In the given circuit,E_1 = E_2 = E_3 = 2 , V | vedclass

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(A) The circuit consists of three parallel branches connected between points $A$ and $B$.Branch $1$ has an $EMF$ $E_1 = 2 \, V$ and resistance $R_1 =

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(A) The circuit consists of three parallel branches connected between points $A$ and $B$.Branch $1$ has an $EMF$ $E_1 = 2 \, V$ and resistance $R_1 = 4 \, \Omega$.Branch $2$ has an $EMF$ $E_2 = 2 \, V$ and zero resistance.Branch $3$ has an $EMF$ $E_3 = 2 \, V$ and resistance $R_2 = 4 \, \Omega$.Since all branches are in parallel,the potential difference $V_{AB}$ across each branch is the same.Using Millman's theorem for parallel branches: $V_{AB} = \frac{\sum (E_i / R_i)}{\sum (1 / R_i)}$.Here,$R_2$ (middle branch) is $0 \, \Omega$ (ideal wire),so $V_{AB} = E_2 = 2 \, V$.The current in branch $1$ is $I_1 = (E_1 - V_{AB}) / R_1 = (2 - 2) / 4 = 0 \, A$.The current in branch $3$ is $I_3 = (E_3 - V_{AB}) / R_2 = (2 - 2) / 4 = 0 \, A$.The current in the middle branch is determined by the external circuit connected to $A$ and $B$. If $A$ and $B$ are open,the total current flowing between $A$ and $B$ is $0 \, A$.

In the given circuit,$E_1 = E_2 = E_3 = 2 \, V$ and $R_1 = R_2 = 4 \, \Omega$. What is the current flowing between points $A$ and $B$ (in $, A$)?

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