Electromagnetic Induction

JEE Main level, single correct, medium

132 of 153

Question 132 of 153

An inductance coil has a reactance of 100Ω100\Omega . When an AC signal of frequency 1000Hz1000Hz is applied to the coil, the applied voltage leads the current by 45∘45^{\circ}. The self-inductance of the coil is

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Answer: (A) 1.1×10−2H1.1\times {10}^{-2}H

tanθ=xLR=tan⁡45∘tan\theta =\dfrac{{x}_{L}}{R}=\tan 45^{\circ}
xL=R{x}_{L}=R
=100=xL2+R2=100=\sqrt{{x}_{L}^{2}+{R}^{2}}
100=R2+R2100=\sqrt{{R}^{2}+{R}^{2}}
2R=100\sqrt{2}R=100
R=502R=50\sqrt{2}
∴xL=502∴{x}_{L}=50\sqrt{2}
Lω=502L\omega =50\sqrt{2}
L=5022π×1000=252πmHL=\dfrac{50\sqrt{2}}{2\pi \times 1000}=\dfrac{25\sqrt{2}}{\pi }mH
=1.1×10−2H=1.1\times {10}^{-2}H

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