JEE Main 2022 · 29 July · morning · Single correct · +4 / −1
An alternating emf E=440 sin 100 π t is applied to a circuit containing an inductance of √2/π H. If an a.c. ammeter is connected in the circuit, its reading will be :
- (a) 4.4 A
- (b) 1.55 A
- (c) 2.2 A
- (d) 3.11 A
Answer and solution
Answer (c) 2.2 A
Solution
Physics sir · 4 steps
An a.c. ammeter shows the rms current, so find the peak current from X_L and then divide by √2.
- Read the emf. Comparing with E = E₀ sin ω t gives the peak emf E₀ = 440 V and ω = 100π rad s⁻¹.
- Inductive reactance. X_L = ω L = 100π × √2/π = 100√2 Ω. The π cancels.
Iᵣₘₛ = I₀/√2 = E₀/(√2 X_L) = 440/(√2 × 100√2) = 440/200
- Peak versus rms. The peak current is I₀ = 440/100√2 ≈ 3.11 A. An a.c. ammeter does not show this peak value. It shows the rms value, 3.11/√2.
- So. The ammeter reads Iᵣₘₛ = 2.2 A, which is option (c).
The idea to keep. A.c. meters always read rms values. Find the peak current from E₀/X, then divide by √2 exactly once.
Answer from the NTA final answer key. Solution by PrepWiser.