In the circuit, $E_1=E_2=E_3=2 V$ and $R_1$ and $\mathrm{R}_2=4 \mathrm{~W}$. Then the current flowing through $\mathrm{E}_2$ is

• A

  Zero
• B

  A from $A$ to $B$
• C

  $4 \mathrm{~A}$ from $\mathrm{A}$ to $\mathrm{B}$
• Correct answer
  D

  $2 \mathrm{~A}$ from $\mathrm{B}$ to $\mathrm{A}$

Solution

$$ \begin{aligned} & E_{e q}=\frac{\frac{E_1}{R_1}+\frac{E_3}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}}=\frac{E_1 R_2+E_3 R_1}{R_1+R_2}=\frac{2 \times 4+2 \times 4}{42+4}=2 V \\ & R_{e q}=\frac{R_1 R_2}{R_1+R_2}=\frac{4 \times 4}{4 \times 4}=2 \Omega \\ & -E_2+i R_{e q}-E_{e q}=0 \\ & i=\frac{E_2+E_{e q}}{R_{e q}}=\frac{2+2}{2}=2 A \end{aligned} $$ Hence, the current flowing through $\mathrm{E}_2$ is $2 \mathrm{~A}$ from $\mathrm{A}$ to $\mathrm{B}$

Hence, the correct answer is option: 'D'.