# Power dissipated in RLC circuit

An RLC series circuit constists of a resistor of 100 ohm, a capacitor of 10.0uF, and an inductor of 0.250 H. The circuit is connected to a power supply of 120 V and 60 Hz. What is the power dissipated in the circuit?

I got 37 W (rounding 2 S.F's)
the solutions manual has the answer as 73 W.
Is 37 watts correct?

Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
Originally posted by fish
Is 37 watts correct?
Edit: I had said your answer was wrong, but I was mistaken. I believe you are correct. ( )

Last edited:
X_L= 94.25= 94 ohms
X_C= 265.26= 265 ohms
Z= 198.10 ohms=198 ohms
I_rms= V_rms/Z = 120V/198ohms = .606 A

-
P= I_rms^2(R) = .606A^2 (100 ohms) = 36.7 W

also tried this way:
tan()=(X_L-X_C)/R = (94ohm-265ohm)/100ohm = -1.71
phase angle =tan^-1(-1.71)= -59.7 degrees
-
P =V_rms^2/Z * cos ()=120V^2/198ohm (cos(-59.7))= 37 W

another way tried:
cos()=R/Z= 100ohm/198ohm= .505
-
P=I_rms*V_rms*cos()= .606A*120V*.505=37 W

I supposed I could round 37 down some and multiply by 2 since it's an average
and get the total number. Maybe question isn't asking for avg.
or maybe use a different equation like

-
P = I_rms*V_rms= I_rms^2*R (equation was in the solution manual)
= .606A*120V=73 W
but I_rms^2*R = 37 W so that doesn't make sense (not equivalent)

Doc Al
Mentor
Sorry fish, but you were right all along. (That's what I get for trying to do things in my head.) More importantly your methods are all correct. (I'll edit my earlier post.)

Originally posted by fish
-
P = I_rms*V_rms= I_rms^2*R (equation was in the solution manual)
= .606A*120V=73 W
but I_rms^2*R = 37 W so that doesn't make sense (not equivalent)
That equation in the book is wrong; P = I_rms*V_rms*cos&theta; (they left out the phase factor).