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 450. The self-inductance of the coil is:

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6.7×10−7H

1.1×10−1H

5.5×10−5H

1.1×10−2H

answer is D.

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Detailed Solution

tanϕ=XLR⇒tan450=XLR⇒XL=RImpedance =Z=R2+XL2⇒Z=2XL⇒100=2XL⇒XL=1002⇒ωL=1002⇒2πfL=1002⇒L=1.1×10−2H

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