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We are most often concerned with average power rather than its fluctuations—that 60-W light bulb in your desk lamp has an average power consumption of 60 W, for example. As illustrated in [link] , the average power P ave size 12{P rSub { size 8{"ave"} } } {} is

P ave = 1 2 I 0 V 0 . size 12{P rSub { size 8{"ave"} } = { {1} over {2} } I rSub { size 8{0} } V rSub { size 8{0} } } {}

This is evident from the graph, since the areas above and below the ( 1 / 2 ) I 0 V 0 size 12{ \( 1/2 \) I rSub { size 8{0} } V rSub { size 8{0} } } {} line are equal, but it can also be proven using trigonometric identities. Similarly, we define an average or rms current     I rms size 12{I rSub { size 8{"rms"} } } {} and average or rms voltage     V rms size 12{V rSub { size 8{"rms"} } } {} to be, respectively,

I rms = I 0 2 size 12{I rSub { size 8{"rms "} } = { {I rSub { size 8{0} } } over { sqrt {2} } } } {}

and

V rms = V 0 2 . size 12{V rSub { size 8{"rms "} } = { {V rSub { size 8{0} } } over { sqrt {2} } } } {}

where rms stands for root mean square, a particular kind of average. In general, to obtain a root mean square, the particular quantity is squared, its mean (or average) is found, and the square root is taken. This is useful for AC, since the average value is zero. Now,

P ave = I rms V rms , size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } } {}

which gives

P ave = I 0 2 V 0 2 = 1 2 I 0 V 0 , size 12{P rSub { size 8{"ave"} } = { {I rSub { size 8{0} } } over { sqrt {2} } } cdot { {V rSub { size 8{0} } } over { sqrt {2} } } = { {1} over {2} } I rSub { size 8{0} } V rSub { size 8{0} } } {}

as stated above. It is standard practice to quote I rms size 12{I rSub { size 8{"rms"} } } {} , V rms size 12{V rSub { size 8{"rms"} } } {} , and P ave size 12{P rSub { size 8{"ave"} } } {} rather than the peak values. For example, most household electricity is 120 V AC, which means that V rms size 12{V rSub { size 8{"rms"} } } {} is 120 V. The common 10-A circuit breaker will interrupt a sustained I rms size 12{I rSub { size 8{"rms"} } } {} greater than 10 A. Your 1.0-kW microwave oven consumes P ave = 1.0 kW size 12{P rSub { size 8{"ave"} } =1 "." 0`"kW"} {} , and so on. You can think of these rms and average values as the equivalent DC values for a simple resistive circuit.

To summarize, when dealing with AC, Ohm’s law and the equations for power are completely analogous to those for DC, but rms and average values are used for AC. Thus, for AC, Ohm’s law is written

I rms = V rms R . size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over {R} } } {}

The various expressions for AC power P ave size 12{P rSub { size 8{"ave"} } } {} are

P ave = I rms V rms , size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } } {}
P ave = V rms 2 R , size 12{P rSub { size 8{"ave"} } = { {V rSub { size 8{"rms"} } rSup { size 8{2} } } over {R} } } {}

and

P ave = I rms 2 R . size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } rSup { size 8{2} } R} {}

Peak voltage and power for ac

(a) What is the value of the peak voltage for 120-V AC power? (b) What is the peak power consumption rate of a 60.0-W AC light bulb?

Strategy

We are told that V rms size 12{V rSub { size 8{"rms"} } } {} is 120 V and P ave size 12{P rSub { size 8{"ave"} } } {} is 60.0 W. We can use V rms = V 0 2 size 12{V rSub { size 8{"rms "} } = { {V rSub { size 8{0} } } over { sqrt {2} } } } {} to find the peak voltage, and we can manipulate the definition of power to find the peak power from the given average power.

Solution for (a)

Solving the equation V rms = V 0 2 size 12{V rSub { size 8{"rms "} } = { {V rSub { size 8{0} } } over { sqrt {2} } } } {} for the peak voltage V 0 size 12{V rSub { size 8{0} } } {} and substituting the known value for V rms size 12{V rSub { size 8{"rms"} } } {} gives

V 0 = 2 V rms = 1 . 414 ( 120 V ) = 170 V . size 12{V rSub { size 8{0} } = sqrt {2} V rSub { size 8{"rms"} } =" 1" "." "414" \( "120"" V" \) =" 170 V"} {}

Discussion for (a)

This means that the AC voltage swings from 170 V to –170 V and back 60 times every second. An equivalent DC voltage is a constant 120 V.

Solution for (b)

Peak power is peak current times peak voltage. Thus,

P 0 = I 0 V 0 = 2 1 2 I 0 V 0 = 2 P ave . size 12{P rSub { size 8{0} } = I rSub { size 8{0} } V rSub { size 8{0} } =" 2" left ( { {1} over {2} } I rSub { size 8{0} } V rSub { size 8{0} } right ) =" 2"P rSub { size 8{"ave"} } } {}

We know the average power is 60.0 W, and so

P 0 = 2 ( 60 . 0 W ) = 120 W . size 12{P rSub { size 8{0} } =" 2" \( "60" "." "0 W" \) =" 120 W"} {}

Discussion

So the power swings from zero to 120 W one hundred twenty times per second (twice each cycle), and the power averages 60 W.

Why use ac for power distribution?

Most large power-distribution systems are AC. Moreover, the power is transmitted at much higher voltages than the 120-V AC (240 V in most parts of the world) we use in homes and on the job. Economies of scale make it cheaper to build a few very large electric power-generation plants than to build numerous small ones. This necessitates sending power long distances, and it is obviously important that energy losses en route be minimized. High voltages can be transmitted with much smaller power losses than low voltages, as we shall see. (See [link] .) For safety reasons, the voltage at the user is reduced to familiar values. The crucial factor is that it is much easier to increase and decrease AC voltages than DC, so AC is used in most large power distribution systems.

Questions & Answers

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Source:  OpenStax, College physics ii. OpenStax CNX. Nov 29, 2012 Download for free at http://legacy.cnx.org/content/col11458/1.2
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