Question

An LCR circuit contains resistance of 110 $\Omega$ and a supply of 220 V at 300 rad/s angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 45°. If on the other hand, only inductor is removed the current leads by 45° with the applied voltage. The rms current flowing in the circuit will be:

• A. 1 A
• B. 1.5 A
• C. 2 A
• D. 2.5 A

Hint:

In an LCR series circuit, when the inductive reactance ($X_L$) equals the capacitive reactance ($X_C$), the circuit is at resonance. At resonance, the impedance is at its minimum value and is equal to the resistance ($Z=R$), and the current is maximum.

Answer 1

The Correct Option is C

Given values: $R = 110 \Omega$, $V_{rms} = 220$ V, $\omega = 300$ rad/s.
Case 1: Capacitance is removed (RL circuit).
The phase angle $\phi$ is given by $\tan\phi = \frac{X_L}{R}$.
The current lags the voltage by 45°, so $\phi = 45^\circ$.
$\tan(45^\circ) = 1 = \frac{X_L}{R} \implies X_L = R = 110 \Omega$.
Case 2: Inductor is removed (RC circuit).
The phase angle $\phi$ is given by $\tan\phi = \frac{-X_C}{R}$.
The current leads the voltage by 45°, so $\phi = -45^\circ$.
$\tan(-45^\circ) = -1 = \frac{-X_C}{R} \implies X_C = R = 110 \Omega$.
Now, consider the full LCR circuit. The total impedance Z is given by:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Since we found $X_L = 110 \Omega$ and $X_C = 110 \Omega$, the circuit is in a state of resonance.
$Z = \sqrt{(110)^2 + (110 - 110)^2} = \sqrt{(110)^2} = 110 \Omega$.
The rms current flowing in the circuit is calculated using Ohm's law for AC circuits:
$I_{rms} = \frac{V_{rms}}{Z} = \frac{220 \text{ V}}{110 \Omega} = 2$ A.