# Finding the linear system of equations of a circuit using Kirchoff's rules

I have the following circuit and I am trying to find the linear system of equations using Kirchoff's rules

For the 2 nodes, I can determine that the equations for both are $$I_1 + I_2 - I_3 = 0$$ For the left loop, I can determine that the equation is $$-R_1*I_2 = E_0$$, but according to my textbook, the overall loop is $$R_2 *I_3= E_0$$. I don't understand this because I thought that the overall loop would be $$-R_2 *I_3= E_0$$ since the current flow is clockwise. And the right loop would be $$-R_3*I_3 + R_1*I_2 = 0$$. Can someone help me understand this? Thanks!

• The convention is that the current goes from a higher voltage potential to a lower voltage potential. So just as $E_0 = - R_1 I_2$, we have $E_0 = R_2 I_3$. Look at the direction of the currents $I_2$ and $I_3$ through each resistor. The equations the text gives are consistent.
– mjw
Jun 13 at 22:17

In this case, as you move clockwise around the outside loop, you've got an increase in potential over the battery of $$\Delta V_1 = E_0$$, and a decrease in potential over the resistor of $$\Delta V_2 = -I_3R_2$$. If you add these up, you have to have no net change in potential:
$$\sum\Delta V = 0$$ $$\Rightarrow \Delta V_1 + \Delta V_2 = 0$$ $$\Rightarrow E_0 - I_3R_2 = 0$$ $$\Rightarrow E_0 = I_3R_2 .$$
\begin{aligned} I_1+I_2-I_3&=0\\ R_1 I_2 &= -E_0 \\R_2 I_3 &= E_0 \end{aligned}