# Power in an AC Circuit

1. Apr 2, 2013

### pious&peevish

Not sure if this is in the right category because circuits are more of an electrical engineering-related area, but this is part of an assignment for a standard second semester calculus-based physics course (i.e. E&M), so I'll leave it here for now. Feel free to move it to another category if it does not belong here.

1. The problem statement, all variables and given/known data

A voltage Δv = (110 V) sin ωt (in SI units) is applied across a series combination of a 2.07 H inductor, a 13.8 μF capacitor, and a 11.0 Ω resistor.

a) Determine the angular frequency, ω0 at which the power delivered to the resistor is a maximum.
b) Calculate the power at that frequency.
c) Determine the two angular frequencies ω1 and ω2 at which the power delivered is one-half the maximum value. [The Q of the circuit is approximately ω0/(ω2 - ω1).] Enter the smaller one first.

2. Relevant equations

P=IV=V^2/R
Z = √((R)^2+(Xl-Xc)^2)
I(rms) = V(rms)/Z

3. The attempt at a solution

I already found the answer to a) - it was 187 rad/s. I used the impedance equation Z = √((R)^2+(Xl-Xc)^2) to find it.
I know b) is referring to the resonant frequency, but I'm not sure how to proceed. For c), I don't know what "the Q of the circuit" refers to.

Thanks in advance for any pointers!

2. Apr 2, 2013

### rude man

So is a).
It tells you what Q is right in the problem!

OK, you have found the frequency of max. power. So you can compute the current. Given that, what is the power in an R-L-C circuit? Is there power dissipated in all three components or ... ?

Then, at what two frequencies is that power halved?

3. Apr 2, 2013

### pious&peevish

Thanks. So let me see if I've got this straight: since I(rms)=V(rms)/Z, and I have V(rms) and the components required to calculate Z, I can solve for I(rms). Then would P=I(rms)*V(rms)? I know in an RLC circuit, only the resistor dissipates energy. So then would the total power be something like P=(I(rms)*V(rms))-(I(rms)^2(R))?

By the way, what I meant about Q was that I don't know what its *physical interpretation* is.

4. Apr 2, 2013

### rude man

If only the resistor dissipates power, which you got right, and you know the current thru the resistor, what is the total power dissipation in the circuit?

You will need to compute current in part d) at frequencies other than resonance, so need to get that right.

By the way, what I meant about Q was that I don't know what its *physical interpretation* is.[/QUOTE]

OK, the physical interpretaton is bandwidth, in the sense that between w2 and w1 the power is at least half the power at resonance.

5. Apr 3, 2013

### pious&peevish

So the total power would be just P=(I(rms))^2*R? For the last part, would I then divide the value for power by 2 and then use the equation for impedance (Z) again to determine the two angular frequencies?

Thanks a lot, by the way; this has been very helpful.

6. Apr 3, 2013

### rude man

You are doing very well!

7. Apr 3, 2013

### pious&peevish

OK... something odd happened. I got the correct answer for 9c, but for some strange reason the corresponding answer to 9b was marked wrong.

I did Z = sqrt[(11^2)+((187*2.07)-(1/(187*(13.8*10^-6)))^2)], and this simplified to Z = 11.007 ohms.

Then I(rms) = V(rms)/Z, and I plugged in 110 V for V (rms) and 11.007 ohms for Z to get I (rms) = 9.993 A.

So then I did P = I(rms)^2 * R, and ended up with (9.993 A)^2 * 11 ohms = 1098.47 W. Since the computer only takes 3 sig figs, I input this as 1.10*(10^3) W but it was wrong. I just don't understand how this answer was marked wrong, while the more complicated part (9c) turned out right.

8. Apr 3, 2013

### rude man

They pulled a fast one on you. They said V = 110V sin(wt). But what is the rms voltage for that?

(Fact: 110V house voltage is not 110V sin(2pi 60t)!

P.S. part c) turned out right due to cancellation effects.

9. Apr 3, 2013

### pious&peevish

So... would V (rms) be 110 V * (sqrt(2))?

10. Apr 3, 2013

### rude man

In a word - no.

11. Apr 3, 2013

### pious&peevish

12. Apr 3, 2013

### rude man

No again.
Please look up rms voltage in your textbook or on the Web.

13. Apr 3, 2013

### pious&peevish

Thanks! I got the right answer.