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

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

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

Science Advisor
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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 ?

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NoahCygnus
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 ?

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"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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Science Advisor
Homework Helper
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## ?

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NoahCygnus
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.

Science Advisor
Homework Helper
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*}

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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.

NoahCygnus
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*}

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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?

NoahCygnus
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?

Staff Emeritus
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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.

NoahCygnus
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.

Science Advisor
Homework Helper
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.

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NoahCygnus
NoahCygnus
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.

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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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Science Advisor
Homework Helper
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.

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NoahCygnus
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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.

NoahCygnus
NoahCygnus
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.

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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?

Homework Helper
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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.