# How to calculate the EMF in a circuit with 2 batteries with current flowing towards each other?

#### Electric Sheep

Joined Mar 8, 2023
19
Hello everyone. I have a question that I already posted elsewhere but I didn't get a clear enough answer for me, and I didn't want to frustrate the kind people over there so I thought I would post it here.

This is not homework and I do not study electrical engineering and I have next to no knowledge of it besides the basics of what current, voltage, and resistance and simple parallel and series circuits is/are.

Below is the question.

The question says:
What is the resultant EMF in the circuit?

This is a standalone question in a practise booklet for an exam set by a private company, that came with answers but no explanation. The type of questions that the company makes are notoriously badly worded or unconventional, for example some questions ask you to calculate things in a certain way that you would never do in real life. Other times, questions requires you to mind read what they actually mean, in other words, you can understand where they came from if you looked first at the answer, but if you tried to answer the question first, there could be more than one way to do so, leading to different answers. There is no more context I can provide for this question because the questions are standalone (ie, the previous question is for a totally unrelated topic and the next question I already know how to answer).

So, for this question, I know the answer is 18 because the booklet said so. And since it is 18, I also know that the method used to get this was 24-6.
My question is why do it this way when the formula for EMF is e = V + Ir where e = the electromotive force, V is the volts, I is the current, and r is the internal circuit.

What happened to the e = V + Ir formula? Why was this not used at all? People who were discussing this question also said that if the batteries were facing the same way, the EMF would simply be 24 + 6 = 30. This makes sense to me logically because the 2 batteries would be working together to power the device they are part of. But again, the formula is not used. Is the EMF simply a sum of the 2 battery's voltages (I don't think it is)?

The booklet does not provide explanations, instead, people post their solutions online and discuss them in videos or comments in forums. One of the explanations for this is that the batteries's positive terminals are facing the each other so the voltage is working against each other, and the 24V one is going to over power the 6V one. That is all logical even if you don't have any electrical circuit knowledge, but it doesn't explain how you could look at this question asking for EMF and know to simply do 24-6.

I searched for some definitions of EMF online and I found:
"Emf is the voltage developed between two terminals of a battery or source, in the absence of electric current."
and
"EMF is the potential difference measured across a power source without a load connected to it".
To me these bold words mean a circuit that is not using any power or does not have any devices connected to it, for example a circuit with a light bulb that is switched off or simply a closed circuit connected to a battery (this circuit in the diagram is closed, right?). In the diagram, the light bulb would be the 4.5 Ohm resistor. If the light bulb is switched off, then there would be no resistance. The fact that the resistance is stated as 4.5 Ohms means it must be on or having a circuit pass through it.

#### ericgibbs

Joined Jan 29, 2010
18,867
hi ES,
Welcome to AAC.
Can you summarize your question regarding this circuit,?
E

#### Electric Sheep

Joined Mar 8, 2023
19
It doesn't let me edit the question anymore so I would like to add something here: To be clear: the question doesn't tell you to use that formula, I looked up how to calculate EMF and found this formula myself. Sometimes the questions provide you with a formula to use, sometimes they don't.

#### Electric Sheep

Joined Mar 8, 2023
19
hi ES,
Welcome to AAC.
Can you summarize your question regarding this circuit,?
E
The question simply says "What is the resultant EMF in the circuit?" but my question is why is it calculated as 24 - 6 when there is a special formula for calculating EMF, which is e = V + Ir.

#### Irving

Joined Jan 30, 2016
3,897
Not quite. The resultant EMF is the difference between the two EMF sources. However the potential difference (PD) around the circuit, across each component, or between any two points, is the result of a calculation that involves V, I & R

#### ericgibbs

Joined Jan 29, 2010
18,867
hi ES.
As I read the question, the Load across which the EMF is measured is the 4.5R resistor.

Around the loop the total voltage is 24v-6v=18v
Total resistance around the loop = 6R , so i= 18v/6R =3 Amps

Across the 4.5R at 3Amps
EMF = 4.5R*3A= 13.5V.

E

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#### WBahn

Joined Mar 31, 2012
30,076
I would agree that the question is badly worded and requires a bit of mind reading if all it asks is what is the EMF around the circuit.

Here's the mind-reading part: The "circuit" is to be interpreted as a black box containing two batteries with their negative terminals connected and their positive terminals brought out to the two posts on the box. The 4.5 Ω resistor is then the load that is connected to this black box. The EMF of the "circuit" is then considered to be the voltage across the terminals of the black box when the load resistor is removed.

If I were evaluating a candidate's answer to this problem, I would not be blindly looking for them to give the "right" answer, but instead looking to see if they showed enough work to see their reasoning, if their reasoning was in any way rational, and if their work and answer was consistent with that reasoning (even if the reasoning was flawed).

#### Electric Sheep

Joined Mar 8, 2023
19
Not quite. The resultant EMF is the difference between the two EMF sources. However the potential difference (PD) around the circuit, across each component, or between any two points, is the result of a calculation that involves V, I & R
Hello and thank you for the reply.
Since apparently EMF and Voltage are not the same, and the question didn't call the batteries "EMF 1" and "EMF 2", and "potential difference (PD) around the circuit" is calculated with V, I and R, how can you say that the resultant EMF of the CIRCUIT is just 24 volts - 6 volts?

#### Electric Sheep

Joined Mar 8, 2023
19
I would agree that the question is badly worded and requires a bit of mind reading if all it asks is what is the EMF around the circuit.

Here's the mind-reading part: The "circuit" is to be interpreted as a black box containing two batteries with their negative terminals connected and their positive terminals brought out to the two posts on the box. The 4.5 Ω resistor is then the load that is connected to this black box. The EMF of the "circuit" is then considered to be the voltage across the terminals of the black box when the load resistor is removed.

If I were evaluating a candidate's answer to this problem, I would not be blindly looking for them to give the "right" answer, but instead looking to see if they showed enough work to see their reasoning, if their reasoning was in any way rational, and if their work and answer was consistent with that reasoning (even if the reasoning was flawed).
Hello, thank you for the reply.
Could you elaborate a bit more? This question is actually a multiple choice question, I just didn't post the answer choices because they were just numbers and I already knew the answer, and our working is not graded (hence just one of the reasons the exam is flawed and unfair).

In your black box example, what part of it would allow me to ignore the load attached and simply calculate the voltage across the terminals? In other words:
1) e = V + Ir is a formula that is used for EMF
2) so far all the examples and answers I have seen, says to simply remove or ignore the load resistor and calculate the EMF as the voltage across the terminals

then does that mean any time I see a circuit with 2 batteries and the question asks me to find the EMF, I can just "remove or ignore the load resistor and calculate the EMF as the voltage across the terminals"? Because if that is the case, what is the point of having the e = V +Ir formula if any time, I can just ignore the load resistor?

#### Electric Sheep

Joined Mar 8, 2023
19
hi ES.
As I read the question, the Load across which the EMF is measured is the 4.5R resistor.

Around the loop the total voltage is 24v-6v=18v
Total resistance around the loop = 6R , so i= 18v/6R =3 Amps

Across the 4.5R at 3Amps
EMF = 4.5R*3A= 13.5V.

E
Hello E,
Thank you for the detailed analysis. I agree that the total voltage is 24V - 6V. But my confusion is that, I am told voltage is not the same as EMF, but for this question, some how, the answer to "what is the resultant EMF" is precisely 24V - 6V = 18 Volts. Why?

#### crutschow

Joined Mar 14, 2008
34,464
what is the resultant EMF" is precisely 24V - 6V = 18 Volts. Why?
The total EMF around that circuit loop is the sum of the two battery (EMF) voltages.
Since the polarity of the two batteries are opposing, the resultant net EMF is their difference or 24V-6V = 18V.

Now that EMF voltage is distributed around the loop as determined the the loop current and the resistance of each resistor.
So the current around the loop is the net EMF (here 18V) divided by the total of all the loop resistances.
Then the volage drop across each resistor is that current times its resistance value.
The total of all those resistive voltage drops, of course, equals the net EMF value.
(I'll leave those calculations as an exercise for the reader).

Make sense?

#### Electric Sheep

Joined Mar 8, 2023
19
The total EMF around that circuit loop is the sum of the two battery (EMF) voltages.
Since the polarity of the two batteries are opposing, the resultant net EMF is their difference or 24V-6V = 18V.

Now that EMF voltage is distributed around the loop as determined the the loop current and the resistance of each resistor.
So the current around the loop is the net EMF (here 18V) divided by the total of all the loop resistances.
Then the volage drop across each resistor is that current times its resistance value.
The total of all those resistive voltage drops, of course, equals the net EMF value.
(I'll leave those calculations as an exercise for the reader).

Make sense?
Thanks for your answer. The only thing that makes sense are the first 2 sentences. (would you be able to revise the third sentence, the wording is a little confusing?)

Definitions on the internet says that EMF is the voltage or potential difference across the cell terminals when there is no current flowing through the circuit.
What confuses me is this:
I think there IS a current flowing through this circuit - we just don't know how big the current is unless you do the calculations. There is a resistor with load 4.5 Ohms. This could be for example, a light bulb.

If we are allowed to ignore loads and calculate as if the load does not exist, then does that mean we can treat all circuits like this one, and just simply find the difference between the two terminals? Is it that easy, just ignore the load, pretend it is not there, and subtract one Voltage from the other?
The other thing is confusing me is that I thought EMF is not the same as voltage.

#### crutschow

Joined Mar 14, 2008
34,464
You seem to be having a basic problem with understanding voltage and current
I thought EMF is not the same as voltage.
An electrical power source (such as a battery) generates an EMF (Electro Motive Force) whose strength is given in volts.
This voltage will appear across any load attached to it, which generates a current inversely proportion to the load's resistance.
So volts is a measure of the EMF of a source.
This voltage can then appear across a load.
would you be able to revise the third sentence
The current around a loop equal the net EMF voltage divided by the total loop resistance.
The voltage drop across each resistor then equals its resistance times the loop current.
I think there IS a current flowing through this circuit - we just don't know how big the current is unless you do the calculations.
Of course.
How else would we know the current value unless it is calculated?
If we are allowed to ignore loads
That's a curious statement.
Why do you think that, as generally we can't?
The load allows current to flow which can cause a voltage drop in the source resistance (no real source has zero resistance).
In your example, that appears to be represented by the 1Ω and 0.5Ω resistors.
So you calculate the loop current by dividing the loop net EMF by the load resistance plus the source resistances in series.
You can then determine the load voltage by multiplying the load resistance times the loop current.
The load voltage is thus the loop net EMF minus the voltage drops across the source resistances.

There's nothing mystical about all this, it's just a simple application of Ohm's law, nothing more, nothing less.

Does that now all make sense?

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#### ericgibbs

Joined Jan 29, 2010
18,867
Definitions on the internet says that EMF is the voltage or potential difference across the cell terminals when there is no current flowing through the circuit.
Hi ES,
That is correct for the NO current condition, but wrong for the Potential Difference when current is flowing in the circuit.

But when current does flow in the circuit, it becomes the Potential Difference Voltage, between the voltage terminals.

In your circuit example, there is current flowing from the higher 24V voltage terminal to the lower 6V voltage terminal, via the 4.5R, so you are not measuring the EMF open circuit voltage, but the Potential Difference voltage across the 4.5R

Do you follow, OK.?
E

https://www.birmingham.ac.uk/teachers/study-resources/stem/physics/electromotive-force.aspx

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#### Electric Sheep

Joined Mar 8, 2023
19
Hi ES,
That is correct for the NO current condition, but wrong for the Potential Difference when current is flowing in the circuit.

But when current does flow in the circuit, it becomes the Potential Difference Voltage, between the voltage terminals.

In your circuit example, there is current flowing from the higher 24V voltage terminal to the lower 6V voltage terminal, via the 4.5R, so you are not measuring the EMF open circuit voltage, but the Potential Difference voltage across the 4.5R

Do you follow, OK.?
E

https://www.birmingham.ac.uk/teachers/study-resources/stem/physics/electromotive-force.aspx
1. We all agree this is a closed circuit.
2. We all agree there is a current. (and that it is coming from the 24V battery via the 4.5 Ohm resistor)
3. BUT: the definition of EMF that says "EMF is the voltage or potential difference across the cell terminals when there is no current flowing through the circuit." apparently does not apply here because points 1 and 2 means there is a load, and there is a current. If this definition does not apply, then EMF cannot be the voltage or potential difference across the cell terminals (the cell terminals being the 24V and 6V batteries), but, the accepted method to get the answer (18V) is precisely found by calculating the voltage or potential difference across the cell terminals, ie by doing 24V - 6V. I see a conflict here. The conflict confuses me.

#### Electric Sheep

Joined Mar 8, 2023
19
That's a curious statement.
Why do you think that, as generally we can't?
I have not studied electrical circuits and this particular question is just one of its type in a practise exam that has questions on various topics including electrical circuits. The real exam I will be doing soon will also have questions about the same various topics but I have no idea if they will or will not ask a similar question/s (for example, in a different practise paper, the electrical circuitry question was about which light bulb will be brightest). Therefore my goal is to understand how the answer to this question was obtained, and what I can do if there was another like this particular one (Asking for EMF when only resistance and voltage are given).

So, I would like to explain the reasoning I am using to try and understand what is happening, and to explain why I thought maybe we can just ignore the load:
1. The only elements labelled in the diagram are the resistance and voltage, but the question asks for EMF.
2. In order for me to calculate the EMF, I searched online for what 'EMF' is.
3. As per my understanding, EMF is not the same as voltage so it never occurred to me that I can calculate it so simply by subtracting 6V from 24V to get 18V. So I looked at how to calculate the EMF and found the formula e = V +Ir. Also, I found 2 definitions of EMF, that said
"EMF is the voltage or potential difference across the cell terminals when there is no current flowing through the circuit."
and also
"EMF is the potential difference measured across a power source without a load connected to it".
BUT:
4. We all agree this is a closed circuit, there is a current and that there is a load. (and that the current is coming from the 24V battery via the 4.5 Ohm load/resistor)

Since, working in reverse, I knew that the answer was 18V (because the booklet said so, albeit without explaining why), and according to some unsatisfactory explanations posted by people online, the reason it is 18V is because you simply calculate it by subtracting 6V from 24V and the formula e = V + Ir is not even used even though that is what it's for. This, in conjunction with the definitions in point 3 above, to me looks like we ignored the load and circuit on purpose, solely for the purpose of solving the question: the definitions apparently do not apply here, because even though we know there is a load and we know there is a current, we still used that method (difference between terminals) so the definition cannot apply. If this definition does not apply, then EMF cannot be the voltage or potential difference across the cell terminals (the cell terminals being the 24V and 6V batteries). But, the accepted method to get the answer (18V) is precisely found by calculating the voltage or potential difference across the cell terminals, ie by doing 24V - 6V. The only way to do that is to purposely ignore the load (resistor of 4.5 Ohm).

So it seems like the load was ignored on purpose in order to facilitate the possibility of calculating the EMF in this particular method. I thought this purposeful ignoring of the load is simply a way to make calculations easier, kind of like how in kinetics you usually ignore air resistance just for the sake of calculation.

If ignoring the load on purpose is a practical method for the sake of making calculations easier, then I was planning on using this method if there was a similar question that asked me to 'find the EMF', because clearly I am still not understanding this.

#### ericgibbs

Joined Jan 29, 2010
18,867
Hi ES.
This EMF versus PD designation has been a cause of many a dispute or misunderstanding.

A very long time ago, my Univ tutor advised us that the simplest way to resolve this definition was to assume that when measuring an open circuit voltage source is to say EMF.
You may not know the actual EMF energy value of the source, unless you intend to carry out measurements.

When measuring voltages in a circuit carrying current, consider the measurement as a potential difference, PD.

Other’s may disagree with this simplistic definition, but it works for me

What your example is trying to show is the effect of the internal resistances of the voltages sources have on the voltage available at the voltage terminals, for that level of load current.

In order to carry out this measurement, you have to add an external load, in order to reveal the internal resistance.

E

#### Electric Sheep

Joined Mar 8, 2023
19
A very long time ago, my Univ tutor advised us that the simplest way to resolve this definition was to assume that when measuring an open circuit voltage source is to say EMF.
You may not know the actual EMF energy value of the source, unless you intend to carry out measurements.

When measuring voltages in a circuit carrying current, consider the measurement as a potential difference, PD.

Other’s may disagree with this simplistic definition, but it works for me
Oh so is that all it is, application of a simplistic definition where we "consider the measurement [of EMF] as a potential difference [ie voltage]" and that is why the answer was just 24V - 6V?

#### ericgibbs

Joined Jan 29, 2010
18,867
Hi ES,
It would be only be 24V -6V between the two voltage sources if the external load was not present, because the internal resistances of the voltage sources would not have the effect of reducing the actual available voltage at the terminals.
You could say without the 4.5R, the EMF o/c voltage is 18V and with an external load of 4.5R, and the internal voltage drops this reduces to 13.5V [which I would call the PD]

BTW: depending upon the battery technology and the current drawn from the battery, the apparent internal resistance would most likely change.

E

#### Electric Sheep

Joined Mar 8, 2023
19
Hi ES,
It would be only be 24V -6V between the two voltage sources if the external load was not present,
"if the external load was not present".
Fact: The answer is 18 V.
Fact: There a load is present.

So how can the answer be 24V - 6V =18?