# Need help applying Kirchhoff's voltage law

1. Feb 22, 2013

### InvalidID

I made a circuit that matches the attached figure. Then, I measured values of voltage and current and I ended up with the attached table. Now I'm applying KVL to both loops.

For the first loop:
$$-23.9+13.981+10.120≅0\\ 0.201≅0$$
For the second loop:
$$10.120+4.118+6≅0$$
What am I doing wrong for the second loop?

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2. Feb 22, 2013

### Staff: Mentor

Mark the polarities of the voltages you measured on the circuit diagram. When you do a "KVL walk" around a loop, take into account these polarities; does the potential drop or rise when you "walk over" a given component along your path?

3. Feb 22, 2013

### tiny-tim

Hi InvalidID!
You haven't drawn any arrows on your diagram, to show the direction of the current.

So how do you know whether to use plus or minus for the voltage drops across each resistor?

Apply KCL to node 2

4. Feb 22, 2013

### InvalidID

I thought resistors don't have polarities?

I assumed that the current flows in clockwise direction.

KCL applied to node 2:

0.601+4.594=5.195
5.195=5.195

Edit: I think I might be mixing up mesh analysis with KVL. You can only apply KVL to the large loop, right? You can't apply it to the individual meshes, correct?

Last edited: Feb 22, 2013
5. Feb 22, 2013

### tiny-tim

Hi InvalidID!
You can apply KVL to any loop.

In this case, you can apply KVL to all three loops (the outside one, and the two small ones), but the KVL equation for the two small ones will add up to the KVL equation for the large one, so you only have two independent KVL equations.

ok, now do KVL for two loops, and draw in those arrows!!

6. Feb 22, 2013

### Staff: Mentor

Resistors themselves do not have polarities. However, when a current flows through a resistor, the potential drops in the direction of current flow. You may have noticed when you were measuring voltages across the resistors that you would see a positive value with the meter leads placed in one orientation, and a negative value (same magnitude) if the leads were reversed. So when you measure the voltage across a resistor, you should take note of the polarity you see since that will also tell you the direction that the current is flowing.
KVL can be applied around any closed path.

7. Feb 23, 2013

### InvalidID

I've applied KVL to all 3 loops but the equations of the smaller loops don't add up to the the equation of the larger loop.

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8. Feb 23, 2013

### Staff: Mentor

KVL states that the sum of the potential changes around a closed path (a loop) is zero. It doesn't say anything about a sum of the equations derived from this property.

9. Feb 23, 2013

### InvalidID

So I suppose you are disagreeing with tiny-tim? Did I setup the equations correctly?

10. Feb 23, 2013

### Staff: Mentor

I don't think I'm disagreeing with tiny-tim; the circuit admits two independent loops, since two loops (chosen appropriately) are sufficient to include every component of the circuit at least once. This means that if you use a third loop, its equation will be linearly dependent (mathematically speaking) on the other two. A straight sum of terms from two of the equations usually will not result in the third equation; Some scaling of the equations might be required (multiplication by constant values).

Last edited: Feb 23, 2013
11. Feb 23, 2013

### InvalidID

Alright, but if I input the values into the KVL equation for the loop on the right, I get:

$$10.120+4.118+6≅0$$

which isn't correct. :S

12. Feb 23, 2013

### Staff: Mentor

Then you have a sign issue with the terms. Did you mark in the polarities of the potential drops due to the currents and take them into account when you wrote the KVL expression?

13. Feb 23, 2013

### tiny-tim

Hi InvalidID!
No you haven't, you've written equations like E - R1 - R2 = 0.

Sorry, but that is nothing like Kirchhoff's law.

KVL requires you to add the potential drops across all the components in the loop (and the emf).

The potential drop is IR, not R, and you multiply it by 1 or -1 depending on the direction of the current.

Write i1 i2 and i3 on your diagram, with arrows specifying a direction for each current, then write out the three loop equations.​
With 3 loops, the sum or difference of 2 KVL equations will always equal the third.

(so long as you don't multiply one by a factor for no particular reason)

14. Feb 23, 2013

### InvalidID

I think I got it now. Is this correct?

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15. Feb 23, 2013

### tiny-tim

yes, that looks ok now

always draw the arrows for KVL like that!

(i think you were getting confused with mesh analysis, where there's one circular arrow for the whole of each loop)

16. Feb 23, 2013

### InvalidID

Kind of embarrassing question, but I forget why VE is negative in the main loop.

17. Feb 23, 2013

### tiny-tim

why shouldn't it be?

if you go clockwise round both the left and the main loop, why would you make the two VEs different?

18. Feb 23, 2013

### InvalidID

Well, we would want to be consistent so they would either both be positive or negative. We're going from the negative end to the positive end of the battery but why does that give negative voltage? Because the positive end is 15V higher than the negative end? So going from negative to positive, it would be -15V?

Also, VR2 is negative in the right loop because we're going in the clockwise direction when doing KVL which is in the opposite direction of i2, right?

19. Feb 23, 2013

### tiny-tim

i can never remember which way round the battery works!

but i'm not taking the exams, so i don't need to!

you'll just have to remember it
(btw, you can use the X2 button twice: VR2: isn't that cute? )

yes, i2 is positive in one loop but negative in the other loop, because the arrows go opposite ways

(you get the same thing in mesh analysis: the arrows always go different ways for any section that's in two loops)