state joules law of heating a bulb is rated 110w 330v calculate resistance what is the energy consumed in kwh by 2 such bulbs operated for 5 hours at a stretch
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Joules heating law is = I^2RT
Power = IR
= V^2 / R ( applying ohms law)
Potential Difference = 330v
Power = 110w
P = (330)^2 / R
R = 330 × 330 / 110 .......... (Power = 110)
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R = 990 ohms
Energy consumed = Power × time
Power of 2 bulbs = 110 × 2
= 220 w
In kW = 220 / 1000 = 0.22 kW .............. ( 1kW = 1000W)
Time = 5 hours.
Energy = 0.22kW × 5h
= 1.1kWh
Since energy is expressed in terms of kilo - Watt hour.
Energy = 1.1 kWh.
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