1.

If the current in an electric bulb increases by 1 %, what will be the change in the power of a bulb ? [Assume that the resistance of the filament of a bulh remains constant]

Answer»

INCREASES by 1%
DECREASES by 1 %
increases by 2%
decreases by 2%

Solution :increases by 2%
Power of a bulbP = `I^(2) ` R , where R = Resistance 0 FORCE
Now, increase or 1% of current , NEW current
I. = I + 1% ,I = I + 0.01 I
I. = 1.01 I
New Power P.= `I.^(2) ` R
P.= `(1.01I)^(2) R = 1.02 I^(2)` R
`(P.)/(P) = (1.02 I^(2)R)/(I^(2) R)`
P.= 1.02 P
= P + 0.02 P = P + 2% P
Change in the power of bulb= 2%


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