In the circuit,E_1 = E_2 = E_3 = 2 text{ V} a | vedclass

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(A) Let $V_A$ and $V_B$ be the potentials at points $A$ and $B$ respectively. Let $V_A - V_B = V$. Applying Kirchhoff's current law at node $A$,the su

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(A) Let $V_A$ and $V_B$ be the potentials at points $A$ and $B$ respectively. Let $V_A - V_B = V$. Applying Kirchhoff's current law at node $A$,the sum of currents leaving the node is zero: $\frac{V_A - V_B - E_1}{R_1} + \frac{V_A - V_B - E_2}{0} + \frac{V_A - V_B - E_3}{R_2} = 0$. Since the branch containing $E_2$ has no resistance,the potential difference across it is $V_A - V_B = E_2 = 2 	ext{ V}$. Now,calculate the current in the top branch $(I_1)$: $I_1 = \frac{V_A - V_B - E_1}{R_1} = \frac{2 - 2}{4} = 0 	ext{ A}$. Calculate the current in the bottom branch $(I_3)$: $I_3 = \frac{V_A - V_B - E_3}{R_2} = \frac{2 - 2}{4} = 0 	ext{ A}$. Applying Kirchhoff's current law at node $A$: $I_{E_2} + I_1 + I_3 = 0 \Rightarrow I_{E_2} + 0 + 0 = 0 \Rightarrow I_{E_2} = 0$. Thus,the current flowing through $E_2$ is zero.

In the circuit,$E_1 = E_2 = E_3 = 2 \text{ V}$ and $R_1 = R_2 = 4 \text{ } \Omega$. Then the current flowing through $E_2$ is

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