Previous Year NEET Question- Equivalent Circuits

• NEET 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