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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Current Electricity
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Two wires of resistivity $\rho 1$ and $ \rho 2$ are connected in parallel. 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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when resistance are connected in series
$ R = \large\frac{R_1R_2}{R_1+R_2}$
$ \rho \large\frac{l}{2A} =\large\frac{ \rho_ 1\large\frac{l}{A} \rho_2 \large\frac{l}{A}}{ \rho_1\large\frac{l}{A}+ \rho_2 \large\frac{l}{A}}$
$ \Rightarrow \rho =2\large\frac{\rho _1\rho_2}{\rho_1+\rho_2}$

 

answered Feb 8, 2014 by thanvigandhi_1
edited Sep 10, 2014 by thagee.vedartham
 

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