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Power dissipated in circuit

  1. Dec 19, 2003 #1
    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?
  2. jcsd
  3. Dec 19, 2003 #2

    Doc Al

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    Staff: Mentor

    Edit: I had said your answer was wrong, but I was mistaken. I believe you are correct. ( )
    Last edited: Dec 19, 2003
  4. Dec 19, 2003 #3
    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)
  5. Dec 19, 2003 #4

    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.)

    That equation in the book is wrong; P = I_rms*V_rms*cosθ (they left out the phase factor).
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