Question

An inductor of $1H$ is connected across a $220V,50Hz$ supply. The peak value of the current is approximately

• A. 0.5 A
• B. 0.7 A
• C. 1 A
• D. 1.4 A

Answer 1

Answer: (C)

Current $\left(i_0\right)=\frac{E_0}{X_L}$
$\Rightarrow i_0=\frac{E_0}{L\omega }$
$\therefore i_0=\frac{220\times \sqrt{2}}{1\times 2\pi \times 50}$
$\Rightarrow i_0=\frac{220\times \sqrt{2}}{100\pi }=\frac{220\times \sqrt{2}}{100\times 3.14}$
$\Rightarrow i_0=1A$