An inductance coil has an impedance of 100 Ω. When an AC signal of… | JEE Main 2020 (2 Sep, Shift 2)

Q47JEE Main 2020 · 2 SEP · SHIFT 2Difficult

RL and RC Series Circuits

An inductance coil has an impedance of 100 $\Omega$ . When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. The self-inductance of the coil is

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Answer: (d)

Voltage leads current by $45^\circ$ : $\tan 45^\circ = \frac{X_L}{R} = 1 \Rightarrow X_L = R$ .
$Z = \sqrt{R^2 + X_L^2} = \sqrt{2}\,X_L = 100\ \Omega \Rightarrow X_L = 70.7\ \Omega$
$L = \frac{X_L}{2\pi f} = \frac{70.7}{2\pi \times 1000} \approx 1.1 \times 10^{-2}\ \text{H}$
So the answer is (d) $1.1 \times 10^{-2}$ H.

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