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RLC circuit determine the voltage across each element

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data
    For the series RLC circuit shown in Fig. Q8, determine the voltage across each element, and draw a complete phasor diagram.

    os69eg.jpg

    2. Relevant equations


    3. The attempt at a solution

    Total Impedance:
    Z = R+Xc+Xl
    = 75 - 60j + 25j
    = 75-j35

    Z = 82.76∠-25.02 (Phasor form)

    Total Current:
    I = V/Z
    = (10∠0)/(82.76∠-25.02)
    = 0.12∠25.02

    Now my question is when finding the voltage across the resistor will it just be the magnitude of the current*resistance ie, 0.12*75 ?

    Also when finding the voltage across the inductor, is this correct ?
    V = (0.12∠25.02)*(75.06∠18.43)......................Z = 75+j25, in phasor form = 75.06∠18.43
    = 9∠43.45

    And the capacitor
    V = (0.12∠25.02)*(96.05∠-38.66)......................Z = 75-j60, in phasor form = 96.05∠-38.66
    = 11.53∠-13.64

    Thanks
     
  2. jcsd
  3. Apr 6, 2015 #2

    mfb

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    Staff: Mentor

    Sure.

    If you add all values you should get the 10 V of the source back. Does that happen?
     
  4. Apr 6, 2015 #3
    Okay, so the voltage across the resistor would be V = 0.12*75 = 9V

    For the voltage across the capacitor, it should be
    V = I*Z
    = (0.12∠25.02)*(60∠-90)
    = 7.2∠-64.98

    Voltage across the inductor
    V = I*Z
    = (0.12∠25.02)*(25∠90)
    = 3∠115.02

    But 9+7.2+3 does not equal the voltage of the source ?
     
  5. Apr 6, 2015 #4

    mfb

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    You have to take the angles into account. Or calculate the real and imaginary part separately.
     
  6. Apr 6, 2015 #5

    gneill

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    Staff: Mentor

    That will give you the magnitude of the voltage across the resistor, but won't give you the phase of that voltage. Use the complex current for the calculation.
    The voltages are all complex values. Add appropriately.
     
  7. Apr 6, 2015 #6
    Okay,

    Voltage of resistor
    Vr = 9∠25.02

    Voltage of inductor
    Vl = 3∠115.02

    Voltage of capacitor
    Vc = 7.2∠-64.98

    So, Vt = Vr+Vl+Vc
    = 3.05-6.52j-1.27+2.72j+8.16+3.81j
    Vt = 9.94+0.01j

    |Vt| = 9.94V
     
  8. Apr 6, 2015 #7

    gneill

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    Staff: Mentor

    Looks good.

    Keep a few extra digits in intermediate values in order to prevent truncation and roundoff errors from creeping into final values. Round results for presentation to the required sig figs after you're done calculating.
     
  9. Apr 6, 2015 #8
    Ok, thanks guys

    One last question, if they where all in parallel then the total impedance would be,
    1/Z = 1/R + 1/Xc + 1/Xl ?
     
  10. Apr 6, 2015 #9

    gneill

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    Staff: Mentor

    Yes.
     
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