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Two conductors have the same resistance at $0^{\circ} C$ but their temperature coefficients of resistance are $ \alpha 1 \: and \: \alpha 2$. The temperature coefficient for the parallel combination is

$\begin {array} {1 1} (A)\; \alpha_1 + \alpha_2 & \quad (B)\;\large\frac{(\alpha_1+ \alpha_2)}{2} \\ (C)\;\large\frac{(\alpha_1 \alpha_2)}{(\alpha_1+ \alpha_2)} & \quad (D)\;\large\frac{2(\alpha_1 \alpha_2)}{(\alpha_1+ \alpha_2)} \end {array}$


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$R_p = \large\frac{(R.R)}{(R + R)} $ $= \large\frac{R}{2}$
Substituting the value of $R_p$, and $R$ in terms of the temperature coefficients
and considering the temperature coefficients as small, one gets the following expression
$ \alpha_p = \large\frac{(\alpha_1 + \alpha_2) }{2}$
Ans : (B)


answered Feb 13, 2014 by thanvigandhi_1
edited Sep 3, 2014 by thagee.vedartham

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