An a.c. source of 15 V, 50 Hz is connected across an inductor (L) and resistance (R) in series R.M.S. current of 0.5 A flows in the circuit. The phase difference between applied voltage and current is (π/3) radian. The value of resistance (R) is (tan 60° = √(3)) (a) 10 Ω (b) 12 Ω (c) 15 Ω (d) 20 Ω

Answer: (c). Z = V/I = 15/0.5 = 30 Ω R = Zcos 60° = 30 × 1/2 = 15 Ω So the answer is (c) 15 Ω.

An a.c. source of 15 V, 50 Hz is connected across an inductor (L) and… | MHT CET 2023 (12 May, Shift 2)

Q36MHT CET 2023 · 12 MAY · SHIFT 2Medium

RL and RC Series Circuits

An a.c. source of $15\text{ V},\ 50\text{ Hz}$ is connected across an inductor (L) and resistance (R) in series R.M.S. current of $0.5\text{ A}$ flows in the circuit. The phase difference between applied voltage and current is $\left(\frac{\pi}{3}\right)$ radian. The value of resistance $(R)$ is $\left(\tan 60^\circ = \sqrt{3}\right)$

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Answer: (c)

$Z = \frac{V}{I} = \frac{15}{0.5} = 30\ \Omega$
$R = Z\cos 60^\circ = 30 \times \frac{1}{2} = 15\ \Omega$
So the answer is (c) $15\ \Omega$ .

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RL and RC Series Circuits question from MHT CET 2023 (12 May, Shift 2)

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