For the circuit given below,the Kirchhoff's l | vedclass

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(C) To apply Kirchhoff's Voltage Law $(KVL)$ to the loop $BCDEB$,we traverse the loop starting from point $B$ in the clockwise direction.$1$. Moving f

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$i_2 R_2 + E_2 - E_3 - i_3 R_1 = 0$

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(C) To apply Kirchhoff's Voltage Law $(KVL)$ to the loop $BCDEB$,we traverse the loop starting from point $B$ in the clockwise direction.$1$. Moving from $B$ to $C$ through resistor $R_2$ in the direction of current $i_2$,the potential drop is $-i_2 R_2$.$2$. Moving from $C$ to $D$ through battery $E_2$,we go from the positive terminal to the negative terminal,so the potential change is $-E_2$.$3$. Moving from $D$ to $E$ through battery $E_3$,we go from the negative terminal to the positive terminal,so the potential change is $+E_3$.$4$. Moving from $E$ to $B$ through resistor $R_1$ against the direction of current $i_3$,the potential change is $+i_3 R_1$.Summing these changes to zero:$-i_2 R_2 - E_2 + E_3 + i_3 R_1 = 0$Multiplying by $-1$ gives:$i_2 R_2 + E_2 - E_3 - i_3 R_1 = 0$

For the circuit given below,the Kirchhoff's loop rule for the loop $BCDEB$ is given by the equation:

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