A 50, W bulb is in series with a room heater | vedclass

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(D) The power consumed by the heater is given by $P_{	ext{heater}} = I^2 R_H = \left( \frac{V}{R_H + R_b} ight)^2 R_H$,where $R_H$ is the resistanc

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(D) The power consumed by the heater is given by $P_{	ext{heater}} = I^2 R_H = \left( \frac{V}{R_H + R_b} ight)^2 R_H$,where $R_H$ is the resistance of the heater and $R_b$ is the resistance of the bulb.To maximize $P_{	ext{heater}}$,the denominator $(R_H + R_b)^2$ must be minimized. Since $R_H$ is constant,we need to minimize $R_b$.We know that for a bulb,$P = \frac{V^2}{R}$,which implies $R = \frac{V^2}{P}$.Therefore,to minimize $R_b$,the power rating $P$ of the bulb must be as large as possible.Comparing the given options,the $200\, W$ bulb has the lowest resistance.Thus,the $50\, W$ bulb should be replaced by a $200\, W$ bulb.

$A$ $50\, W$ bulb is in series with a room heater and the combination is connected across the mains. To get maximum heater output,the $50\, W$ bulb should be replaced by a bulb of: (in $, W$)

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