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Q.

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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a

6.7×10−7H

b

1.1×10−1H

c

5.5×10−5H

d

1.1×10−2H

answer is D.

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Detailed Solution

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