In the circuit shown here,E_1 = E_2 = E_3 = 2 | vedclass

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(A) Let the potential at point $B$ be $V_B = 0 \, V$. Then the potential at point $A$ is $V_A = V$. Applying Kirchhoff's Current Law $(KCL)$ at node $

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zero

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(A) Let the potential at point $B$ be $V_B = 0 \, V$. Then the potential at point $A$ is $V_A = V$. Applying Kirchhoff's Current Law $(KCL)$ at node $A$:$\frac{V_A - E_1}{R_1} + \frac{V_A - E_2}{0} + \frac{V_A - E_3}{R_2} = 0$ Since the branch containing $E_2$ has no resistance, the potential difference across it is fixed by the battery $E_2$. Thus, $V_A - V_B = E_2 = 2 \, V$. Now, calculate the currents in the other branches:$I_1 = \frac{V_A - E_1}{R_1} = \frac{2 - 2}{4} = 0 \, A$$I_3 = \frac{V_A - E_3}{R_2} = \frac{2 - 2}{4} = 0 \, A$Applying $KCL$ at node $A$: $I_1 + I_2 + I_3 = 0$, where $I_2$ is the current flowing from $A$ to $B$ through $E_2$.$0 + I_2 + 0 = 0 \implies I_2 = 0 \, A$. Therefore, the current flowing through battery $E_2$ is zero.

In the circuit shown here,$E_1 = E_2 = E_3 = 2 \, V$ and $R_1 = R_2 = 4 \, \Omega$. The current flowing between points $A$ and $B$ through battery $E_2$ is

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