Question

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