Q: Find the maximum value of current when a coil of inductance 2 H is connected to 150 V, 50 cycles / sec supply.

**Click to See Solution : **

Sol: Here L = 2 H, E

_{rms}= 150 V, f = 50 HzX_{L} = ω L = ( 2 π f ) L = 628 ohm

RMS value of current through the inductor,

$\large i_{rms} = \frac{E_{rms}}{X_L} = \frac{150}{628} $

= 0.24 A

$\large i_{rms} = \frac{i_0}{\sqrt{2}} $

$\large i_0 = \sqrt{2} i_{rms}$

= 1.414 × 0.24 = 0.339 A