# Homework Help: Loop Current Method/Loop Analysis Confusion

1. Sep 5, 2016

### Marcin H

1. The problem statement, all variables and given/known data

2. Relevant equations
Loob Analysis
V=IR

3. The attempt at a solution
So I think I have the right equation here, but I am a bit confused on some parts. I am not too sure why i2=i1-ix. The way my professor explains it doesn't make much sense. Also (TYPO) in the second equation, the +2(i2) term should be 2(-i2) but I am not sure what the reasoning behind that is. This is my first time with loop analysis and it's not clicking yet, so could someone break down this equation bit by bit and explain it? The main part I am confused about is the branch in the middle with the 2ohm resistor and finding the current equation like i2=i1-ix. The rest makes sense.

2. Sep 5, 2016

### phinds

This is extremely straight forward, nothing strange about it. You have defined i1 as being a current that goes down the middle path and i2 as the total current going down the middle path. Doesn't seem like there should be anything confusing about the fact that i1 therefore makes a positive contribution to i2. ix on the other had goes up the middle path so makes a negative contribution to i2.

Your loop equations are correct, so just solve them and be done with it.

3. Sep 5, 2016

### Marcin H

In the second equation should it be -2V -3ix -2(i2)+3ix=0 or should that be + (2i2)?

4. Sep 5, 2016

### phinds

Your loop equations are correct, so just solve them and be done with it.

5. Sep 5, 2016

### Marcin H

Can you explain why it is positive then? In equation 2 when you go around the loop and get to 2ohm resistor its polarities are - to + and the way the circuit is drawn you will hit the minus sign first making it negative. I don't understand why it's positive?

6. Sep 5, 2016

### phinds

Resistors don't HAVE polarities. You assign polarity of the voltage in the loop equation based on the designated current direction. Current always enters a resistor at the positive side of the assigned voltage. If your calculated current turns out to be negative then the polarity of the voltage is opposite to what you assigned it in the loop. That is, the + or - assigned to the voltage on a resistor in the loop equation is a fiction until you have solved the equations at which point you know whether it is correct or the opposite of what you have assigned.

You have essentially multiplied your second loop equation by -1 which is irrelevant to the solution, thus my statement that your loop equations are correct.

7. Sep 5, 2016

### Staff: Mentor

Error in eqn (2): in your loop the contribution of the independent source is +2V

8. Sep 5, 2016

### Marcin H

I started my loop in the top right corner, so wouldn't that be -2V? - to +?

9. Sep 5, 2016

### Staff: Mentor

Top right is an odd place to start, but is manageable.

....and then you summed voltage rises proceeding clockwise or CCW?

Going CW: +2 + (3ix + 2i2) + -3ix = 0 ⏪ grouping central branch voltages for clarity

(There is room for confusion because there are two different 3ix terms in the loop. It may have been clearer had the examiner made the 3Ω resistor into a 4Ω.)

Last edited: Sep 5, 2016
10. Sep 5, 2016

### Marcin H

I went like the arrow in my picture. From top right and down CW. And it's going - to + for the 2V source which is a voltage rise right? Voltage rise is negative right?

11. Sep 5, 2016

### Staff: Mentor

When adding voltages going CW they are all considered to be "voltages rises", even the negative ones.

When you go from - to +2V this is a voltage rise of +2V.

12. Sep 5, 2016

### Marcin H

Are you sure? Then what is the reason the 5V is negative in the first equation? Assuming I started bottom left corner... Is that also wrong? I have always been taught if you are going in a loop and hit a minus sign first then that element in the circuit will be negative in your loob equation.

13. Sep 5, 2016

### Staff: Mentor

Eqn (1) is what you'd write if you were summing the voltage drops, going clockwise. This also would be valid, though you do need to be scrupulously consistent in keeping to voltage rises or voltage drops.

14. Sep 5, 2016

### Staff: Mentor

∑ voltage rises going counter-clockwise: -2 + 3ix + (-2i2 - 3ix) = 0
the bracketted terms represent the central vertical branch

15. Sep 5, 2016

### Marcin H

If you go clockwise in the loop on the left (starting in the bottom left hand corner going up) then you would have a voltage rise, - to + which is NEGATIVE (EDIT*). I am just adding these all up and setting them equal to 0. Voltage rises (- to +) is negative and voltage drop (+ to -) is positive. Maybe @phinds can chime in on this. He said that my equations are correct. But I am still a bit confused about the 2(i2) part in my second equation. I still think that it is negative because i2 has to be travelling towards ground, but because of my (CCW) loop i am going against the current in that branch which would make it negative.

Last edited: Sep 5, 2016
16. Sep 5, 2016

### Staff: Mentor

This seems inconsistent.

You need to state your starting point, the loop's direction (CW or CCW), and whether you are summing voltage rises or voltage drops.

17. Sep 5, 2016

### Marcin H

Sorry that was a typo. Fixed now.

And I am summing both. The sum of Vrise has to equal the sum of Vdrop. Or in other wards the sum of V in the circuit has to be 0.

18. Sep 5, 2016

### phinds

OOPS ... you are correct. I confused myself by failing to recognize that I was using a polarity rule based on the direction of Ix, but doing it for a resistor with a current flowing in the opposite direction and adding in a voltage based on that current, not the loop current. I never use something like I2 but rather always express it as Ix - I1 so as to avoid this unneeded complication with its additional possibilities for errors such as the one I just made. If you avoid anything like I2 and just ALWAYS use only the loop currents in each loop you can't fall into this trap.

19. Sep 5, 2016

### Marcin H

Ok good. This was driving me crazy. Can you elaborate on the reasoning for that branch again. So if you are looking at the left loop i1 you can say i1-ix because ix is opposing it. For loop 2 it would be ix -i1 right? Is that the correct way to think about it.

20. Sep 5, 2016

### phinds

Yes, but really, the way to think about everything is to just use the loop currents in each loop and don't get hung up on the kind of issue you're having here. Forget trying to merge the currents. If a branch has two of the loop currents flowing through it, then use the correct sign for each element based on the current direction, as follows:

The important thing to do is always consider all voltages to be positive if their + is where the current enters. For resistors this is trivial but for sources you have to use the - sign if the current enters at the - terminal. Further, if you have, for example, a second loop current involved in an element such as is the case here with the 2ohm resistor, you consider the contribution of that loop current to be the same sign as that of the current in the loop you are writing the equation for if it flows in the same direction but opposite if it is flowing opposite. Thus in this example, your loop 1 summation for the 2 ohm resistor would be 2I1 - 2Ix and for your loop X summation, it would be 2Ix - 2I1

EDITED to get that last bit with the correct terminology

Last edited: Sep 5, 2016
21. Sep 5, 2016

### Staff: Mentor

Quite so. The arrow labelled i2 would be better labelled (i1 - ix), eliminating the unwanted clutter of unnecessary variables.