Q129 / 150 · MHT CET 2021 (20 Sep Shift 1) · Physics
An inductive coil has a resistance of $100 \Omega$. When an a.c. signal of frequency $1000 \mathrm{~Hz}$ is applied to the coil the voltage leads the current by $45^{\circ}$. The inductance of the coil is
- A$\frac{0.25}{2 \pi} \mathrm{H}$
- Correct answerB$\frac{0.05}{\pi} \mathrm{H}$
- C$\frac{0.25}{\pi} \mathrm{H}$
- D$\frac{0.5}{\pi} \mathrm{H}$
Solution
$$
\tan \phi=\tan 45^{\circ}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}
$$
or $\mathrm{X}_{\mathrm{L}}=\mathrm{R}$
$$
2 \pi \mathrm{fL}=\mathrm{R}
$$
Or $L=\frac{\mathrm{R}}{2 \pi \mathrm{f}}=\frac{100}{2 \pi \times 1000}=\frac{0.05}{\pi} \mathrm{H}$
Hence, the correct answer is option: 'B'.