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(B) Applying Kirchhoff's Voltage Law $(KVL)$ to the loop $E B C D E$ starting from point $E$ and moving in the direction $E \rightarrow B \rightarrow

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$E_1 - (I_1 + I_2) R - I_1 r_1 = 0$

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(B) Applying Kirchhoff's Voltage Law $(KVL)$ to the loop $E B C D E$ starting from point $E$ and moving in the direction $E \rightarrow B \rightarrow C \rightarrow D \rightarrow E$:$1$. Moving from $E$ to $B$ through the branch containing $E_1$ and $r_1$,we encounter the negative terminal of $E_1$ first,so we gain potential $+E_1$. Then,moving in the direction of current $I_1$ through $r_1$,we have a potential drop of $-I_1 r_1$.$2$. Moving from $C$ to $D$ through the resistance $R$,we move in the direction of the current $(I_1 + I_2)$,resulting in a potential drop of $-(I_1 + I_2) R$.$3$. Summing these potential changes around the closed loop gives:$E_1 - I_1 r_1 - (I_1 + I_2) R = 0$Thus,the correct equation is $E_1 - (I_1 + I_2) R - I_1 r_1 = 0$.

An electrical circuit consists of the resistances $R, r_1, R_2$ and voltage sources $E_1$ and $E_2$ as shown in the figure. The correct equation for the loop $E B C D E$ is

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