RLC circuit phasor addition gives zero impedance

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sdusheyko
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the problem statement is attached.

i've begun by converting everything into phasor notation but I'm not quite sure where to go from here. when i try to add the C and the L phasors, i get zero. this doesn't seem right.
another problem is i have no idea what to do when i get the impedances of the components added together. do i just add that to the input voltage for the answer?

any help appreciated.
 
Attachments
  • Circuit schematic with a voltage source x(t) on the left in series with resistor R, then inductor L in parallel with capacitor C, with output voltage y(t) across the capacitor; problem text above gives R = 1, L = 10^-3, C = 10^-3 and x(t) = 2cos(ωt - π/4).
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so I've converted the circuit to it's current source equivalent ( i can't remember if that's called thevenin or norton) and am able to add the impedances.

what to do from here?

i think I'm supposed to add the component impedances to the voltage input signal impedance and plug into v=ir for the voltage. but maybe not... I'm lost like i said. am i on the right track or far from it?
 
Yes, you have to go into the frequency domain. From here, you can treat all the components like they're resistors with their respective impedances

So, yeah. For your source, you have 2 at an angle -pi/4. Impedance of Res, cap and inductor are 1, -j, and j, respectively (i.e., Zc = -j/(wC))

Now you just have to do circuit analysis from here. How can you get y(t)? Since voltages are the same in parallel, what if we combine the inductor and cap?

-j // j = ([-j]*[j])/(j-j) = undefined.

Well that's weird. I'm not sure what that means physically. I was going to combine those two components and then use voltage division across the equivalent impedance there. Ask your teacher about this. Maybe try to do a node equation?
 
What you're finding is that when ω=1000, the LC combination has infinite impedance because that's the resonance frequency of the circuit. No current can flow, so the entire input voltage appears at the output.