Two wires of resistance R_1 and R_2 have temp | vedclass

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(C) Let the resistances at temperature $t$ be $R_{t1}$ and $R_{t2}$.$R_{t1} = R_1(1 + \alpha_1 t)$ and $R_{t2} = R_2(1 + \alpha_2 t)$Since they are co

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$\frac{R_1 \alpha_1 + R_2 \alpha_2}{R_1 + R_2}$

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(C) Let the resistances at temperature $t$ be $R_{t1}$ and $R_{t2}$.$R_{t1} = R_1(1 + \alpha_1 t)$ and $R_{t2} = R_2(1 + \alpha_2 t)$Since they are connected in series,the equivalent resistance $R_{eq}$ at temperature $t$ is:$R_{eq} = R_{t1} + R_{t2} = R_1(1 + \alpha_1 t) + R_2(1 + \alpha_2 t)$$R_{eq} = (R_1 + R_2) + (R_1 \alpha_1 + R_2 \alpha_2)t$We can write this as $R_{eq} = R_{eq,0}(1 + \alpha_{eff} t)$,where $R_{eq,0} = R_1 + R_2$.$R_{eq} = (R_1 + R_2) \left[ 1 + \left( \frac{R_1 \alpha_1 + R_2 \alpha_2}{R_1 + R_2} \right) t \right]$Comparing this with the standard form,the effective temperature coefficient is $\alpha_{eff} = \frac{R_1 \alpha_1 + R_2 \alpha_2}{R_1 + R_2}$.

Two wires of resistance $R_1$ and $R_2$ have temperature coefficients of resistance $\alpha_1$ and $\alpha_2$,respectively. These are joined in series. The effective temperature coefficient of resistance is

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