Electric Power

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An incandescent bulb rated 60 W is designed to operate on a 240 V supply. What is the resistance of its filament at operating conditions?

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Electrical power dissipated by a resistive device relates voltage and resistance through $P = V^2/R$, since the energy delivered per second to the filament is converted into heat and light. This form is convenient when the operating voltage and power rating are known but the current is not. Rearranging gives $R = V^2/P$. Substituting $V = 240$ V and $P = 60$ W yields $R = (240)^2/60 = 57600/60 = 960\;\Omega$. The value 240 $\Omega$ wrongly equates resistance with the voltage. The value 480 $\Omega$ results from using $V$ instead of $V^2$ and then doubling. The value 120 $\Omega$ comes from dividing the voltage by the power without squaring. This applies the NCERT power-resistance relationship for ohmic devices. A plausibility check supports the answer: the operating current would be $I = P/V = 60/240 = 0.25$ A, and $V/I = 240/0.25 = 960\;\Omega$, in exact agreement with the computed resistance.

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