A coil of resistance 450 Ω and self-inductance 1.5 henry is connected to an a.c. source of frequency (150)/(π) Hz. The phase difference between voltage and current is (a) tan ⁻¹(0.5) (b) tan ⁻¹(1) (c) tan ⁻¹(1.5) (d) tan ⁻¹(2.0)

Answer: (b). XL = 2π f L = 2π × (150)/(π) × 1.5 = 450 Ω tan φ = XL/R = (450)/(450) = 1 So the answer is (b) tan ⁻¹(1).

A coil of resistance 450 Ω and self-inductance 1.5 henry is connected… | MHT CET 2025 (19 Apr, Shift 2)

Q16MHT CET 2025 · 19 APR · SHIFT 2Easy

RL and RC Series Circuits

A coil of resistance $450\ \Omega$ and self-inductance 1.5 henry is connected to an a.c. source of frequency $\frac{150}{\pi}\text{ Hz}$ . The phase difference between voltage and current is

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Answer: (b)

$X_L = 2\pi f L = 2\pi \times \frac{150}{\pi} \times 1.5 = 450\ \Omega$
$\tan\phi = \frac{X_L}{R} = \frac{450}{450} = 1$
So the answer is (b) $\tan^{-1}(1)$ .

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