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Physics
Two wires of same dimensions but resistivities ρ1 and ρ2 are connected in series. The equivalent resistivity of the combination is
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Q. Two wires of same dimensions but resistivities $\rho_1$ and $\rho_2$ are connected in series. The equivalent resistivity of the combination is
KCET
KCET 2003
Current Electricity
A
$2(\rho_1 + \rho_2)$
7%
B
$\sqrt{\rho_1\rho_2}$
5%
C
$\frac{\rho_1+\rho_2}{2}$
66%
D
$1/2(\rho_1+\rho_2)$
22%
Solution:
$R _{1}=\frac{\rho_{1} l }{ A } R _{2}=\frac{\rho_{2} l }{ A } R + R =\frac{2 \rho l }{ A } \\ \therefore 2 R = R _{1}+ R _{2} \\ 2 \rho=\rho_{1}+\rho_{2} \quad \therefore \rho=\frac{1}{2}\left(\rho_{1}+\rho_{2}\right)$