The resistance of a bulb filament is 100 Ome | vedclass

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(B) The resistance at temperature $t$ is given by the formula $R_t = R_0(1 + \alpha \Delta T)$,where $R_0$ is the resistance at $0 \ ^oC$ and $\Delta

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$400$

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(B) The resistance at temperature $t$ is given by the formula $R_t = R_0(1 + \alpha \Delta T)$,where $R_0$ is the resistance at $0 \ ^oC$ and $\Delta T$ is the temperature change.Given: $R_1 = 100 \ \Omega$ at $t_1 = 100 \ ^oC$,$R_2 = 200 \ \Omega$ at $t_2 = T$,and $\alpha = 0.005 \ ^oC^{-1}$.Using the relation $R_t = R_{ref}[1 + \alpha(t - t_{ref})]$,we can write:$100 = R_0(1 + 0.005 	imes 100) \implies 100 = R_0(1 + 0.5) \implies 100 = 1.5 R_0 \implies R_0 = \frac{100}{1.5} = \frac{200}{3} \ \Omega$.Now,for $R_2 = 200 \ \Omega$ at temperature $T$:$200 = R_0(1 + 0.005 	imes T) = \frac{200}{3}(1 + 0.005T)$.Dividing both sides by $200$:$1 = \frac{1}{3}(1 + 0.005T) \implies 3 = 1 + 0.005T \implies 2 = 0.005T$.$T = \frac{2}{0.005} = \frac{2000}{5} = 400 \ ^oC$.

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 $(^oC)$ will its resistance become $200 \ \Omega$?

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