# If the current directions are arbitrarily picked, then why am I getting different results? [closed]

So I have the following circuit, and I have to determine each current going through it $$(I_1, I_2, I_3)$$ using Kirchhoff's laws.

But I'm confused as to how I should pick the current directions. In the diagram above, which was taken from my professor's notes, the direction of $$I_2$$ is opposite to the direction of $$I_1$$. So I thought let me try analyzing the same circuit but changing the direction of $$I_2$$ to be from left to right. Turns out if done this way, the results are not the same as my professor's. So my question is, shouldn't the current values be the same since I was told that the current directions are chosen arbitrarily and they do not make a difference?

• Please, give details about your computation, otherwise it is hard to determine what exactly went wrong. You may also guess it yourself when you try to explain it
– OON
Commented Feb 21, 2023 at 15:48
• You must have made a mistake! For example, if you wanted to say $I_2$ is going the other way, you could just work with the variable $I_2'=-I_2$, and make this replacement in every equation in your professor's solution. Commented Feb 21, 2023 at 15:51
• @David so the mistake doesn't lie in the way I chose the directions? Commented Feb 21, 2023 at 15:52

It doesn't make any physical difference. However, it does make a mathematical difference. Changing the sign of $$I_2$$ will generally require changing several of your formulas. And of course it will change the sign of the answer for $$I_2$$. If you do not make the change carefully you can easily get a mistake.

Here is the solution with the configuration shown:

Voltage Loops:

$$U_1 - I_1 R_1 - I_3 R_3 = 0\\ -U_2 + I_3 R_3 + I_2 R_2 = 0$$

Current Networks:

$$I_1+I_2-I_3=0$$

Solving this gives $$I_1 = \frac{R_2U_1+R_3(U_1-U_2)}{R_1(R_2+R_3)+R_2R_3}$$, $$I_2 = \frac{R_1U_2+R_3(U_2-U_1)}{R_1(R_2+R_3)+R_2R_3}$$, $$I_3=\frac{R_1U_2+R_2U_1}{R_1(R_2+R_3)+R_2R_3}$$.

Now we look at the system again, but with $$I_2$$ going the other direction:

Voltage Loops:

$$U_1 - I_1 R_1 - I_3 R_3 = 0\\ -U_2 + I_3 R_3 - I_2 R_2 = 0$$

Current Networks:

$$I_1-I_2-I_3=0$$

Solving this gives $$I_1 = \frac{R_2U_1+R_3(U_1-U_2)}{R_1(R_2+R_3)+R_2R_3}$$, $$I_2 = -\frac{R_1U_2+R_3(U_2-U_1)}{R_1(R_2+R_3)+R_2R_3}$$, $$I_3=\frac{R_1U_2+R_2U_1}{R_1(R_2+R_3)+R_2R_3}$$.

The difference between the two solutions is the sign on $$I_2.$$ Of course, because we drew $$I_2$$ pointing in different directions, the answer is the same just as expected.