Q. A thin wire of resistance 4$\ \Omega$ is bent to form a circle. The resistance across any diameter is
A) 4$\ \Omega$
B) 2$\ \Omega$
C) 1$\ \Omega$
D) 8$\ \Omega$
Explanation
(c) Given that the resistance of the total wire is 4$\Omega$.
Here, ACB(2$\Omega$) and ADB (2$\Omega$) are in parallel.
So, the resistance across any diameter is
$\Longrightarrow \ \ \ \frac{1}{R} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$
$\Longrightarrow \ \ \ R = 1\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.