An inductive coil has a resistance of 100 Ω. When an a.c. signal of frequency 1000 Hz is applied to the coil the voltage leads the current by 45°. The inductance of the coil is (tan 45° = 1) (a) (0.25)/(2π) H (b) (0.05)/(π) H (c) (0.25)/(π) H (d) (0.5)/(π) H

Answer: (b). tan 45° = XL/R = 1 ⇒ XL = 100 Ω L = (XL)/(2π f) = (100)/(2π × 1000) = (0.05)/(π) H So the answer is (b) (0.05)/(π) H.

An inductive coil has a resistance of 100 Ω. When an a.c. signal of… | MHT CET 2021 (20 Sep, Shift 2)

Q39MHT CET 2021 · 20 SEP · SHIFT 2Medium

RL and RC Series Circuits

An inductive coil has a resistance of $100\ \Omega$ . When an a.c. signal of frequency $1000\text{ Hz}$ is applied to the coil the voltage leads the current by $45^\circ$ . The inductance of the coil is $(\tan 45^\circ = 1$ )

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Answer: (b)

$\tan 45^\circ = \frac{X_L}{R} = 1 \Rightarrow X_L = 100\ \Omega$
$L = \frac{X_L}{2\pi f} = \frac{100}{2\pi \times 1000} = \frac{0.05}{\pi}\text{ H}$
So the answer is (b) $\frac{0.05}{\pi}$ H.

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