Question

A pure inductor of $26\text{mH}$ is connected to an ac source of $220\text{V}$. Given frequency of the source as $50\text{Hz}$, the rms current in the circuit is

​​​​​​​$[$1 Mark$]$

• A. $7\text{A}$
• B. $14\text{A}$
• C. $28\text{A}$
• D. $742\text{A}$

Answer 1

The correct option is C $28\text{A}$

Given,
$L=25\text{mH},f=50\text{Hz},V_{rms}=220\text{V}$

The inductive reactance is

$X_L=2\pi fL=2\times \frac{22}{7}\times 50\times 25\times {10}^{-3}\Omega$

The rms current in the circuit is

$I_{rms}=\frac{V_{rms}}{X_L}=\frac{220}{2\times \frac{22}{7}\times 50\times 25\times {10}^{-3}}$

$\Rightarrow I_{rms}=\frac{7\times 1000}{2\times 5\times 25}=28\text{A}$