Question

A pure inductor of inductance $0.1H$ is connected to an AC source (of rms voltage) $220V$ angular frequency of $300Hz.$ The rms current is

• A. $\frac{3}{22}$ A
• B. $\frac{22}{3}$ A
• C. $\frac{11}{150}$ A
• D. $\frac{150}{11}$ A
• E. $\frac{11}{3\pi }$ A

Answer 1

Answer: (E)

Given, pure inductor of inductance $L=0.1H$
$AC$ source voltage $V=220V$ and angular frequency, $f=300Hz$
As, impedance, $Z=\omega L$
$=2\pi \times 300\times 0.1=30\Omega (\therefore \omega =2\pi f)$
So, the rms current $I_{rms}=\frac{V_{rms}}{Z}$
$=\frac{220}{30\times 2\pi }=\frac{11}{3\pi }A$