Q: Find the maximum value of current when a coil of inductance 2 H is connected to 150 V, 50 cycles / sec supply.
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Sol: Here L = 2 H, Erms = 150 V, f = 50 Hz
XL = 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