Computing the voltage across a parallel RLC circuit at resonance?

by Micrified   Last Updated October 09, 2019 17:25 PM


For the circuit illustrated below, I am trying to compute the sinusoidal voltage across resistor \$R_{1}$ when the parallel LC component is in resonance:

I setup my problem with the following information:

  1. $$V = 8\times sin(\omega t)$$

  2. The voltage across \$R_{1}$ is a voltage divider. I just need to define it in terms of impedances. I do that below:

$$V_{R} = V \times \frac{Z_{R}}{Z_{R} + (Z_{C} + Z_{L})}$$

The Circuit

enter image description here

So I know that at resonance, angular frequency \$\omega_{0} = \frac{1}{\sqrt{LC}}$. This should make the parallel LC component of the circuit get a coefficient of zero:

$$V_{R} = V \times \frac{R}{R + (\frac{1}{j(\omega C - \frac{1}{\omega L})})} = V \times \frac{R}{R - j(\frac{1}{(\omega C - \frac{1}{\omega L})})} = V \times \frac{R}{R - j(\frac{1}{0})}$$

This makes my answer undefined though. However, I also know that if there is no impedance, than the voltage is simply determined by the resistor. In which case won't the voltage across the resistor simply be the same as the voltage produced by the source?

I'm trying to reconcile these two explanations I have been given but I am confused as to why I cannot obtain something like $V+{R} = V$, or if that even is the correct conclusion to make when the circuit is at a resonant frequency.

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