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Parallel circuit with resonant branch

  1. Oct 29, 2008 #1
    Here's a question my mate asked me to help out with, but I'm a bit stuck, any help would be good. Apologies for not being able to show the circuit (don't know how) so I'll explain it.

    A parallel circuit has a supply of 1 Vrms and has 2 branches. The first branch has a single capacitor of 2.2 microFarads. The second branch has a resistor of 5 Ohms, and inductor of 0.2 milliHenrys and a capacitor of 30 nanoFarads.

    The first part was to work out the sesonant frequency of the branch with the 3 components, we both got 2054.7 Hz.

    The second part was to find the magnitude the magnitude and the phase angle of the current drawn for the entire of the circuit at the frequency of part a (i.e 2054.7Hz). So we worked out that the impedance of the lone capacitor was 35.2 Ohms.

    From this point we are at a bit confused. Are we right up to this point and where do we go from here?

    Cheers.
     
  2. jcsd
  3. Oct 29, 2008 #2
    How did you get the resonant frequency of the 3 component branch?
    The equation for resonance, if i'm not mistaken is 1/sqrt(LC)

    So given C = 30nF and L = .2mH, resonant frequency should be 408 krads, or about 65 kHz.

    Your approach is correct for the capacitor impedance. Remember that a capacitor is purely reactive.

    Resonance implies that the L and C cancels out at that frequency, so on the second branch, you only have the resistor left. So you're left with a simple RC parallel circuit connected to a voltage source.
     
  4. Oct 29, 2008 #3
  5. Oct 30, 2008 #4
    Am I wrong in thinking the formula for resonance was 1 over 2 x pi x sqrt(LC)
     
  6. Oct 30, 2008 #5
    2pi if you need the freq in Hz.

    If you did that, you still should get around 64.974 kHz
     
  7. Oct 30, 2008 #6
    OK sorry, I got the figures wrong. The RLC branch is made of a 5 Ohm resistor, a 30nF capacitor and a 0.2H inductor. My bad,

    Can anyone shed any more light on this now.....?


    Thanks.
     
  8. Oct 30, 2008 #7
  9. Oct 30, 2008 #8
    Right got that, so a parallel circuit, the resonant branch impedance is 5 Ohm at angle 0, the lone capacitor branch is 35.2 angle -90 deg.

    Thats product/sum = 4.9 at angle -8.1 or 4.9+j0.7.

    With a source of 1Vrms, how do you work out the current drawn, what is with the "rms" part?

    Cheers, almost there.
     
  10. Oct 31, 2008 #9
    Z = 4.9 - j0.7 Still capacitive.. so minus.


    Phase angle is about 8.1 deg. tan-1 0.7/4.9 Xc/R

    You have to work out the modulus of the impedance . Sq rt of the sums of the squares of R and Xc and then I = V/Z
     
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