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Power dissipated in RLC circuit

  • Thread starter fish
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49
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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?
 

Answers and Replies

Doc Al
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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:
49
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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
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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*cosθ (they left out the phase factor).
 

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