Direction of current in a "middle" branch?

[DoobleD]

Consider the following circuit (arrows represent current direction) :

How does one determine the direction of the current in the middle branch ?

Would it be correct to assume that :

- if emf 1 > emf 2, current directionin middle branch is up,
- if emf 2 > emf 1, current direction in middle branch is down,
- if emf 1 = emf 2, current in middle branch is 0 (neither up or down) ?

 

[.Scott]

Since R1 = R3, that is correct.

 

[DoobleD]

Since R1 = R3, that is correct.

Thanks !

 

[CWatters]

You can't always predict which direction the current is going in the outer loops.

What you do is define a direction as +ve. Then write KVC and/or KVL equations (simultaneous equations). Then you solve them to find the current in each branch or loop. If one or more answers turn out to be negative then the current is actually flowing the other way.

 

[DoobleD]

You can't always predict which direction the current is going in the outer loops.

Even in the case above where the batteries are in "opposite" directions (emf 1 has positive down and emf 2 has positive up) ? How could current flow clockwise (in the outer loop) in such a case ?

 

[Dale]

Even in the case above where the batteries are in "opposite" directions (emf 1 has positive up and emf 2 has positive down) ? How could current flow clockwise (in the outer loop) in such a case ?

Write down KVL or KCL and you can actually solve for the conditions where that happens.

 

[DoobleD]

Write down KVL or KCL and you can actually solve for the conditions where that happens.

Ok, I obtain that a clockwise current requires that emf 2 is negative and that its absolute value is superior to the value of emf 1. If that's correct, isn't a negative emf simply mean that the battery has been drawn in the wrong direction on the figure ? Implying that if the figure is correctly drawn correctly, you could actually know the direction of current in the outer loop just by looking at the figure.

Or I simply made maths mistakes. :D

I ended up with the current at R_3 being equal to (emf 1 + emf 2) / (3R). R being the value of one resistor (with R_1 = R_2 = R_3 = R).

 

[Dale]

I obtain that a clockwise current requires that emf 2 is negative and that its absolute value is superior to the value of emf 1.

Excellent, that is the correct general approach.

If that's correct, isn't a negative emf simply mean that the battery has been drawn in the wrong direction on the figure ? Implying that if the figure is correctly drawn correctly, you could actually know the direction of current in the outer loop just by looking at the figure.

Often circuits are drawn wrong. In fact, many times you draw a circuit where you KNOW that at least one of the things you drew was wrong and you just have to do the math to figure out which one is the wrong one.

 

[DoobleD]

Excellent, that is the correct general approach.

Often circuits are drawn wrong. In fact, many times you draw a circuit where you KNOW that at least one of the things you drew was wrong and you just have to do the math to figure out which one is the wrong one.

Wonderful, thanks !