Question

A pure inductor of 25 mH is connected to an ac source of 220 V Given the frequency of the sour ce as 50 Hz, the rms current in the circuit is

• A. 7 A
• B. 14 A
• C. 28 A
• D. 42 A

Answer 1

Answer: (C)

Here, $L=25mH,v=50Hz,V_{rms}=220V$
The inductive reactance is
$X_L=2\pi vL=2\times \frac{22}{7}\times 50\times 25\times 10^{-3}\Omega$
The rms current in the circuit is
$I_{rms}=\frac{V_{rms}}{X_L}=\frac{220}{2\times \frac{22}{7}\times 50\times 25\times 10^{-3}}$
$=\frac{7\times 1000}{2\times 5\times 25}A=28A$