Q117 / 155 · WBJEE 2016 · Physics
Two wires of same radius having lengths $l_{1}$
and $I_{2}$ and resistivities $\rho_{1}$ and $\rho_{2}$ are connected in series. The equivalent resistivity will be
- A$\frac{\rho_{1} l_{2}+\rho_{2} I_{1}}{\rho_{1}+\rho_{2}}$
- Correct answerB$\frac{\rho_{1} l_{1}+\rho_{2} /_{2}}{l_{1}+l_{2}}$
- C$\frac{\rho_{1} l_{1}-\rho_{2} /_{2}}{l_{1}-l_{2}}$
- D$\frac{\rho_{1} l_{2}+\rho_{2} l_{1}}{l_{1}+l_{2}}$
Solution
$R=\frac{\rho l}{A}$
where, $\rho=$ resistivity $l=$ length of the resistance wire $A=$ area of cross-section
When the wires are connected in senes, then
$$
\begin{aligned}
R &=R_{1}+R_{2} \\
R &=\frac{\rho_{1} l_{1}}{A_{1}}+\frac{\rho_{2}l_{2}}{A_{2}} \\
R &=\frac{\rho_{1} l_{1}}{A}+\frac{\rho_{2}l_{2}}{A} \\
\end{aligned}
$$
$\mathrm{\rho}_{\mathrm{eq}}=$ resistivity of combination
$$
\begin{array}{l}
I=I_{1}+I_{2} \\
A=A
\end{array}
$$
Thus, $\rho_{\mathrm{eq}} \frac{I_{1}+I_{2}}{A}=\frac{\rho _{1}l_{1}+\rho_{2} l_{2}}{A}$
$$
\rho_{e q}=\frac{\rho_{1}l_{1}+\rho_{2} l_{2}}{I_{1}+I_{2}}
$$
Hence, the correct answer is option: 'B'.