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(D) The power consumed by a bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance of the bulb.For a bulb

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Equally with both

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(D) The power consumed by a bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance of the bulb.For a bulb,the resistance $R$ is a constant property of the filament.When connected to a $DC$ source of voltage $V$,the power consumed is $P_{DC} = \frac{V^2}{R}$.When connected to an $AC$ source of the same $RMS$ voltage $V$,the power consumed is $P_{AC} = \frac{V_{rms}^2}{R} = \frac{V^2}{R}$.Since the power consumed in both cases is identical,the brightness of the bulb will be the same.

$A$ bulb is connected first with $DC$ and then $AC$ of the same voltage. In which case will it shine more brightly?

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