JEE Main 2016 — Alternating Current Question with Solution

Given that the lamp draws a current of $10 \mathrm{A}$ upon a $80 \mathrm{V}$ D.C. supply, the resistance is,

$R = \frac{80}{10} = 8 \mathrm{\Omega }$

When connected to an AC supply with an inductor in series, as shown in the figure above, the current,

 $I = 10 \mathrm{A}$

and the voltage drop over the lamp is $80 \mathrm{V}$ and $V_L$ over the inductance.

Applying the rules of $L-R$ circuit,

${\mathrm{V}}_{\mathrm{L}}^2+{80}^2={220}^2$

$\Rightarrow {\mathrm{V}}_{\mathrm{L}}=\sqrt{48400-6400}=\mathrm{210}$ V

The Inductive reactance of the given inductor of inductance $L$, for the frequency of supply $f$,

${\mathrm{X}}_{\mathrm{L}}=2\mathrm{\pi fL} ...\left(1\right)$

The voltage drop across the inductance,

${\mathrm{V}}_{\mathrm{L}}= I\mathrm{ }{\mathrm{X}}_{\mathrm{L}}=210$

$\Rightarrow {\mathrm{X}}_{\mathrm{L}}=2\mathrm{\pi }\mathrm{f}\mathrm{L}=\frac{V_L}{I}=\frac{210}{10}=21 \mathrm{\Omega }$

Substituting values in equation $\left(1\right)$

$\mathrm{L}=\frac{21}{2\pi f}=\frac{21}{2\pi \times 50}=\frac{21}{100\mathrm{ }\mathrm{\pi }}=0.065\mathrm{ }\mathrm{H}$