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

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(C) The resistance at temperature $T$ is given by $R_T = R_0(1 + \alpha \Delta T)$,where $R_0$ is the resistance at $0\,^{\circ}C$.Given:$R_1 = 100\,\

Vedclass · Accepted Answer

$400$

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(C) The resistance at temperature $T$ is given by $R_T = R_0(1 + \alpha \Delta T)$,where $R_0$ is the resistance at $0\,^{\circ}C$.Given:$R_1 = 100\,\Omega$ at $T_1 = 100\,^{\circ}C$$R_2 = 200\,\Omega$ at $T_2 = T$$\alpha = 0.005\,^{\circ}C^{-1}$Using the formula $R = R_0(1 + \alpha T)$:$100 = R_0(1 + 0.005 	imes 100) = R_0(1 + 0.5) = 1.5 R_0$$R_0 = \frac{100}{1.5} = \frac{200}{3}\,\Omega$Now,for $R_2 = 200\,\Omega$:$200 = R_0(1 + 0.005 	imes T)$$200 = \frac{200}{3}(1 + 0.005 T)$$3 = 1 + 0.005 T$$2 = 0.005 T$$T = \frac{2}{0.005} = 400\,^{\circ}C$.

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

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