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Alternating source driving RLC circuit

  1. Jul 30, 2008 #1
    1. The problem statement, all variables and given/known data

    An alternating source drives a series RLC circuit with an emf amplitude of 6.00 V, at a phase angle of +30.0 degrees. When the potential difference across the capacitor reaches its maximum positive value of +5.00 V, what is the potential difference across the inductor (sign included)?

    2. Relevant equations

    I know for series RLC circuits the following equations are relevant

    [tex]I= \frac{E_m}{ \sqrt{R^2+(\omega_dL-1/\omega_dC)^2}}[/tex]

    [tex]\tan{\phi}=\frac{X_L-X_C}{R}[/tex]

    The "E" used in the above equation is supposed to be the symbol for emf

    3. The attempt at a solution

    It seems to me that there isn't enough information to make use of the above equations since there are too many unknowns. We don't know what the driving angular frequency is for one and to calculate R we need to know I etc. I believe that there is something I've missed or am not aware of. Perhaps there's a connection somewhere I'm not making.

    Please point me in the right direction.

    Thanks!
    phyz
     
  2. jcsd
  3. Jul 30, 2008 #2

    alphysicist

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    Homework Helper

    Hi phyzmatix,

    The relationship you have listed under "relevant equations" is not what you want to be thinking about here.

    Here's the thing: the equation [itex]I_m = V_m/Z[/itex] relates the maximum current and maximum voltage (between two points that have impedance Z). However, the time when the circuit voltage is at [itex]V_m[/itex] is not the time when the circuit current is at [itex]I_m[/itex] (unless the phase constant is zero, which means unless [itex]X_L - X_C =0[/itex]).

    Instead, for this problem you need to consider how currents and voltages are related at the same time. Since the [itex]R[/itex],[itex] L[/itex], and [itex]C[/itex] are in series, how are the currents through them related at any instant? Since the capacitor voltage is at a maximum, what does that tell you about the capacitor current (because you know the current voltage phase relationship for the capacitor by itself)?

    Once you answer that, you can determine the voltages across two other parts of the circuit at that time (the same way), so you'll have the voltages across everything except the inductor at that particular time. What do you get? Do you then see how to get the inductor voltage?
     
  4. Jul 31, 2008 #3
    Hi alphysicist!

    Thanks for your reply. :smile:

    So, let's see if I'm getting this right. When the capacitor voltage is at a maximum, the capacitor current is zero (right?) and since [tex]R[/tex],[tex]L[/tex] and [tex]C[/tex] are in series, then the current through [tex]R[/tex] and [tex]L[/tex] will also be zero?
     
  5. Jul 31, 2008 #4

    alphysicist

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    Homework Helper

    Yes, and also the current through the AC voltage source will be zero. So you'll need to find the voltage across the resistor and voltage source at that moment in time, and then you can apply Kirchoff's loop rule. Just be sure to keep track of the signs. (For example, for the voltage source, what are the general time dependent functions for the current and voltage? Once you write those down with the given information, you'll have two equations in two unknowns.)

    I won't be able to respond any more for the next several days; I'm going out of town in just a few hours from now. (So if you do have any questions I'm not ignoring you or anything.)
     
  6. Jul 31, 2008 #5
    You'll probably only get this once you're back now, but thank you for your help, it made a huge difference!

    Hope your trip is/was very enjoyable :smile:
     
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