The Series RLC Circuit — Question 8

Question 8

A shorted capacitor leaves an inductive circuit at the former resonance frequency

A series RLC circuit is operating at its original resonance frequency. The capacitor is then replaced by an ideal wire, while source frequency, amplitude, R, and L stay fixed. Define $\mathcal Q=\omega_0L/R$. Compare steady states.

1. Find the new-to-old current amplitude ratio.

2. Find the new current lag and resistor power ratio.

Question 8 - Solution

A shorted capacitor leaves an inductive circuit at the former resonance frequency

(a) Method and calculation

$$Z_{\mathrm{old}}=R,\quad Z_{\mathrm{new}}=\sqrt{R^2+(\omega_0L)^2}.$$$$\boxed{I_{\mathrm{new}}/I_{\mathrm{old}}=1/\sqrt{1+\mathcal Q^2}}.$$

(b) Interpretation and further calculation

$$\boxed{\phi_{\mathrm{new}}=\tan^{-1}\mathcal Q\quad\text{lag}}.$$$$\boxed{P_{R,\mathrm{new}}/P_{R,\mathrm{old}}=1/(1+\mathcal Q^2)}.$$

Physical check

Removing the capacitor removes reactive cancellation; it does not make the series impedance vanish.