The resistance of a bulb filament is 100, Ome | vedclass

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(C) Given: Initial temperature $	heta_1 = 100\, ^oC$,Initial resistance $R_{	heta_1} = 100\, \Omega$,Final resistance $R_{	heta_2} = 200\, \Omega$,

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(C) Given: Initial temperature $	heta_1 = 100\, ^oC$,Initial resistance $R_{	heta_1} = 100\, \Omega$,Final resistance $R_{	heta_2} = 200\, \Omega$,Temperature coefficient $\alpha = 0.005\, (^oC)^{-1}$.The formula relating resistance and temperature is $R_{	heta_2} = R_{	heta_1} [1 + \alpha (	heta_2 - 	heta_1)]$.Rearranging the formula to solve for $	heta_2$:$R_{	heta_2} - R_{	heta_1} = \alpha R_{	heta_1} (	heta_2 - 	heta_1)$$\frac{R_{	heta_2} - R_{	heta_1}}{\alpha R_{	heta_1}} = 	heta_2 - 	heta_1$$	heta_2 = \frac{R_{	heta_2} - R_{	heta_1}}{\alpha R_{	heta_1}} + 	heta_1$Substituting the values:$	heta_2 = \frac{200 - 100}{0.005 	imes 100} + 100$$	heta_2 = \frac{100}{0.5} + 100$$	heta_2 = 200 + 100 = 300\, ^oC$.

The resistance of a bulb filament is $100\, \Omega$ at a temperature of $100\, ^oC$. If its temperature coefficient of resistance is $0.005\, (^oC)^{-1}$,at what temperature will its resistance become $200\, \Omega$?

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