Question

A current I=2.82cos377t (I in amps, t in seconds, and the "angle" is in radians) flows in a series LR circuit in which L = 3.85 mH and R = 1.16kΩ . What is the average power dissipation in kW?

Answer 1

I thought 377 isomega so X_L is omega*L=.231

Answer 2

Only the resistor dissipates power

Answer 3

$$
<I^2\times R>
$$

Answer 4

I tried that and 9.22kW is not the answer

Answer 5

Not \(I^2\times R\), but the time average power \(<I^2\times R>\): http://www.wolframalpha.com/input/?i=average+power&lk=4&num=1

Answer 6

What's the average of the cos function? Multiply that times 9.22 kW

Answer 7

Divide by 2pi, a typical period to find average

Answer 8

How do you get the average of cos?

Answer 9

I meant \(\cos^2x\):
$$
<I^2\times R>=\cfrac{1}{2\pi}\int_{-\pi}^{\pi}(2.82\cos(377~t))^2\times 1160~dt
$$

Answer 10

$$
\int_{-\pi}^{\pi}\cos^2(377~t)~dt=\pi
$$

Answer 11

$$
<I^2\times R>=\cfrac{2.82^2\times 1,160}{2\pi}\times \pi\\
=\cfrac{2.82^2\times 1,160}{2}\\
$$

Answer 12

Oh ok I see, thank you! I have been trying this one for an hour!

Answer 13

So, because we have a sinusoidal function, the dissipated power is actually 1/2 of that of a DC circuit, which makes sense that its less, but that it's 1/2 is surprising.

Answer 14

Oh, btw, that is why dimmer switches use variable inductors and not variable resistors -- less power dissipated.

Answer 15

Thank you :)