# 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?

• 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. Commented Sep 14, 2012 at 20:18

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

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.

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:

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

In fact you can get the average power with the peak power. I mean the average power is $$1/2$$ of the peak power. But this only can be used on symmetric graphics.