Q. A uniform resistance R and length L is cut into for equal parts, each of length L/4, which are then connected in parallel combination. The effective resistance of the combination will be
A) R
B) 4R
C) $\frac{R}{4}$
D) $\frac{R}{16}$
Explanation
(d) $\ \ \ \ \ \ \ R = \frac{\rho l}{A}$
$or\ \ \ \ R \propto l\ \ $
$\therefore\ \ \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} = \frac{L}{L/4} = 4$
Or $R_{2} = \frac{R}{4}\ \ \ \ \ \ \ \ \left( \because R_{1} = R \right)$
In parallel combination of such four resistances.
$\ \ \ \ \ \ \ \frac{1}{R^{'}} = \frac{1}{R_{2}} + \frac{1}{R_{2}} + \frac{1}{R_{2}} + \frac{1}{R_{2}}$
$or\ \frac{1}{R^{'}} = \frac{1}{R/4} + \frac{1}{R/4} + \frac{1}{R/4} + \frac{1}{R/4}$
$or\ \frac{1}{R^{'}} = \frac{4}{R} + \frac{4}{R} + \frac{4}{R} + \frac{4}{R}$
$or\ \ \frac{1}{R^{'}} = \frac{16}{R}$
$or\ \ R^{'} = \frac{R}{16}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.