Q. In the circuit shown here, $E_{1} = E_{2} = E_{3} = 2V$ and $R_{1} = R_{2} = 4\ ohm.$ The current flowing between points $A$ and $B$ through battery $E_{2}$ is
A) Zero
B) $2\ amp$ from $A$ to $B$
C) $2\ amp\ $from $B$ to $A$
D) None of the above
Explanation
(b) The given circuit can be redrawn
$E_{eq} = \frac{E_{1}R_{2} + E_{3}R_{1}}{R_{1} + R_{2}} = \frac{2 \times 4 + 2 \times 4}{4 + 4} = 2V$ and
$R_{eq} = \frac{4}{2} = 2\Omega$. Current $i = \frac{2 + 2}{2} = 2A$ from $A$ to $B$ through $E_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.