Q. What is the equivalent resistance between $A\ and\ B$ in the given circuit?

A) $4\Omega$
B) $2\Omega$
C) $\frac{8}{3}\Omega$
D) $\frac{3}{8}\Omega$
Explanation

(c) $R_{1}$ and $R_{2}$ are in series
$\therefore\ \ R_{12} = R_{1} + R_{2} = 4\Omega$
$R_{12}and\ R_{3}$ are in parallel
$R_{123} = \frac{R_{3} \times R_{12}}{R_{3} + R_{12}}$
$\ \ \ \ \ \ \ \ \ \ = \frac{4 \times 4}{4 + 4}$
$\ \ \ \ \ \ \ \ \ \ = 2\Omega$
$R_{123}$and $R_{4}$ are in series
$\therefore\ R_{1234} = R_{123} + R_{4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\Omega + 2\Omega$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4\Omega$
$R_{1234}$ and $R_{5}$ are in parallel
$\therefore R_{12345} = 2\Omega$
$R_{12345}$ and $R_{6}$ are in series
$2\Omega + 2\Omega = 4\Omega$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = R_{123456}$
Now, $R_{123456}$and $R_{7}$are in parallel
$\therefore R_{comb} = \frac{4 \times 8}{4 + 8}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{32}{12} = \frac{8}{3}\Omega$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.