One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has self inductance of value (f=50Hz)

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0.052 H

2.42 H

16.2 mH

1.62 mH

answer is A.

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Detailed Solution

Current through the bulb i=PV=6010=6AV=VR2+VL2(100)2=(10)2+VL2⇒VL=99.5Volt Also VL=iXL=i×(2πV)⇒99.5=6×2×3.14×50×L⇒L=0.052H

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