Q. The resistance is connected as shown in the figure below. Find the equivalent resistance between the points A and B.
A) 205$\Omega$
B) 10$\ \Omega$
C) 3.5$\ \Omega$
D) 5$\ \Omega$
Explanation
(d) Net resistance of 3$\Omega$ and 7$\Omega$ resistors (in series)
R’=3+7=10$\Omega$
R’ and 10$\Omega$ are in parallel, so
$R^{'}\mathbf{=}\frac{10 \times 10}{10 + 10} = 5\Omega$
$R^{'}and\ 5\Omega\ $are in series, so
R’=5+5=10$\Omega$
Now, R’ and 10$\Omega$ are in parallel, so
$R = \frac{10 \times 10}{10 + 10} = 5\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.