Question

Find the maximum value of current when an inductance of 2 henry is connected to an a.c. source of 200 volts, 50 Hz. ​

Answer 1

Answer:

0.450 or $\frac{\sqrt{2} }{\pi }$

Explanation: I max = $\frac{Vmax}{Z}$

where Z is reactance of inductor

Z= W x L

w=2 $\pi$* f

L=Inductance

Vmax = 200$\sqrt{2}$