An inductance of (300)/(π) mH, a capacitance of 1/π mF and a resistance of 20 Ω are connected in series with an a.c. source of 240 V, 50 Hz. The phase angle of the circuit is (a) tan ⁻¹(0) (b) tan ⁻¹(1/√(3)) (c) tan ⁻¹(1) (d) tan ⁻¹(√(3))

Answer: (c). XL = 2π × 50 × (0.3)/(π) = 30 Ω XC = 12π × 50 × 10⁻³π = 10 Ω tan φ = (XL - XC)/(R) = 20/20 = 1 So the answer is (c) tan ⁻¹(1).

An inductance of (300)/(π) mH, a capacitance of 1/π mF and a… | MHT CET 2024 (11 May, Shift 1)

Q28MHT CET 2024 · 11 MAY · SHIFT 1Medium

Series LCR Circuit

An inductance of $\frac{300}{\pi}\text{ mH}$ , a capacitance of $\frac{1}{\pi}\text{ mF}$ and a resistance of $20\ \Omega$ are connected in series with an a.c. source of $240\text{ V},\ 50\text{ Hz}$ . The phase angle of the circuit is

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Answer: (c)

$X_L = 2\pi \times 50 \times \dfrac{0.3}{\pi} = 30\ \Omega$
$X_C = \dfrac{1}{2\pi \times 50 \times \frac{10^{-3}}{\pi}} = 10\ \Omega$
$\tan\phi = \dfrac{X_L - X_C}{R} = \dfrac{20}{20} = 1$
So the answer is (c) $\tan^{-1}(1)$ .

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Series LCR Circuit question from MHT CET 2024 (11 May, Shift 1)

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