An inductance coil has a reactance of 100Ω. When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. The self-inductance of the coil is:
An inductance coil has a reactance of 100Ω. When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. The self-inductance of the coil is:
Option 1 -
1.1 * 10⁻² H
Option 2 -
1.1 * 10⁻¹ H
Option 3 -
6.7 * 10⁻⁷ H
Option 4 -
5.5 * 10⁻⁵ H
Asked by Shiksha User
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1 Answer
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Correct Option - 1
Detailed Solution:R = 100Ω. tanφ = (X_L-X_C)/R. tan45° = X_L/R (since it's an inductor). X_L=R=100Ω.
Lω=100 ⇒ L (2πf)=100 ⇒ L = 100/ (2π1000) ≈ 1.59 × 10? ² H. The closest option is (A).
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