Q. There resistances of 4$\ \Omega$ each are connected as shown in figure. If the point D divides the resistance into two equal halves, the resistance between points A and D will be
A) 12$\ \Omega$
B) 6$\ \Omega$
C) 3$\ \Omega$
D) $\frac{1}{3}\Omega$
Explanation
(c) The equivalent circuit is given by
4$\Omega$ and 2$\Omega$ resistances are in series on both sides.
$\therefore 4\Omega + 2\Omega = 6\Omega$
Then 6$\Omega$ and6$\Omega$ resistances are in parallel on both sides
$\frac{1}{R} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$
R=3$\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.