An AC source V(t) = 300 sin(400t) V is connected to each load in List-I. Match each load with the corresponding current-time graph in List-II.

Question

G1 Physics · Accepted Answer

ω = 400 rad s⁻¹.

P: R = 30 Ω. I₀ = 300/30 = 10 A, in phase with V → graph 3.

Q: Xᴸ = ωL = 40 Ω. |Z| = √(30²+40²)=50 Ω, so I₀=6 A and current lags → graph 5.

R: Xᶜ = 1/(ωC)=50 Ω and Xᴸ=10 Ω. Net reactance = −40 Ω, |Z|=50 Ω, so I₀=6 A and current leads → graph 2.

S: Xᴸ = 50 Ω and Xᶜ = 50 Ω, so resonance occurs. Z = 60 Ω, I₀ = 5 A, in phase → graph 1.

Hence P→3, Q→5, R→2, S→1: option A.