How to use Kirchhoff's voltage law on this circuit?

  • #1
NoahCygnus
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Homework Statement:
Calculate the potential at point O in the section of the circuit shown in the diagram.
Relevant Equations:
##\Sigma\Delta V = 0##
kirchoff.jpg

In this circuit, I have to find the potential at point O. I tried using Kirchhoff's voltage law for the three open loops AOC, AOB and BOC to arrive at potential at O. According to my calculations the potential at O should be 0, but that is not the case according to the source. So I must be using Kirchhoff's voltage rule wrong. So can anyone guide me how to work this problem using KVL?

Here's my calculation:

Using KVL on AOC:

##V_1 +I_1R_1 +V_O -I_3R_3 = V_3##

Manipulating this equation to arrive at:

##I_1R_1 +V_O -I_3R_3 = V_3 - V_1## (I)

KVL on AOB:

##V_1 +I_1R_1 +V_O - I_2R_2 = V_2##

Manipulation this to arrive at:

##I_1R_1 +V_O -I_2R_2 = V_2 - V_1## (II)

KVL on BOC:

##V_2 +I_2R_2 + V_O -I_3R_3 = V_3##

Manipulating this equation to arrive at:

##I_2R_2 + V_O-I_3R_3 = V_3 - V_2## (III)

Then using the three equations above to arrive at ##V_O##

## III-(I-II) ##
Do the math and we get ##V_O=0##

But the answer in the textbook is ##V_O=[V_1/R_1 +V_2/R_2 +V_3/R_3][1/R_1 +1/R_2 +1/R_3]^-1##.

What am I doing wrong?
 

Answers and Replies

  • #2
BvU
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well, for instance:$$V_1 +I_1R_1 +V_O -I_3R_3 = V_3$$ doesn't sound like KVL. Can you put KVL in words and check ?

##\ ##
 
  • #3
NoahCygnus
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well, for instance:$$V_1 +I_1R_1 +V_O -I_3R_3 = V_3$$ doesn't sound like KVL. Can you put KVL in words and check ?

##\ ##
"The algebraic sum of changes in potential around any 'closed path' of an electric circuit is zero."
I know the law mentions closed paths but hear me out. I have seen people use the law on open circuits treating the end points of the open path as closed.

I will take an example where I have seen people using KVL on open circuits:

Here we have to calculate potential difference between A and B,

I have seen people do this;

$$V_A - 10 - 2(5) + 5 +15-10(2) = V_B$$

I don't understand how they can do that, so I just followed the same thing on the circuit mentioned in my original problem.
 

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  • #4
BvU
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I just followed the same thing
Fair enough. In that case, what would you think of e.g. $$
V_1 +I_1R_1 -I_3R_3 = V_3$$ i.e. without the ##V_O## :smile: ?

##\ ##
 
  • #5
NoahCygnus
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Homework Statement:: Calculate the potential at point O in the section of the circuit shown in the diagram.
Relevant Equations:: ##\Sigma\Delta V = 0##

Using KVL on AOC:

$$V_1+I_1R_1+V_O−I_3R_3=V_3$$
But I did consider ##V_O## in the first equation. I am sure I didn't miss it.
 
  • #6
BvU
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I am sure I didn't miss it.
You certainly didn't. But you should have! Do you agree with $$\begin {align*} V_1 + i_1 R_1 &= V_O \tag{1}\\
V_2 + i_2 R_2 &= V_O \tag 2 \\
V_3 + i_3 R_3 &= V_O \tag 3 \qquad ?\end {align*}$$If so, then you should also agree with 1 + 3 yielding
$$\begin {align*} V_1 + i_1 R_1 -i_3R_3 &= V_3 \tag{1+3}\qquad !\qquad\quad\ \end {align*}$$
:smile:

##\ ##
 
  • #7
Orodruin
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In other words, ##V_O## is not an emf in the circuit. It is just the value that the potential takes at O.
 
  • #8
NoahCygnus
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You certainly didn't. But you should have! Do you agree with $$\begin {align*} V_1 + i_1 R_1 &= V_O \tag{1}\\
V_2 + i_2 R_2 &= V_O \tag 2 \\
V_3 + i_3 R_3 &= V_O \tag 3 \qquad ?\end {align*}$$If so, then you should also agree with 1 + 3 yielding
$$\begin {align*} V_1 + i_1 R_1 -i_3R_3 &= V_3 \tag{1+3}\qquad !\qquad\quad\ \end {align*}$$
:smile:

##\ ##
Oh yes, you used the loop law on AO, BO and CO instead of taking AOC, AOB and BOC like I did. I agree with that! But what's wrong with considering AOC, AOB, and BOC in my equations?
 
  • #9
NoahCygnus
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Oh yes, you used the loop law on AO, BO and CO instead of taking AOC, AOB and BOC like I did. I agree with that! But what's wrong with considering AOC, AOB, and BOC in my equations? How does that change anything?
 
  • #10
Orodruin
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Oh yes, you used the loop law on AO, BO and CO instead of taking AOC, AOB and BOC like I did. I agree with that! But what's wrong with considering AOC, AOB, and BOC in my equations?
Again, you used ##V_O## as an emf, not as just the potential value at O.
 
  • #11
NoahCygnus
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Again, you used ##V_O## as an emf, not as just the potential value at O.
Pardon my lack of knowledge, I know EMF is the potential of a cell when current is not drawn from it but I don't quite understand how I am using EMF of point O instead of its potential value.
 
  • #12
BvU
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No need to apologise ! We're all here (me too!) to make mistakes and have others help us to fix them somewhat.

taking AOC

If I take AOC I get
$$ \begin {align*} V_1 + i_1 R_1 -i_3R_3 -V_3 &= 0 \end {align*}
$$or, alternatively:
$$\begin {align*} V_1 -V_3 & = -i_1 R_1 +i_3R_3 \end {align*}
$$which, of course is all the same mathematically.

@Orodruin (and I) are trying to make clear to you that O is an intermediate point and that ##V_O## has no place in the loop equation.

##\ ##
 
  • #13
NoahCygnus
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No need to apologise ! We're all here (me too!) to make mistakes and have others help us to fix them somewhat.



If I take AOC I get
$$ \begin {align*} V_1 + i_1 R_1 -i_3R_3 -V_3 &= 0 \end {align*}
$$or, alternatively:
$$\begin {align*} V_1 -V_3 & = -i_1 R_1 +i_3R_3 \end {align*}
$$which, of course is all the same mathematically.

@Orodruin (and I) are trying to make clear to you that O is an intermediate point and that ##V_O## has no place in the loop equation.

##\ ##
Thank you. You are very kind. Reading your last statement and Orodruin's comment on how I am treating point O like it has an emf, I had a realization. There's just a wire at point O, there's no potential drop or gain there like in a resistor or a cell whose potential drop/gain we include in the equations. It's just a point on the wire, we don't include the potential at a point in the wire in the equations. Am I correct to assume that?

Also the equations where you used KVL on just AO, you included the term for ##V_O## because you considered it to be attached to a hidden cell not shown in the circuit with ##V_A## being the positive terminal and ##V_O## being the negative terminal. Is that assumption correct as well?
 
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  • #14
BvU
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It's just a point on the wire, we don't include the potential at a point in the wire in the equations. Am I correct to assume that?
I think so, yes

Also the equations where you used KVL on just AO, you included the term for ##V_O## because you considered it to be attached to a hidden cell not shown in the circuit with ##V_A## being the positive terminal and ##V_O## being the negative terminal. Is that assumption correct as well?
It's an intermediate point. Three ways to get there and all three should end up at the same potential.

I don't think it is very useful to imagine an extra voltage source connected to that point (it should provide zero current anyway).
But if you do(*), O should be the positive terminal and ground the negative. (The same ground as the negative terminals of ##V_1## etc.)

Simple check: All ##V_i## equal. What is ##V_O## ?

(*) And that way you have 'upgraded' a three-point star to a four-point star.

##\ ##
 
  • #15
Orodruin
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It's just a point on the wire, we don't include the potential at a point in the wire in the equations
Exactly! Otherwise we could just add an arbitrary number of intermediate points and get a different result.
 
  • #16
NoahCygnus
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I think so, yes


It's an intermediate point. Three ways to get there and all three should end up at the same potential.

I don't think it is very useful to imagine an extra voltage source connected to that point (it should provide zero current anyway).
But if you do(*), O should be the positive terminal and ground the negative. (The same ground as the negative terminals of ##V_1## etc.)

Simple check: All ##V_i## equal. What is ##V_O## ?

(*) And that way you have 'upgraded' a three-point star to a four-point star.

##\ ##
Yes, O should be positive as conventional current goes from O to A. If I do imagine a voltage source connected to O, could you elaborate how why it should provide zero current?
 
  • #17
Steve4Physics
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On a related point, may I point out that the Post #1 diagram shows 3 currents leaving point O, violating Kirchoff's first law! Of course, that can be resolved if one or two of the current-values is/are meant to be negative.
 

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