JEE Main 2023 · 6 April · evening · Single correct · +4 / −1
A capacitor of capacitance 150.0 μ F is connected to an alternating source of emf given by E=36 sin (120 π t) V. The maximum value of current in the circuit is approximately equal to :
- (a) 1/√2 A
- (b) 2 √2 A
- (c) √2 A
- (d) 2 A
Answer and solution
Answer (d) 2 A
Solution
Physics sir · 3 steps
The 36 in front of the sine is already the peak voltage, so peak current is just E₀ ω C.
- Read the source. Compare E = 36 sin(120π t) with E = E₀ sin ω t. The peak emf is E₀ = 36 V and ω = 120π rad s⁻¹.
- Capacitive reactance. A capacitor opposes AC with reactance X_C = 1/(ω C). Here ω C = 120π × 150 × 10⁻⁶ = 0.0565 Ω⁻¹, so X_C ≈ 17.7 Ω.
I₀ = E₀/X_C = E₀ ω C = 36 × 0.0565 ≈ 2.04 A
- So. The maximum current is about 2 A, which is option (d).
The idea to keep. In E = E₀ sin ω t, the number in front is the peak value. Peak voltage divided by reactance gives peak current. Use √2 only when the question asks for rms.
Answer from the NTA final answer key. Solution by PrepWiser.