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An electric bulb marked 40 W and 200 V is used in a circuit of supply voltage 100 V. Now its power is .... .

Answer»

<P>10 W
20 W
40 W
100 W

Solution :10 W
P = `(V^(2))/(R) ` , where R is CONSTANT (bulb is same )
P `prop V^(2)`
`THEREFORE (P_(2))/(P_(1)) = ((V_(2))/(V_(1)))^(2)`
`therefore P_(2) = P_(1) xx ((V_(2))/(V_(1)))^(2)`
= ` 40 xx ((100)/(200))^(2) = 40 xx (1)/(4) = 10` W


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