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 $V_A = E_2 = 2 \, V$.Applying Kirchhoff's Current Law at node

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(A) Let the potential at point $B$ be $V_B = 0 \, V$. Then the potential at point $A$ is $V_A = E_2 = 2 \, V$.Applying Kirchhoff's Current Law at node $A$:$\frac{V_A - E_1}{R_1} + \frac{V_A - E_3}{R_2} + \frac{V_A - E_2}{0} = 0$ is not applicable directly as $E_2$ is a branch.Let $I$ be the current through $E_2$ from $A$ to $B$. The branches with $E_1, R_1$ and $E_3, R_2$ are in parallel between $A$ and $B$.The equivalent $EMF$ of the two parallel branches is $E_{eq} = \frac{E_1/R_1 + E_3/R_2}{1/R_1 + 1/R_2} = \frac{2/4 + 2/4}{1/4 + 1/4} = \frac{1}{0.5} = 2 \, V$.The equivalent resistance is $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{4 	imes 4}{4 + 4} = 2 \, \Omega$.Now,the circuit simplifies to a loop with $E_{eq} = 2 \, V$ and $R_{eq} = 2 \, \Omega$ in series with $E_2 = 2 \, V$.The current $I$ flowing from $A$ to $B$ through $E_2$ is given by $I = \frac{E_{eq} - E_2}{R_{eq}} = \frac{2 - 2}{2} = 0 \, A$.

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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