bulb emitting a dim, yellowish white color. The image on the right is a photograph of a sixty watt incandescent bulb emitting a brighter white light. Part b is a single photograph of a compact fluorescent lightbulb glowing in bright pure white color.
(a) Which of these lightbulbs, the 25-W bulb (upper left) or the 60-W bulb (upper right), has the higher resistance? Which draws more current? Which uses the most energy? Can you tell from the color that the 25-W filament is cooler? Is the brighter bulb a different color and if so why? (credits: Dickbauch, Wikimedia Commons; Greg Westfall, Flickr) (b) This compact fluorescent light (CFL) puts out the same intensity of light as the 60-W bulb, but at 1/4 to 1/10 the input power. (credit: dbgg1979, Flickr)

Electric energy depends on both the voltage involved and the charge moved. This is expressed most simply as PE = qV size 12{"PE"= ital "qV"} {} , where q size 12{q} {} is the charge moved and V 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 = PE t = qV t . size 12{P = { { ital "PE"} over {t} } = { { ital "qV"} over {t} } "."} {}

Recognizing that current is I = q / t size 12{I = q/t} {} (note that Δ t = t size 12{Δt=t} {} here), the expression for power becomes

P = IV. size 12{P = ital "IV."} {}

Electric power ( P 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 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 W . In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes ( 1 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 = IV size 12{P = ital "IV"} {} . Substituting I = V/R size 12{I = ital "V/R"} {} gives P = ( V / R ) V = V 2 / R size 12{P = \( V/R \) V=V rSup { size 8{2} } R} {} . Similarly, substituting V = IR size 12{V= ital "IR"} {} gives P = I ( IR ) = I 2 R 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 size 12{P = ital "IV"} {}
P = V 2 R size 12{P = { {V rSup { size 8{2} } } over {R} } } {}
P = I 2 R . size 12{P = I rSup { size 8{2} } R"."} {}

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, P size 12{P} {} can be the power dissipated by a single device and not the total power in the circuit.)

Questions & Answers

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º20.0º with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate an
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Practice Key Terms 1

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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