CUET Preparation Today
The value of current through an inductor of inductance 2.0 H and negligible resistance when connected to an AC source of 157 V, 50 Hz will be: |
78.5 A 1.57 A $\frac{1}{2}$ A $\frac{1}{4}$ A |
$\frac{1}{4}$ A |
The correct answer is Option (4) → $\frac{1}{4}$ A Using Ohm's law, Current, $I=\frac{V}{Z_L}=\frac{V}{2\pi vL}$ $=\frac{157}{2\pi×50×2.0}$ $=\frac{157}{628.3}=0.25A$ |
