Q. A uniform wire of resistance $9\ \Omega$ is cut into 3 equal parts. They are connected in the form of equilateral triangle $ABC.$ A cell of $e.m.f.$ $2\ V$ and negligible internal resistance is connected across $B$ and $C.$ Potential difference across $AB$ is
A) $1\ V$
B) $2\ V$
C) $3\ V$
D) $0.5\ V$
Explanation
(a) The circuit can be drawn as follows
Equivalent resistance $R = \frac{3 \times (3 + 3)}{3 + (3 + 3)} = 2\Omega$
Current $i = \frac{2}{2} = 1A$. So, $i_{1} = 1 \times \left( \frac{3}{3 + 6} \right) = \frac{1}{3}A$
Potential difference between $A$ and $B = \frac{1}{3} \times 3 = 1volt$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.