JEE Main 2022 · 25 June · evening · Single correct · +4 / −1
A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 μF 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 and solution
Answer (a) tan⁻¹(0.17)
Solution
Physics sir · 3 steps
Find both reactances first, then the phase angle is just the net reactance over R.
- Read off omega. Compare V = 210 sin 3000t with V = V₀ sin ω t. So ω = 3000 rad s⁻¹.
- Both reactances. X_L = ω L = 3000 × 10 × 10⁻³ = 30 Ω. X_C = 1/(ω C) = 1/(3000 × 25 × 10⁻⁶) = 1/0.075 = 13.33 Ω.
tan Φ = (X_L - X_C)/R = (30 - 13.33)/100 = 16.67/100 ≈ 0.17
- So. Φ = tan⁻¹(0.17), which is option (a). Since X_L > X_C, the voltage leads the current by this angle.
The idea to keep. In a series LCR circuit, tan Φ = (X_L - X_C)/R. The inductor and capacitor push the phase in opposite directions, so always subtract their reactances before dividing by R.
Answer from the NTA final answer key. Solution by PrepWiser.