JEE Main · 2016 · 03 Apr 2016 · Offline · Alternating Current

The lamp behaves as a resistor fixed by its DC rating, $R=\frac{80}{10}=8\Omega$, and to run normally it must again carry 10 A. On the AC supply the lamp and series inductor together must drop the full 220 V rms while passing 10 A, so the circuit impedance is $Z=\frac{220}{10}=22\Omega$. Since $Z^2=R^2+(\omega L{)}^2$, $(\omega L{)}^2=22^2-8^2=420$, so $\omega L=\sqrt{420}\approx 20.5\Omega$. With $\omega =2\pi \times 50$, $L=\frac{20.5}{2\pi \times 50}\approx 0.065$ H.