# Average power dissipated by a resistor on AC current

So let's say we have an AC current of 120 V at 60 Hz. Then i's waveform would be $$f(t) = 120 \sqrt{2} \cos(2 \pi 60 t)$$ Or rather the amplitude times $\sqrt{2}$ times $\cos(2 \pi \times\text{frequency}\times t)$, right?

And if so, then the "real voltage" would be 120 V, and the spikes would be at $120\sqrt{2}$.

And so evidently the peak power dissipated would be the peak voltage divided by the resistance.

But what about the average power? It said to integrate the power over one cycle of the waveform. I tried both $$\int_{0}^{1/60} \frac{f(t)^2}{R} dt$$ and from 0 to 1/120 but I've gotten the wrong answers. What am I doing wrong?

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You should show what you are doing if you want to know what you are doing wrong. To do these integrals with no effort, note that the integral of sin squared (or cos squared) is always found by replacing the squared trigonometric function by it's average value, which is 1/2. Then you can't get confused, since there is nothing complicated to do. – Ron Maimon Sep 14 '12 at 20:18

The peak power is the peak voltage squared divided by the resistance. For average power, you must take the time average of the squared voltage and divide by the resistance.

Step by step:

$p_R(t) = v^2_R(t) / R$

$v^2_R(t) = (120 \sqrt{2})^2 \cdot \frac{1}{2}[1 + \cos(2 \pi 120 t)] = (120 )^2 \cdot [1 + \cos(2 \pi 120 t)]$

$\bar p_R = \dfrac{1}{T} \int_0^T p_R(t) dt$

$T = \dfrac{1}{60}$

Since the integral of a sinusoid over a period is zero, we have:

$\bar p_R = 60 \ \dfrac{(120)^2}{R} \int_0^{\frac{1}{60}} dt = \dfrac{(120)^2}{R}$

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$$\int_{0}^{\tau} \frac{V(t)^2}{R} dt$$

is the energy dissipated in $\tau$ seconds, so the average power is:

$$W = \frac{\int_{0}^{\tau} \frac{V(t)^2}{R} dt}{\tau}$$

or just integrate for a second and don't bother dividing by 1. Either way you should get the correct answer. If it still won't work have a look at this Hyperphysics article.

Note that the power calculation is only correct when the circuit is purely resistive i.e. there are no capacitors or inductors present.

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The factor of square root of two should not be present. I would also avoid using f to represent the voltage waveform, as that symbol is commonly reserved for frequency in Hertz. When we say that we have AC of 120 V at 60 Hz, the voltage alluded to is called the peak voltage which is just another name for the amplitude. The average power over one cycle is then (1/2)120^2/R where the factor of one-half is the result of integrating cos(2 pi frequency t) over a complete cycle. You are correct that a complete cycle is 1/60 s.

The root-mean-square voltage is the peak voltage divided by sqrt(2). This is a contrivance that allows one to use the familiar power relationship P = V^2/R directly without resorting to integration over cycles. It's my guess that you were confusing this with the peak voltage in your problem statement.

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This is incorrect information. The peak voltage of 120VAC is not 120V, it is about 170V. See www.compwest.com/Library/hudhipotpaper.pdf for an O'scope trace to confirm. – Alfred Centauri Sep 5 '12 at 22:11