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Q.
Two wires of same radius having lengths $l_{1}$ and $l_{2}$ and resistivities $\rho_{1}$ and $\rho_{2}$ are connected in series. The equivalent resistivity will be
Current Electricity
Solution:
In series equivalent resistance $R=R_{1}+R_{2}$
$R_{1}=\frac{\rho_{1} l_{1}}{A}, R_{2}=\frac{\rho_{2} l_{2}}{A}$
Hence, $R=\frac{\rho\left(l_{1}+l_{2}\right)}{A}$
$\therefore \frac{\rho\left(l_{1}+l_{2}\right)}{A}=\frac{\rho_{1} l_{1}}{A}+\frac{\rho_{2} l_{2}}{A}$
$\Rightarrow \rho=\frac{\rho_{1} l_{1}+\rho_{2} l_{2}}{l_{1}+l_{2}}$