- JEE Mains 2019: A coil of inductance 0.5 H and resistance 100 $\Omega$ is connected to a 220 V, 50 Hz AC supply. The maximum current in the coil is:
(A) 1.1 A (B) 2.2 A (C) 3.3 A (D) 4.4 A
The answer is (A).
The maximum current in the coil is given by:
$I_{max} = \frac{V_{max}}{\sqrt{R^2 + (2\pi f L)^2}}$
Substituting the values, we get:
$I_{max} = \frac{220}{\sqrt{100^2 + (2\pi * 50 * 0.5)^2}} = 1.1 A$
- JEE Mains 2018: A series LCR circuit with L = 100 mH, C = 10 $\mu$F and R = 100 $\Omega$ is connected to a 220
