When a d.c. voltage of 200 V is applied to a coil of self-inductance ((2√(3))/(π)) H, a current of 1 A flows through it. But by replacing d.c. source with a.c. source of 200 V, the current in the coil is reduced to 0.5 A. Then the frequency of a.c. supply is (a) 100 Hz (b) 60 Hz (c) 75 Hz (d) 50 Hz

Answer: (d). DC: R = 200/1 = 200 Ω. AC: Z = (200)/(0.5) = 400 Ω. XL = √(400² - 200²) = 200√(3) Ω 2π f × (2√(3))/(π) = 200√(3) ⇒ 4√(3) f = 200√(3) ⇒ f = 50 Hz So the answer is (d) 50 Hz.

When a d.c. voltage of 200 V is applied to a coil of self-inductance… | MHT CET 2021 (20 Sep, Shift 1)

Q43MHT CET 2021 · 20 SEP · SHIFT 1Medium

RL and RC Series Circuits

When a d.c. voltage of $200\text{ V}$ is applied to a coil of self-inductance $\left(\frac{2\sqrt{3}}{\pi}\right)\text{ H}$ , a current of $1\text{ A}$ flows through it. But by replacing d.c. source with a.c. source of $200\text{ V}$ , the current in the coil is reduced to $0.5\text{ A}$ . Then the frequency of a.c. supply is

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Answer: (d)

DC: $R = \frac{200}{1} = 200\ \Omega$ . AC: $Z = \frac{200}{0.5} = 400\ \Omega$ .
$X_L = \sqrt{400^2 - 200^2} = 200\sqrt{3}\ \Omega$
$2\pi f \times \frac{2\sqrt{3}}{\pi} = 200\sqrt{3} \Rightarrow 4\sqrt{3} f = 200\sqrt{3} \Rightarrow f = 50\text{ Hz}$
So the answer is (d) 50 Hz.

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