An electric bulb of resistance 200 Ω draws a current of 2 ampere. The energy in kWh consumed by burning it for 5 hours is
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Answer:4kWh
Explanation:
Power=V*i=(iR)*i
=2*2*200
=800 W =800/1000kW
P=4/5kW
E=P*t
=4/5kW*5hr
=4kWh
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