1.

A potential difference of 100 V is applied across an electric bulb marked 40 W. 200 V. The power consumed in the bulb is . …………….. .

Answer»

<P>100W
40W
20W
10W

Solution :`P = V^2/R`
`:.`Resistance of the FILAMENT of the bulb,
`R = V^2/P = ((200)^2)/(40) = 1000Omega`
Now, POWER consumed in the bulb,
`P = V^2/R = ((100)^2)/1000 = 10 W`


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