A 200 watt bulb is connected in secondary coil. Its primary is connected at 400 volt. If efficiency of transformer is 80% and the bulb is running with its peak power then, the current in A in primary coil is
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a
3/2
b
4/3
c
5/8
d
12
answer is C.
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Detailed Solution
as bulb is glowing with maximum power 200 W i.e, output power with 80% efficiency input power is 200×10080= 250 W input current i=pV=250400 =58
A 200 watt bulb is connected in secondary coil. Its primary is connected at 400 volt. If efficiency of transformer is 80% and the bulb is running with its peak power then, the current in A in primary coil is