An AC circuit contains resistance of 12 Ω and inductive reactance 5 Ω. The phase angle between current and potential difference will be (a) sin ⁻¹(12/13) (b) cos ⁻¹(5/12) (c) sin ⁻¹(5/12) (d) cos ⁻¹(12/13)

Answer: (d). Z = √(R² + XL²) = √(144 + 25) = 13 Ω cos φ = R/Z = 12/13, so φ = cos ⁻¹(12/13) (equivalently sin ⁻¹5/13). So the answer is (d) cos ⁻¹(12/13).

An AC circuit contains resistance of 12 Ω and inductive reactance 5… | MHT CET 2020 (19 Oct, Shift 2)

Q45MHT CET 2020 · 19 OCT · SHIFT 2Easy

RL and RC Series Circuits

An AC circuit contains resistance of $12\ \Omega$ and inductive reactance $5\ \Omega$ . The phase angle between current and potential difference will be

Show answer and solution

Answer: (d)

$Z = \sqrt{R^2 + X_L^2} = \sqrt{144 + 25} = 13\ \Omega$
$\cos\phi = \frac{R}{Z} = \frac{12}{13}$ , so $\phi = \cos^{-1}\left(\frac{12}{13}\right)$ (equivalently $\sin^{-1}\frac{5}{13}$ ).
So the answer is (d) $\cos^{-1}\left(\frac{12}{13}\right)$ .

Report a mistake

RL and RC Series Circuits question from MHT CET 2020 (19 Oct, Shift 2)

More questions on RL and RC Series Circuits