JEE Physics 2023 Question with Solution

Given: $L = 0.5 \, \text{mH}$, $C = 20 \, \mu\text{F}$, $R = 20 \, \Omega$, and source voltage $V = 220 \, \text{V}$.

Find: The value for which the current amplitude is written as $\sqrt{K}$, so the required answer is the number inside the square root.

When current is in phase with emf in a series RLC circuit, the circuit is at resonance. Therefore,

and the impedance becomes purely resistive.

So,

The solution gives the current as

Hence the current amplitude is

Therefore, the required value is $242$.

Given: A series RLC circuit with current in phase with emf.

Find: The number $K$ such that current amplitude is $\sqrt{K}$ A.

For an AC circuit, the reactances are

and

Since current is in phase with the emf,

Therefore the net reactance is zero, so the impedance is only resistance:

Then the current value obtained from the source voltage is

Using the extracted solution conclusion, the amplitude is written as

Thus, $K = 242$.