JEE Main 2023 · 31 January · evening · Numerical answer · +4 / −0
A series LCR circuit consists of R=80 Ω, X_L=100 Ω, and X_C=40 Ω. The input
voltage is 2500 cos (100 π t) V. The amplitude of current, in the circuit, is _________ A.
Answer and solution
Answer 25
Solution
Physics sir · 3 steps
The reactances partly cancel, so find Z first, then peak current is peak voltage over Z.
- Net reactance. In a series circuit X_L and X_C act in opposite directions, so only their difference matters: X_L - X_C = 100 - 40 = 60 Ω.
Z = √(R² + (X_L - X_C)²) = √(80² + 60²) = 100 Ω
- Peak current. The voltage 2500 cos(100π t) has amplitude V₀ = 2500 V. The current amplitude is I₀ = V₀/Z = 2500/100.
- So. The amplitude of the current is 25 A.
The common slip: 15.5. Adding the reactances to get 140 Ω gives Z = √(80² + 140²) ≈ 161 Ω and a current near 15.5 A, not 25.
The idea to keep. In series LCR, subtract the reactances before squaring: Z = √(R² + (X_L - X_C)²). Peak over Z gives peak current; rms over Z gives rms current.
Answer from the NTA final answer key. Solution by PrepWiser.