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An electric bulb of resistance 500Omega,...

An electric bulb of resistance `500Omega`, draws a current of 0.4 A. Calculate the power of the bulb and the potential difference at its end.

Text Solution

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Given `R = 500 Omega`
`I = 0.4 A,`
`P = I ^(2) R`
`= (0.4) ^(2) xx 500`
`= 80` Watts
Also, V = IR
`= 200` volts
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ICSE-SAMPLE QUESTION PAPER 5 -SECTION B
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