What is the direction of the currents in each side?

[horsedeg]

Homework Statement

(a) Determine the current in each branch of the circuit.

Homework Equations

V=IR
Summation of the voltages in a loop = 0

The Attempt at a Solution

I have this problem figured out in general, but the only thing confusing me is the current in the middle.
What exactly determines the direction of the current in the middle branch? It screwed me over so bad in a similar problem on a test. The current in the middle is pointing down, but why?

 

[Student100]

Homework Statement

(a) Determine the current in each branch of the circuit.

Homework Equations

V=IR
Summation of the voltages in a loop = 0

The Attempt at a Solution

I have this problem figured out in general, but the only thing confusing me is the current in the middle.
What exactly determines the direction of the current in the middle branch? It screwed me over so bad in a similar problem on a test. The current in the middle is pointing down, but why?

When you write out the system of equations from each branch and solve, you should be able to determine the direction of current from the magnitude (whether they're positive or negative.) A negative magnitude means you've chosen the wrong direction of current initially.

So what were your system of equations?

 

[horsedeg]

Wait, so if I do the systems of equations it'll just be negative?
So the current is down, but let's assume I didn't know and I just guessed that it was up.

For the right loop: 16 - I₁(R₂+3) + 6I₂ - 4 = 0
For the left loop: 16 - I₁(R₂+3) - R₁I₁

Is this correct? Not 100% sure about the sign on the 4.00 V one. So hypothetically, let's say I got both currents from this equation. Would both of them be "right" but just negative?

 

[Student100]

Wait, so if I do the systems of equations it'll just be negative?
So the current is down, but let's assume I didn't know and I just guessed that it was up.

Yes, nothing more tricky about it in these cases. Assuming you set it up correctly. An intial wrong guess will give you the same magnitude of current, with the sign being negetive.

For the right loop: 16 - I₁(R₂+3) + 6I₂ - 4 = 0
For the left loop: 16 - I₁(R₂+3) - R₁I₁

Is this correct? Not 100% sure about the sign on the 4.00 V one. So hypothetically, let's say I got both currents from this equation. Would both of them be "right" but just negative?

You have a junction, so you would need three equations for this system (there are currents $i_1,i_2,i_3$)

Take a look at the example on Wikipedia and then try again:

https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

 

[horsedeg]

Yes, nothing more tricky about it in these cases. Assuming you set it up correctly. An intial wrong guess will give you the same magnitude of current, with the sign being negetive.
You have a junction, so you would need three equations for this system (there are currents $i_1,i_2,i_3$)

Take a look at the example on Wikipedia and then try again:

https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

Right, I forgot I₁=I₂+I₃. That's just kind of in the back of my mind.

So I could use this to get ALL the values? Because on the test it wasn't working. I was doing something wrong. I was trying all these combinations of currents. Maybe I just was doing it completely wrong. Considering V in the diagram is 16V, could you confirm that I set it up right? Because if I didn't then that might be why. One thing I'm not sure about is how the sign of that voltage (4V) in the middle would change if the currents were in different directions.

 

[Student100]

Right, I forgot I₁=I₂+I₃. That's just kind of in the back of my mind.

So I could use this to get ALL the values? Because on the test it wasn't working. I was doing something wrong. I was trying all these combinations of currents. Maybe I just was doing it completely wrong. Considering V in the diagram is 16V, could you confirm that I set it up right? Because if I didn't then that might be why.

So it probably didn't work out because you'd set up the system of equations wrong. I won't do it for you, but I help get you started:

In this configuration, $i_1+i_2=i_3$, now do your loops, and ensure you're using $i_3$ for $R_1$ and the directions according to the picture.

After you've found the system, plug it into a matrix or solve the system by substitution. $i_2$ should be negative, if $V = 16V$ and $R_1$ and $R_2$ satisfy the equations in such a way that this is true.

Edit: To help further, you should get something like the sign for $i_2$ depends on $R_2-3R_1+3$ assuming I did the math right.