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Two wires of resistivity $\rho 1$ and $\rho 2$ are connected in series. Equivalent resistivity of the combination is

$\begin {array} {1 1} (A)\;\rho_1+\rho_2 & \quad (B)\;\large\frac{1}{2} (\rho_1+\rho_2) \\ (C)\;2\large\frac{\rho_1\rho_2}{\rho_1+\rho_2} & \quad (D)\;\large\frac{1}{2}\large\frac{\rho_1\rho_2}{\rho_1+\rho_2} \end {array}$

 

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1 Answer

$R = R1 + R2$
$ \rho \large\frac{2l}{A} = \rho_1 \large\frac{l}{A} + \rho_2\large\frac{l}{A}$
$ \Rightarrow \large\frac{1}{2} (\rho_1+\rho_2)$
Ans : (B)

 

answered Feb 8, 2014 by thanvigandhi_1
edited Aug 23, 2014 by thagee.vedartham
 

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