Question

An LCR circuit contains resistance of $110\Omega$ and a supply of $220V$ at $300\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by ${45}^0$. If on the other hand, only inductor is removed the current leads by ${45}^0$ with the applied voltage. The $rms$ current flowing in the circuit will be:

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

Answer 1

The correct option is B

$2A$

Step1: Given data.

Resistance, $R=110\Omega$

Supply voltage, $V=220V$

Angular frequency, ${\omega }_r=300\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Current lags behind the voltage, $\theta ={45}^0$

Current leads the voltage by removing inductor, $\theta ={45}^0$

Step2: Find the $rms$ current flowing in the circuit.

Since we know that when $L$ and $C$ are connected with $R$ in series the circuit will come in resonance.

Therefore,

$I_{rms}=\frac{V_{rms}}{R}$

Where $I_{rms}$ and $V_{rms}$ is $rms$ current and voltage through the circuit.

$I_{rms}=\frac{220}{110}$

$I_{rms}=2A$

Hence, option B is correct.