KVL Polarity Help Please


by Physicist3
Tags: polarity
Physicist3
Physicist3 is offline
#1
Nov29-12, 11:26 AM
P: 72
http://www.flickr.com/photos/5351295...n/photostream/

With regards to the circuit in the image linked above, I have been told to start in the bottom left hand corner and then work clockwise around the circuit, adding up all the voltages to see if they follow the KVL principle. The thing is, the arrows and polarity is confusing me a bit. Should it be -Vs + V1 + V2 = 0, or Vs - V1 - V2 = 0? Could someone explain which one is correct and what the arrows represent please? Thanks :)
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Jony130
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#2
Nov29-12, 12:03 PM
P: 389
The Arrow show that voltage is nominally positive at Node 'b' relative to Node 'a'.



Hence


Vba = (Vb - Va)


and

Vab = (Va - Vb)

Therefore

Vba = -Vab



And this two equation are correct and the result will be the same

-Vs + V1 + V2 = 0 Is the same think Vs - V1 - V2 = 0

-Vs + V1 + V2 = Vs - V1 - V2 = 0.

Or

Vs = V1 + V2

And another example
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Runei
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#3
Nov29-12, 12:58 PM
P: 134
Vs - V1 - V2 = 0 ⇔ Vs = V1 + V2 ⇔ 0 = -Vs + V1 + V2

Simple algebra :)

Ratch
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#4
Nov29-12, 12:58 PM
P: 315

KVL Polarity Help Please


Physicist3,

Could someone explain which one is correct and what the arrows represent please?
It does not matter how you designate voltage polarities or current directions. I personally like arrows for current direction and signs to designate the polarity of the voltages. It also does not matter whether you go CW or CCW around the circuit for KVL. As long as you are consistent with everything, the results should be the same.

Ratch
Physicist3
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#5
Nov29-12, 07:20 PM
P: 72
Forgot to add, in the circuit i linked, the current has been said to flow clockwise around the circuit, i.e. following the direction of the arrow across the cell. Because of this current direction, am i right in saying that KVL will look like:

Vs + -V1 + -V2 = 0

because if you start in the bottom left corner, when you pass through the battery going clockwise, you go from -ve to +ve therefore moving from a lower to higher potential, meaning Vs will be positive, where as the resistors have an opposite polarity and you move from +ve to -ve, therefore moving from a higher to a lower potential, making v1 and v2 negative?

is this correct?

Thanks for your help :)
jim hardy
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#6
Nov29-12, 11:35 PM
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When first starting out it's more important that you be consistent in your method than anything else.

I was taught to place signs for voltages and arrows for current, as somebody else said above.
Then imagine yourself inside the circuit walking around the loop. Write down each voltage with the sign you encounter as you enter that circuit element.

So in your diagram, i would add signs as follows:
On battery, minus at bottom and plus at top 'cause that's how it's drawn (memory aid - the long line can be broken in half to make a + sign, the short one is already a - sign).
On upper resistor, plus on left ('cause it's to + of battery), and minus on right of course
On right-hand resistor minus on bottom ('cause it's to - of battery, ) and plus on top, of course.

now starting on lower left and going CW, i'd write:
-VS +V1 +V2 = 0

observe if you go CCW instead you get
-V2 -V1 +VS = 0
which when multiplied on both sides by -1 is the same equation, as you know it should be.

You can show yourself that even if you start someplace else in the loop you'll get the same equation.

That's a simple technique that'll keep you out of trouble and is easily remembered.

old jim
Physicist3
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#7
Nov30-12, 09:54 AM
P: 72
Quote Quote by jim hardy View Post
When first starting out it's more important that you be consistent in your method than anything else.

I was taught to place signs for voltages and arrows for current, as somebody else said above.
Then imagine yourself inside the circuit walking around the loop. Write down each voltage with the sign you encounter as you enter that circuit element.

So in your diagram, i would add signs as follows:
On battery, minus at bottom and plus at top 'cause that's how it's drawn (memory aid - the long line can be broken in half to make a + sign, the short one is already a - sign).
On upper resistor, plus on left ('cause it's to + of battery), and minus on right of course
On right-hand resistor minus on bottom ('cause it's to - of battery, ) and plus on top, of course.

now starting on lower left and going CW, i'd write:
-VS +V1 +V2 = 0

observe if you go CCW instead you get
-V2 -V1 +VS = 0
which when multiplied on both sides by -1 is the same equation, as you know it should be.

You can show yourself that even if you start someplace else in the loop you'll get the same equation.

That's a simple technique that'll keep you out of trouble and is easily remembered.

old jim
Thanks. Thats really helpful. From what youve written I can sort of assume that you use the sign you get to first. E.g. you get to the negative of the supply first, hence -VS
jim hardy
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#8
Dec1-12, 07:51 AM
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Yes, write the sign that's on the end you entered,, or as you said , the one you encounter first.

I see that today they use more formal language in teaching than when i studied. Perhaps that's good, for after all our thought processes are tied to the way we speak.

Practice your KVL it's a tool you'll use your whole life.
For now, as you gain confidence in KVL you'll soon see the validity of Jony's arrows and other related concepts. It's more important today that you come to believe in Kirchoff and a simple method is best for that..

old jim


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