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(C) Given that no current flows through the battery with emf $E_1 = 2 \ V$,we have $I_1 = 0 \ A$. In the circuit,the branch containing $E_2 = 5 \ V$ a

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$V_2 = -3 \ V, I_3 = 1 \ A$

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(C) Given that no current flows through the battery with emf $E_1 = 2 \ V$,we have $I_1 = 0 \ A$. In the circuit,the branch containing $E_2 = 5 \ V$ and $R_2$ is in parallel with the branch containing $R_3 = 2 \ \Omega$. Since $I_1 = 0$,the potential difference across the parallel combination is equal to the emf $E_1 = 2 \ V$. Thus,the voltage across $R_3$ is $V_{R3} = 2 \ V$. The current $I_3$ through $R_3$ is given by $I_3 = \frac{V_{R3}}{R_3} = \frac{2 \ V}{2 \ \Omega} = 1 \ A$. Now,applying Kirchhoff's voltage law in the middle branch,the potential difference across the branch is $V_{AB} = 2 \ V$. We have $E_2 - I_3 R_2 = V_{AB}$,where $E_2 = 5 \ V$. $5 \ V - (1 \ A) 	imes R_2 = 2 \ V \Rightarrow R_2 = 3 \ \Omega$. The voltage $V_2$ across $R_2$ is the potential drop across the resistor $R_2$,which is $V_2 = -I_3 R_2 = -1 \ A 	imes 3 \ \Omega = -3 \ V$. Therefore,$V_2 = -3 \ V$ and $I_3 = 1 \ A$.

If the resistances are chosen for the circuit shown in the figure in such a way that no current flows through the battery with emf $E_1 = 2 \ V$,the voltage $V_2$ across $R_2$ and the current $I_3$ flowing through $R_3$ are respectively,

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