Question

Two wires of equal diameters, of resistivities $\rho _{1}$ and $\rho _{2}$ and length $l_{1}$ and $l_{2}$ , respectively are joined in series. The equivalent resistivity of the combination is

• A. $\frac{\rho _{1} l_{1} + \rho _{2} l_{2}}{l_{1} + l_{2}}$
• B. $\frac{\rho _{1} l_{2} + \rho _{2} l_{1}}{l_{1} - l_{2}}$
• C. $\frac{\rho _{1} l_{2} + \rho _{2} l_{1}}{l_{1} + l_{2}}$
• D. $\frac{\rho _{1} l_{1} + \rho _{2} l_{2}}{l_{1} - l_{2}}$

Answer 1

Answer: (A)

Wires having frame diameter, hence area of cross-section will be equal,
i.e. $A_{1}=A_{2}=A$
$R_{1}=\rho _{1}.\frac{l_{1}}{A}$
and $R_{2}=\rho _{2}.\frac{l_{2}}{A}$
Since, both resistance are joined in series.
Hence, $R_{e q}=R_{1}+R_{2}$
$=\rho_1 \cdot \frac{l_1}{A}+\rho_2 \cdot \frac{l_2}{A} $
$ R_{e q}=\frac{1}{A}\left(\rho_1 l_1+\rho_2 l_2\right) $
$ \rho_{e q} \cdot \frac{\left(l_1+l_2\right)}{A}=\frac{1}{A}\left(\rho_1 l_1+\rho_2 l_2\right) $
$ \Rightarrow \rho_{e q}=\frac{\rho_1 l_1+\rho_2 l_2}{l_1+l_2}$