1.

An electric bulb of 200 Omega draws a current of 1 ampere. Calculate the power of the bulb. The potential diffrence at is ends and the energy in kwh consumed in burining it for 5h.

Answer»

Solution :POWER of the bulb
Power of the bulb `P = PR = (1)^(2)x 200`
P = 200W
Energy CONSUMED by bulb in 5H in burning
= Power x time
= 200 x 5
= 1000wh = 1 kwh.


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