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RLC Circuit

  1. Apr 9, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img255.imageshack.us/img255/3048/56893213.jpg [Broken]
    Find the voltage of the applied source.

    2. Relevant equations

    Kirchoff's Voltage Law

    3. The attempt at a solution

    Applying KVL.
    V= 40 + 50 +20

    Is this question that simple?
    Or am I missing something?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 9, 2010 #2
    The voltages across the capacitor and the inductor don't have the same phase, so you can't add them.
  4. Apr 9, 2010 #3
    But the values of resistance and reactance aren't given for capacitor and inductor, so how should I calculate their voltages?
    Or I just can't add them anyway? So how to reach the answer?
  5. Apr 9, 2010 #4
    What if I took them as vectors?
    VR = 40
    VC = -20i
    VL = 50i

    The magnitude of this is sqrt (402 + 302) = sqrt (1600 + 900) = sqrt(2500) = 50 volts
    Is this correct?
  6. Apr 9, 2010 #5
    You don't know what the direction of those vectors is.

    Use the equation for a voltage divider

    [tex] V_R = \frac {R} {X} [/tex]

    [tex] V_L = \frac {i \omega L } { X} [/tex]

    [tex] V_C = ... [/tex]

    where X is the total reactance of the circuit, a complex number that is a
    function of [itex] \omega [/itex] R, C and L

    now compute the magnitude of V_L/V_R and V_C/V_R

    this will give a relation between R and [itex] \omega L [/itex] and [itex] \omega C [/itex]

    wich you can substitute in the equation for V_R above.
  7. Apr 13, 2010 #6
    But the values for R and [itex] \omega L [/itex] and [itex] \omega C [/itex] are not given.
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