One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has self inductance of value (f = 50 Hz)
0.052 H
2.42 H
16.2 mH
1.62 mH
Current through the bulb, I=PV=6010=6 A
V=VR2+VL2
(100)2=(10)2+VL2⇒VL=99.5 V
Also, VL=iXL=i×(2πvL)
⇒99.5=6×2×3.14×50×L⇒L=0.052 H