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An electric bulb is rated at 60W, 240V. Calculate its resistance. lf the voltage drops to 192 V, calculate the power consumed and the current drawn by the bulb. (Assume that the resistance of the bulb remains unchanged).

Answer»

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Solution :`P=(V^(2))/(R) :. R (V^(2))/(P) = ((240)^(2))/(60W)=960Omega`
`P= (V^(2))/(R) = ((192)^(2))/(960) = 38.4 W `
`I =(V)/(R)= (192V)/(960Omega)=0.2 A`


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