JEE Main 2020 · 2 September · evening · Single correct · +4 / −1
An inductance coil has a reactance of 100 Ω. When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45ᵒ. The self-inductance of the coil is
- (a) 6.7 × 10 –7 H
- (b) 1.1 × 10 –1 H
- (c) 5.5 × 10 –5 H
- (d) 1.1 × 10 –2 H
Answer and solution
Answer (d) 1.1 × 10 –2 H
Solution
Physics sir · 3 steps
A 45° lead means R equals X_L, so the 100 Ω is the coil's whole impedance. Split it before finding L.
- Read the phase. A pure inductor makes the voltage lead by 90°. A lead of only 45° means the coil also has resistance, and tan 45° = X_L/R = 1, so X_L = R. The 100 Ω is therefore the coil's total opposition, its impedance Z.
- Split the impedance. Z = √(R² + X_L²) = √2 X_L, so X_L = 100/√2 = 70.7 Ω. Taking X_L = 100 Ω directly gives 1.6 × 10⁻² H, which is not among the options.
L = X_L/(2π f) = 70.7/(2π × 1000)
- So. L = 70.7/6283 = 1.13 × 10⁻² H ≈ 1.1 × 10⁻² H, which is option (d).
The idea to keep. A phase angle smaller than 90° tells you the coil has resistance too. Use tan φ = X_L/R to split the impedance first, then L = X_L/(2π f).
Answer from the NTA final answer key. Solution by PrepWiser.