An electric bulb rated 50 , W-200 , V is conn | vedclass

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Question

(A) The resistance $R$ of the bulb is determined by its rated power $P_{rated}$ and rated voltage $V_{rated}$:$R = \frac{V_{rated}^2}{P_{rated}} = \fr

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(A) The resistance $R$ of the bulb is determined by its rated power $P_{rated}$ and rated voltage $V_{rated}$:$R = \frac{V_{rated}^2}{P_{rated}} = \frac{(200)^2}{50} = \frac{40000}{50} = 800 \, \Omega$When the bulb is connected to a new supply voltage $V_{applied} = 100 \, V$, the actual power dissipated $P_{actual}$ is given by:$P_{actual} = \frac{V_{applied}^2}{R} = \frac{(100)^2}{800} = \frac{10000}{800} = 12.5 \, W$Thus, the power dissipation of the bulb is $12.5 \, W$. Hence, option $A$ is correct.

An electric bulb rated $50 \, W-200 \, V$ is connected across a $100 \, V$ supply. The power dissipation of the bulb is: (in $ \, W$)

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