Question

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 $2V$ and negligible internal resistance is connected across $B$ and $C$. Potential difference across $AB$ is

• A. $1V$
• B. $2V$
• C. $3V$
• D. $0.5V$

Answer 1

The correct option is A $1V$

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 $B=\frac{1}{3}\times 3=1$ volt.