Question

In the circuit shown here, $E_{1}=E_{2}=E_{3}=2 \, V$ and $R_{1}=R_{2}$ $=4$ ohms. The current flowing between points $A$ and $B$ through battery $E_{2}$ is

• A. zero
• B. $2 A$ from $A$ to $B$
• C. $2 A$ from $B$ to $A$
• D. None of the above

Answer 1

Answer: (B)

The given circuit can be redrawn as:

$E_{ eq }=\frac{E_{1} R_{2}+E_{2} R_{1}}{R_{1}+R_{2}}=\frac{2 \times 4+2 \times 4}{4+4}=2 \, V$ and
$R_{ eq }=\frac{4}{2}=2 \, \Omega$
Current, $i=\frac{2+2}{2}=2 \,A$ from $A$ to $B$ through $E_{2}$.