1.

What will be the resistance of a bulb of power 40 W which is connected to a voltage source of 220V?

Answer»

SOLUTION : Data : P= 40 W, V = 220 , V , R = ?
` P = VI = V( V/R)= V^(2)/R`
`R= V^(2)/P = ((200 V)^(2))/( 40 W) = ( 220 xx 220)/40 Omega= (220 xx 220)/ ( 2 xx 20) = Omega`
` = 11 xx110 Omega = 1210 Omega`
The resistanceof the bulb = `1210 Omega`


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