A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 μ F and R = 100 Ω. The phase difference ( Φ) between the applied voltage and resultant current will be: (a) tan ⁻¹ (0.17) (b) tan ⁻¹ (9.46) (c) tan ⁻¹ (0.30) (d) tan ⁻¹ (13.33)

Answer: (a). ω = 3000 rad/s XL = 3000 × 10×10⁻³ = 30 Ω XC = 13000 × 25×10⁻⁶ ≈ 13.33 Ω tan Φ = (XL - XC)/(R) = (16.67)/(100) ≈ 0.17 So the answer is (a) tan ⁻¹(0.17).

A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a… | JEE Main 2022 (25 Jun, Shift 2)

Q44JEE Main 2022 · 25 JUN · SHIFT 2Medium

Series LCR Circuit

A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 $\mu$ F and R = 100 $\Omega$ . The phase difference ( $\Phi$ ) between the applied voltage and resultant current will be:

Show answer and solution

Answer: (a)

$\omega = 3000\text{ rad/s}$
$X_L = 3000 \times 10\times10^{-3} = 30\ \Omega$
$X_C = \dfrac{1}{3000 \times 25\times10^{-6}} \approx 13.33\ \Omega$
$\tan\Phi = \dfrac{X_L - X_C}{R} = \dfrac{16.67}{100} \approx 0.17$
So the answer is (a) $\tan^{-1}(0.17)$ .

Report a mistake

Series LCR Circuit question from JEE Main 2022 (25 Jun, Shift 2)

More questions on Series LCR Circuit