# Multiloop circuit/Kirchoff

1. Nov 12, 2013

### PhizKid

Multiloop circuit/Kirchhoff

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

2. Relevant equations
Kirchhoff's rules

3. The attempt at a solution
These are the 3 paths I decided to take (at random):

So for the first battery, I get:

$2V - 1i_1 - 2i_1 - 4V - 1i_1 = 0$
$-4i_1 = 2V$
$i_1 = -\frac{1}{2}$

2nd:

$4V - 2i_2 - 1i_2 - 4V - 1i_2 = 0$
$- 2i_2 - 1i_2 - 1i_2 = 0$
$- 4i_2 = 0$
$i_2 = 0$

3rd:

$4V - 1i_3 - 2i_3 - 4V - 1i_3 = 0$
$- 1i_3 - 2i_3 - 1i_3 = 0$
$- 4i_3 = 0$
$i_3 = 0$

But these answers are wrong. What did I do incorrectly?

Last edited: Nov 12, 2013
2. Nov 12, 2013

### tiny-tim

Hi PhizKid!

(two h's in kirchhoff! )
sorry, i don't understand what you're doing

(and what 3 paths? i can see 2 loops, and that's it )

3. Nov 12, 2013

### PhizKid

Battery 1, 2, and 3 all have currents that meet at point 'a,' but I can't tell what happens after that when they intersect at that junction.

4. Nov 12, 2013

### tiny-tim

so apply KCL to point 'a'

5. Nov 12, 2013

### PhizKid

So will the junction rule then be $i_1 + i_2 + i_3 = 0$ because all 3 currents are going into this junction, and none are going out?

6. Nov 12, 2013

### tiny-tim

yup!

ok, now do KVL for the left loop and for the right loop (separately)

7. Nov 12, 2013

### PhizKid

The issue I am then having with applying KCL is when I choose my loops. I don't know how many currents there will be, because there are only 3 batteries but 5 resistors. Usually the number of currents are the number of resistors but it's apparently not the case here. (The problem I did before had 2 batteries, 3 resistors, and 3 currents.) What is wrong with the KCL loops I chose in my original post?

8. Nov 12, 2013

### tiny-tim

(you mean KVL … KVL is for loops, KCL is for junctions)

Currents only change at junctions, so you need to assign a different current for each stretch of wire that starts and ends in a junction.

Here, there are only 2 junctions, and 3 wires between them …*so 3 currents.
I can't understand what they are.

Please do KVL for the left loop and for the right loop (separately)

9. Nov 12, 2013

### PhizKid

Is there a reason why I have to choose the left and right loops? Would it be possible to choose the one large loop that goes all the way around the entire circuit?

Anyway, for the left loop, starting at the bottom-left-most corner, going clockwise:

$+2V - 1i_1 + (- 2i_1 - 2i_3 + 2i_2) - 4V - 1i_1 !! = 0$

The part in parenthesis is the part I think I am having trouble with. Here, I assumed that the currents from battery 1 and 3 both go down through the resistor, so they are negative but the current from battery 2 is going up the resistor, so it's positive. Also, I'm not sure which current goes through the resistor closest to the bottom-left.

Right side, starting at the bottom-right-most corner, counter-clockwise:

$-1i_3 !! + 4V - 1i_3 + (- 2i_1 - 2i_3 + 2i_2) - 4V = 0$

The same issue I have here is in the parenthesis, and also the resistor closest to the bottom-right (where I put !!). I'm not sure which current is going through that one, because the 3rd battery is ahead of that resistor.

10. Nov 12, 2013

### tiny-tim

Yes, you can choose any 2 of the three loops.

(there's no point in choosing all 3, since the third equation will be the sum or difference of the other 2)
ah, that's where you're going wrong …

you're assuming that the currents are connected with the batteries, they aren't!

For KVL, you have to add the IR for each resistor.

The I for that resistor is only the i that you marked on the diagram … in this case, it's i2.

11. Nov 12, 2013

### PhizKid

Okay, so if I think I am getting this, the equations should be:

Left loop:
$+2V - 1i_1 - 2i_2 - 4V - 1i_1 = 0$
$-2i_1 - 2i_2 = 2$
$-i_1 - i_2 = 1$

Right loop:
$-1i_3 + 4V - 1i_3 - 2i_2 - 4V = 0$
$-2i_3 - 2i_2 = 0$
$- i_2 - i_3 = 0$

and $i_1 + i_2 + i_3 = 0$

So solving this gives me:

$i_1 = 0$
$i_2 = -1$
$i_3 = 1$

Which doesn't agree with the answers. What did I do incorrectly now?

12. Nov 12, 2013

### tiny-tim

your diagram should always have the currents marked with an arrow, to remind you which way round you chose them.

When you go round the loop, you'll find you're going with some arrows, but against other arrows , and you include them in KVL as minus or plus accordingly.

13. Nov 12, 2013

### PhizKid

I've tried drawing a diagram here:

The signs seem correct to me

14. Nov 12, 2013

### tiny-tim

ah, with that diagram, yes your KVL equations are correct

but your KCL equation needs to be i1 + i3 = i2

15. Nov 12, 2013

### PhizKid

Oh, right

So solving this, I get $i_1 = -\frac{2}{3}$, $i_2 = -\frac{1}{3}$, and $i_3 = \frac{1}{3}$.

I forgot what conclusions I can make if the currents are negative...two of them are negative, but the last one is positive. Does that mean I chose some sort of wrong direction? How do I obtain the correct answers without going through the entire work again trying to guess the correct direction?

16. Nov 12, 2013

### tiny-tim

It doesn't matter …

you're allowed to choose the wrong direction!

everything still works out fine, except of course that the current come out negative instead of positive (or vice versa)

in this case, the question asks you to specify up or down, so work out whether (with your chosen arrows) each positive or negative current corresponds to up or down

(btw, that's why i would have chosen all three arrows to point up … so that i could instantly convert into the correct directions for the answer )

17. Nov 12, 2013

### PhizKid

Intuitively, I don't understand how $i_2$ can be pointing up if $i_1$ and $i_3$ are going into this intersection. It would seem to me that $i_1$ and $i_3$ would combine into some current $i_2$ and go downwards at the junction. Or, if $i_2$ was pointing up, then $i_1$ and $i_3$ would both be going out of the junction, towards the left and right of the circuit drawing, respectively. Would you be able to clarify the analysis on having all 3 currents going into the junction?

Oh wait, I think I understand...this is why battery #2 is there...if it wasn't there, then you couldn't have $i_2$ pointing upwards, right?

Last edited: Nov 12, 2013
18. Nov 12, 2013

### tiny-tim

it can be if i2 is negative
well, you're right of course, and you can do it that way

(in fact, you did do it that way! )

i only said that i would prefer the other way, because of the way this question is worded …

i think it would minimise the chance of my making a mistake at the last moment