Two wires of equal diameters, of resistivities \( \rho_1 \) and \( \rho_2 \) and lengths \( l_1 \) and \( l_2 \) respectively, are joined in series. The equivalent resistivity of the combination is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \frac{\rho_1 l_1+\rho_2 l_2}{l_1+l_2} \)
Wires of equal diameter have the same cross-sectional area \( A \). In series their resistances add, and the combination behaves as one wire of length \( l_1+l_2 \), area \( A \) and an equivalent resistivity \( \rho \).
The sum of the two resistances is:
\( \displaystyle R_1+R_2=\frac{\rho_1l_1}{A}+\frac{\rho_2l_2}{A} \)
The single equivalent wire has:
\( \displaystyle R_{\mathrm{eq}}=\frac{\rho\,(l_1+l_2)}{A} \)
Equating the two:
\( \displaystyle \frac{\rho\,(l_1+l_2)}{A}=\frac{\rho_1l_1+\rho_2l_2}{A} \)
\( \displaystyle \Rightarrow\; \rho=\frac{\rho_1l_1+\rho_2l_2}{l_1+l_2} \)