In a series LCR circuit, the inductance, capacitance and resistance are L = 100 mH, C = 100 \mu F and R = 10 \Omega respectively. They are connected to an AC source of voltage 220 V and frequency of 50 Hz. The approximate value of current in the circuit will be \underline\hspace2cm A.

Alternating Current PYQ — JEE Main 2022 · 25 Jun · Shift 2 · Q72 with Solution

View written solutionFree

Correct answer: 22

Given data

$L = 100\text{ mH} = 0.1\text{ H}$ $C = 100\,\mu\text{F} = 100 \times 10^{-6}\text{ F} = 10^{-4}\text{ F}$ $R = 10\,\Omega$ $V = 220\text{ V}$ $f = 50\text{ Hz}$
Angular frequency

$\omega = 2\pi f = 2\pi(50) = 100\pi \approx 314\,\text{rad/s}$
Inductive reactance

$X_L = \omega L = 314 \times 0.1 = 31.4\,\Omega$
Capacitive reactance

$X_C = \frac{1}{\omega C} = \frac{1}{314 \times 10^{-4}} = \frac{1}{0.0314} \approx 31.8\,\Omega$
Net reactance

$X = X_L - X_C = 31.4 - 31.8 = -0.4\,\Omega$

This is very small, so the circuit is approximately at resonance.
Impedance of series LCR circuit

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

$Z = \sqrt{10^2 + (-0.4)^2} = \sqrt{100 + 0.16} = \sqrt{100.16} \approx 10.0\,\Omega$
Current in the circuit

$I = \frac{V}{Z} = \frac{220}{10.0} \approx 22\text{ A}$
Final answer

$\boxed{22}$

The approximate current is $22\,$ A.