- JEE Mains 2018, Question 26:
A series LCR circuit is connected to an ac source. At resonance, the current amplitude is $I_{0}$ and the average power delivered by the source is $P_{0}$. The frequency of the source is now changed so that the current amplitude is reduced to $I_{0}/2$. The average power delivered by the source now is:
The answer is: $P_{0}/4$
The average power delivered by a series LCR circuit at resonance is given by:
$$P = \frac{1}{2}I_{0}^{2}R$$
where $I_{0}$ is the current amplitude and $R$ is the resistance of the circuit.
When the frequency of the source is changed, the current amplitude is reduced to $I_{0}/2$. The average power delivered by the source now is:
$$P = \frac{1}{2}\left(\frac{I_{0}}{2}\right)^{2}R = \frac{1}{4}I_{0}^{2}R = \frac{P_{0}}{4
