Q. Two wires of the same material and equal length are joined in parallel combination. If one of them has half the thickness of the other and the thinner wire has a resistance of $8\ ohms,$ the resistance of the combination is equal to
A) $\frac{5}{8}\ ohm$
B) $\frac{8}{5}\ ohm$
C) $\frac{3}{8}\ ohm$
D) $\frac{8}{3}\ ohm$
Explanation
(b) $\rho -$ same, $l -$ same, $A_{2} = \frac{1}{4}A_{1}\left\lbrack as\ r_{2} = \frac{r_{1}}{2} \right\rbrack$
By using $R = \rho\frac{l}{A} \Rightarrow \frac{R_{1}}{R_{2}} = \frac{A_{2}}{A_{1}} \Rightarrow \frac{R_{1}}{8} = \frac{1}{4} \Rightarrow R_{1} = 2\Omega$
Hence, $R_{eq} = \frac{R_{1}R_{2}}{R_{1} + R_{2}} = \frac{2 \times 8}{(2 + 8)} = \frac{8}{5}\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.