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(A) Let us assume the current distribution in the given circuit is as shown.Now,we apply Kirchhoff's loop rule in loop $1$ and loop $2$ to get the fol

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$-\frac{3 E}{4 R}$

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(A) Let us assume the current distribution in the given circuit is as shown.Now,we apply Kirchhoff's loop rule in loop $1$ and loop $2$ to get the following equations:In loop $1$,$-i R - (i + i_1) R - 2 E + E = 0$$-2 i R - i_1 R = E$$i_1 R + 2 i R = -E$ ... $(i)$In loop $2$,$-3 E + i_1 R + \frac{i_1}{2} R + i_1 R + 2 E + (i + i_1) R = 0$$\frac{7}{2} i_1 R + i R = E$$7 i_1 R + 2 i R = 2 E$ ... (ii)Now,$7 	imes$ eq. $(i)$ - eq. (ii) gives:$7(i_1 R + 2 i R) - (7 i_1 R + 2 i R) = 7(-E) - 2 E$$7 i_1 R + 14 i R - 7 i_1 R - 2 i R = -9 E$$12 i R = -9 E$$i = -\frac{9 E}{12 R} = -\frac{3 E}{4 R}$

The current $i$ in the circuit given below is

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