## How to calculate the EMF in a circuit with 2 opposing batteries?

## How to calculate the EMF in a circuit with 2 opposing batteries?

(OP)

EDIT:

Just realised now I didn't post the actual question. So sorry everyone.

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

Hello,

I am not an electrical engineering student/engineer, but I have a related question I was hoping someone could help answer.

Here is a picture of the question:

This is not homework.

I would like to know, the formula for EMF (from what I have found online) is e = V + Ir where e = the electromotive force, V is the volts, I is the current, and r is the internal circuit. But the answer for this is simply "24 - 6 = 18", and the explanation 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 what I don't understand is 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. Again, the formula is not used. Is the EMF simply a sum of the 2 battery's voltages?

Thank you in advance.

Just realised now I didn't post the actual question. So sorry everyone.

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

Hello,

I am not an electrical engineering student/engineer, but I have a related question I was hoping someone could help answer.

Here is a picture of the question:

This is not homework.

I would like to know, the formula for EMF (from what I have found online) is e = V + Ir where e = the electromotive force, V is the volts, I is the current, and r is the internal circuit. But the answer for this is simply "24 - 6 = 18", and the explanation 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 what I don't understand is 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. Again, the formula is not used. Is the EMF simply a sum of the 2 battery's voltages?

Thank you in advance.

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

However the simplest explanation is that at any point in the circuit one can calculate the voltage relative to some other point in the circuit the EMF between those two points.

Typically some point is chosen as 0V, which is arbitrary. This circuit could be hanging on a wire that is 10,000V attached to the connection between the two batteries, but the change in EMF would be the same around the circuit.

Pick a direction, such as counter-clockwise, and call that the positive direction. Start at the connection between the cells. Going to the top of the 24V cell is considered +24V; going the other way the +6V in the negative direction is -6V, so the difference in EMF between the tops of the cells is 24V+(-6V) or 18V.

Likewise if you start at the top of the 24V battery and go in the negative direction you see -(-24V) + -(6V) which again is 18V.

Had you been given the 24V and a current though the resistor then you would see +24V - I*4.5 Ohms to see the EMF at the top of the other battery.

See https://www.khanacademy.org/science/in-in-class-12...

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

These question booklets have answers but do not come with explanations about their answers so "suggested answers" are simply what other people have worked out and posted video tutorials of online and people discuss ways to get the answers in the comments.

Sometimes the questions allow you to make real world assumptions, sometimes you are not allowed, but they never tell you when, and the only way you know is by "cheating" and checking the answer key first, then you will know what you were supposed to have done to get the "right" answer. And that is the only context I can provide unfortunately.

So, since the answer is 18, I know it is because they wanted you to do 24 - 6, but my question is, why do it this way when the formula is e = V + Ir? 24 - 6 written as a formula seems to be V2 - V1, which is NOT the same formula as e = V + Ir.

I had a look at the Khan academy link you posted about Kirchhoff's Law. Admittedly, I have not had the chance to look deeply into this yet, but, is that related to the formula e = V + Ir?

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

You would then create a loop equation +24V -I*R2 - I*R1 -6V -I*R3 = 0 and solve for I, which would then allow you to use 24V -I*R1 to solve for the actual voltage across the battery. The sign of the current will get corrected by the loop equation solution.

TTFN (ta ta for now)

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## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

+24 Volts plus -6 Volts = 18 Volts.

You have an EMF of 18 Volts driving current through (0.5 Ohm + 1 Ohm + 4.5 Ohm) = 6 Ohms, 3 Amps.

The 6 Volt battery will be charging.

I have encountered this circuit in Central American Lobster boats.

The engines were 24 Volt, start and charge.

The electronics were 12 Volts, fed from a 12 Volt battery.

The load was 12 Volt engine room lights. (In place of your example's 4.5 Ohm load.)

The lights were always on in the engine room and the 12 Volt battery for the electronics was kept charged.

Question for your instructor;

"Were would we see this circuit in the field?"

Suggestion:

"Would this work?"

--------------------

Ohm's lawNot just a good idea;

It's the LAW!## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

Be careful of battery polarity and current polarity when you add voltages together. One battery will have less than it's ideal voltage at it's terminals, the other more voltage than it's ideal voltage.

It's just a series circuit, pick your convention for current flow direction and start calculating the current and voltage across each component. Again, watch the polarity...

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

As I use the terms, EMF is the force available to drive the current, or the open circuit voltage.

Terminal voltage is the actual voltage under load.

Thus EMF minus internal voltage drop equals terminal voltage.

A rigorous solution is surprisingly difficult.

Consider, at 3 Amps, the internal voltage drop in the 24 Volt battery is 1.5 Volts and the terminal voltage is 22.5 Volts.

With 22.5 Volts the current will be slightly less. A true solution may be moving out of the realm of simple arithmetic.

But, battery actual voltage seldom equals terminal voltage depending on the state of charge.

When the voltages are given as nominal voltages, it is common to neglect the effect of internal voltage drop on the current or to consider only the first iteration of the reduced voltage.

One last important factor: The IR drop across the 6 Volt battery will be a voltage rise rather than a voltage drop.

--------------------

Ohm's lawNot just a good idea;

It's the LAW!## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

--------------------

Ohm's lawNot just a good idea;

It's the LAW!## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

To answer Waross' question :

Question for your instructor;

"Were would we see this circuit in the field?"

and

"Would this work?"

These questions are not done in a classroom, it is from a practise booklet created by the examiners. You do the questions yourself and there is no guidance on what why the correct answer is what it is, so there is no one to ask. As such, I don't even know if this circuit would work in real life or if you would ever have a circuit like this, the only thing I recognised was that it might be charging since both positive terminals are pointed towards each other. Whether the 24V battery is charging the 6 volt one (like a powerbank) or whether they are both contributing to the charging of a device (with resistance 4.5 Ohm) connected to it, I have no idea. "Would this work?" - I would imagine this could be say a toy or electrical device that is being charged simultaneously by 2 batteries? Like a bucket being filled simultaneously by 2 water sources pouring into the same bucket?

I tried some calculations based on what Lionel has suggested. I know that voltage in series is additive, so you can just add all of the voltages. Since we don't know the total voltage, but we do know that the total voltage subtracted from itself is 0, we can do what IRStuff said, which is to subtract all of the voltages and make them equal to zero.

So what I did was, starting at the 24 V battery and going counter clockwise for current flow, I said the negative terminal to positive terminal direction is positive current flow, but then since we are subtracting, I reversed the signs. Since voltage is equal to I*R, I did:

I*R2 - 24V - 4.5*I - (6V - 1*I) = 0

I*R2 - I*4.5 - 6 + I*R1 - 24V = 0

I*R2 - I*4.5 + I*R1 - 6V - 24V = 0

and R2 = 0.5, R1 = 1

then

I(0.5 - 4.5 +1) - 6V - 24V = 0

I(0.5 - 4.5 +1) - 18V = 0

I(0.5 - 4.5 +1) = 18V

I(-3) = 18V

I = 18V/-3

= -6 amps?

(and 6v -I*1 is negative and in their own brackets because it is its own battery and the current flows in the opposite direction for them, from the positive to the negative terminal).

This is slightly different to waross' suggestion in that I used negative 24 whereas waross used positive 24, (+24V -I*R2 - I*R1 -6V -I*R3 = 0 and solve for I). Not sure why 24 is positive here while everything else is negative?

Regardless, do I now use current = -6 in the e = V + Ir formula? r = internal resistance so it is 0.5 and 1, for a total of 1.5, but we still don't have V for the overall circuit. If Voltage is current x internal resistance, and current is -6 and internal internal resistance is 1 + 0.5 + 4.5 = 6, then is the voltage -6*6 = -36V? What about the 4.5Ohms?

This question is supposed to be done in around 1.5 minutes and the formula way seems to be too much effort for what its worth, and even if -36V is correct, how do I get the answer of "18V" from -36V?

Then I saw a definition of EMF (on a tutoring site, so not sure how reliable it is): "Emf is the voltage developed between two terminals of a battery or source, in the absence of electric current." and another one "EMF is the potential difference measured across a power source

withouta load connected to it".For the first definition, we don't know if there is a current or not. Maybe it is not switched on, or the battery, despite being connected, is flat so there is no current. The second definition could have explained why the answer to this question is simply 24V-6V but there

ISa load connected to it, it has a resistance of 4.5Ohms, so I still don't know why it is as simple as 24-6.## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

As 3DD says, there's 18V and 6 ohms in the circuit, so there's actually 3 amps.

TTFN (ta ta for now)

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## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

My question is how do I know that is what I am supposed to do, since, it asks for the EMF and when searching for EMF calculations, I am told to use the e = V + Ir which to me, bears no resemblance to "24 - 6 = 18", plus I will have to find I first.

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

The usual convention places the positive supply voltage to the left and calculates clockwise - swapping the two

+24V - i*4.5 -6V -i*1 -i*.5 = 0

+24-6V -i*(4.5+1+.5) = 0

18V = i*6.

Or, by inspection of the circuit, realize you'll be subtracting the 6V eventually, so do it up front and add the resistances.

+18V = i*6.

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

Okay thank you, I understand this equation and thanks for telling me about the convention.

But my other question remains, which is

whyuse this equation: This calculation gives us the voltage. If the question were asking for the total voltage or even the current, I suppose then this is how I could go about doing it. But the question is asking forEMF. So, does that mean EMF is the same thing as voltage? I didn't think they were the same. As I mentioned I looked up some definitions of them and neither of the understandable definitions were helpful. According to one definition, EMF is the same as voltage when there is no load on the current but there is a load here and it has the resistance of 4.5 Ohms.## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

So the question here is what is the unloaded battery voltage for each cell.

That will be the measured voltage plus the voltage drop / gain due to the internal resistance.

The confusion is that batteries are designed for open circuit voltages of 6 and 24 volts so it makes the question look like those are the open circuit voltages. With that much internal resistance (crappy batteries at best). I can't recall a need to find EMF this way - it's just easily measured. Typically engineers are concerned with voltage drop due to internal resistance.

By way of comparison a car battery may be as low as 0.022 Ohms. Most are less than 1 Ohm.

In this case the 24 V battery has an open circuit voltage higher than 24 because of the internal resistance is dragging it down and the 6 V has an open circuit voltage lower than 6 Volts because the internal resistance is pushing it up.

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

"EMF is the unloaded voltage and is taken as a constant ... So the question here is what is the unloaded battery voltage for each cell."

What is "unloaded voltage"? I looked up "unloaded voltage" and all the results talk about "voltage dividers" and of course I have no idea what that is. To help me, I would like to refer for a moment to the definitions of EMF I found online:

"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. 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.

Is the "unloaded voltage" for each cell, the voltage that the battery has, if it isn't powering anything? I think of load as a burden so an unloaded voltage or "unloaded battery voltage" sounds like a battery connected to a device but the device is not used so no current is flowing. Like a battery in a remote control that is not being used. (I also looked up "unloaded battery voltage" and the results were all about flat batteries).

"That will be the measured voltage plus the voltage drop / gain due to the internal resistance."

Just want to make sure I understand this: The "measured voltage" is the 6V and the 24 V and the voltage drop or gain would be due to R1 and R2, so the unloaded battery voltage for each cell is calculated as "6V + I*1" and "24 + I*0.5". Doing it this way means we don't know the value for current I at the moment, so we also don't know the exact values for "the unloaded battery voltage for each cell". But suppose at this point we calculated the the unloaded battery voltage for each cell. How does that get the EMF?

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

You didn't watch the signs. Draw the current loop clockwise or counterclockwise. I'd go counter-clockwise myself and it appears you started that way. Either can work as a starting point. Follow the loop around and put a + sign on the correct end of each component. Then add or subtract the voltage of each device to sum to zero.

Another way to handle this is to re-draw it with all the components in a straight vertical line so the components being in a loop don't mess with you getting the signs right.

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

--------------------

Ohm's lawNot just a good idea;

It's the LAW!## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

In any case, each battery has an "EMF" which is the voltage shown next to the battery. If there is no current, i.e., the circuit is open, then the EMF = battery voltage as shown. If, however, the battery is connected in a circuit with significant current draw, then the voltage of the battery = EMF + IR drop across its internal resistance.

TTFN (ta ta for now)

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

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## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

EMF is what adds energy to the system and that energy is added at 18V.

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

There are numerous good advice. I am confused too. I try to look at it as following:

1. Remove the 4.5 ohm load , the LHS battery

EMF is 6 Vand the RHS batteryEMF is 24 V.Note: (a)This is the

EMF voltageacross the battery whenNOTconnected to any load =open circuit voltage.(b) the battery internal resistances of 1 Ohm and 0.5 Ohm respectively, are irrelevant.

2. When the 4.5 Ohm load is connected:

2.1 the loop resistance R=4.5 + 1 + 0.5 = 6 Ohm.

Note: Both batteries internal resistance of 1 Ohm and 0.5 Ohm are considered connected

in series withthe load 4.5 Ohm.2.2 The current I = E/R ...A . i.e. 18/6 = 3 A

Note: (a) E = 24-6 = 18 V

(b) battery internal resistances are irrelevant.

2.3 the RHS battery is

discharging 3 ANote: (a) voltage drop across the battery internal resistance vd = 0.5 Ohm x 3 A = 1.5 V .

(b) the voltage across the battery would be EMF - vd = 24 V - 1.5 V = 22.5 V.

3. Summery:

The voltage across RHS battery = 22.5 V.Che Kuan Yau (Singapore)

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

But, The respective terminal voltages are now LHS 9 Volts, RHS 22.5 Volts.

The voltage across the 4.5 Ohm load is 13.5 Volts.

The internal voltage drop on charging is the opposite sine to the voltage drop on discharge.

Better to stick to EMF and KISS.

Final proof:

13.5 Volts/4.5 Ohms = 3 Amps.

--------------------

Ohm's lawNot just a good idea;

It's the LAW!## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

Further to my earlier post, I would like to add the following:

1. The loop current is 3 A.

2. The RHS battery is discharging 3 A (i.e. current flow out of the RHS battery

terminal +to LHS battery).3. The voltage across the load resistor is V=I x R = 3A x 4.5 Ohm =

13.5 V(i.e. current flow from RHS towards LHS).4. The voltage across the load resistor would be 13.5 V (i.e.

RHS +andLHS -)Che Kuan Yau (Singapore)

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

I would like to check that I understood something by asking a few questions and answering them myself.

1. What kind of circuit would you say this is, closed or open? I think it is closed.

2. Would you say there is a load on the circuit or there is no load? I would say there is a load and it has the resistance of 4.5 Ohms.

Consequences:

Since the batteries are connected to something (a resistor with resistance of 4.5 Ohms), and the circuit is closed, then "the voltage of the battery = EMF + IR". And since we are given the voltage already (24V and 6V), if we rearrange the formula to have EMF on its own, then it would be voltage of the battery - IR = EMF.

I am going to call the voltage of the battery Voltage B.

I*R gives you Voltage. I will call this Voltage Y.

Therefore for this question at least, EMF = Voltage B - Voltage Y = EMF.

Voltage B would be 24 because it is the biggest. Is that the correct reason? And because we are looking at the difference between the 2 terminals, the R in I*R = Voltage Y would be of the other battery, with voltage 6.

We do not know the current, I.

Am I correct so far?

So now we have:

EMF = Voltage B - Voltage Y where Voltage Y is I * 6.

=

EMF = 24 - I*6.

We still don't know I.

As an alternative, I tried to do the above in conjunction with what 3DDave said which was

so that I = 3.

But substituting 3 into the EMF equation gives me 24 - 3*6 = 6, not 18.

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

I completely agree with most of the things you have said. It is clear to me that the EMF is the voltage across a battery when it is NOT connected to a load. However the circuit in the question has a load. You can see it, it has a resistance of 4.5 Ohms.

So then my question is, whenever a question asks for the EMF, can I just mentally remove or completely ignore any load and pretend it is an open circuit, then find the difference between the two voltages? Is that how simple it is?

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

Since conventional current rules make the incoming end of a component positive, the signs will correspond to the voltages measured across each resistor, assuming the current direction is correct. This corresponds to the equation I originally posted. When the equation is solved, the current is positive 3A, which means that the CCW direction was correct. If we had asserted the current was CW, then the solution to the that loop equation would have been -3 amps, thus correcting the flow direction to CCW.

If the two battery voltages are swapped, then the conventional current would correctly flow in the CW direction.

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## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

No, that's only the answer to that specific question, for that specific circuit. What would you have done had there been a current source in the circuit? Or, had the circuit had multiple loops?

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## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

EMF is a constant and is given in the diagram.

What isn't constant is the voltage across the internal resistance. It doesn't matter what is hooked up, at least until the chemistry inside the cell is consumed.

If they want the EMF it is what is stated on the diagram

If they want the battery terminal voltage for that circuit, that's also been calculated here but is not EMF.

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

".....So then my question is, whenever a question asks for the EMF, can I just mentally remove or completely ignore any load and pretend it is an open circuit, then find the difference between the two voltages? Is that how simple it is?......"You may answer as following:

1. The

EMFof the LHS battery is 6 V and the RHS battery is 24 V ;when the circuit isOpenwithout may current flowing through.2. When the circuit is

closedwith a load of 4.5 Ohm, with current of 3 A flowing; thevoltage ( NOT EMF)or call it[potential difference]at each point:(a) across RHS battery is 22.5 V,

(b) across load is 13.5 V , (i.e.

RHS + and LHS -),(c) .....

In this way you answer the question in full, with no ambiguity; on the terminology used between

EMFandpotential difference.Che Kuan Yau (Singapore)

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

Some people seem to say EMF is not the same as voltage (and I would agree as my research told me so, but I do not understand the difference). For example, Che12345 wrote:

(here it is claimed that voltage is not the same as EMF)

but then at times it seems that EMF

ISthe same as voltage for example:by Che12345:

(here EMF was used instead of voltage)

and by 3DDave:

(nowhere in the diagram is anything labelled as 'EMF', thus this to me says EMF must be the same as voltage, as voltage and resistance are the only things labelled and it sure is not resistance so must be voltage)

My goal is to figure out a rule here so I know what to do if I come across a similar question in the exam (as I mentioned, this is from a practise paper. The practise paper has questions on various topics including electrical circuits. The real exam will have questions about the same topics but may or may 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). At the moment, it seems like EMF is treated as the voltage (even though they are not the same)

on purpose, and the rule is simply to ignored the loadon purpose, in order to get the answer of 24V - 6V = 18V (I say 'on purpose' because despite definitions saying EMF is only the voltage when there is NO current nor load, and there obviously is a current and load, the solution just went 24V - 6V = 18V and the only way to do that despite the definitions and violations of these definitions is to ignore the load and current on purpose).And thank you IRStuff for the annotated diagram.

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

So, the first question should be what does the practice exercises done before the practice exam define EMF as?

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

They are trying to see if you know what EMF is and are adding information to get you to calculate voltage drops and in-circuit voltages and waste your time.

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

https://www.physicsforums.com/threads/internal-res...

TTFN (ta ta for now)

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

Forget voltage drops for now.

Forget the resistor for now.

Forget charging and discharging for now.

The question was concerning EMF.

The answer is 18 Volts EMF.

+24 Volts EMF - 6 Volts EMF = 18 Volts EMF.

Keep working on it until you are able to successfully subtract 6 V from 24 V and get 18 V.

Don't try our patience any longer.

--------------------

Ohm's lawNot just a good idea;

It's the LAW!## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

-----

The figure below shows circuit in which the positive terminals of two batteries are connected t0 each other via a 4.5 Ohm resistor: The batteries are indicated by dashed lines. The EMFs of the batteries are 6 V and 24 V and their respective internal resistances are 1 Ohm and 0.5 Ohm.

10) What is the resultant EMF in the circuit? 6V 18 V 24 V 30 V

11) What is the current flowing in the circuit? 5A 4A 3A 1A

---

The amusing answer transcript:

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

This is because you're NOT supposed to use the test to learn; you use the test to verify your learning.

I did actually explain it, EMF -- electromotive force; it's whatever "thing" there is in a circuit that supplies current flow. Since resistors aren't active devices, they cannot supply current, which leaves the two batteries.

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

This exam is called the gamsat, interested parties can look it up. Anyone can do it regardless of your background. The exam covers biology, chemistry and physics but not in the conventional factual information way. It can ask questions from a mix of any topics in those three areas in any ratio. It often asks convoluted and flawed questions that use scientific premises to ask "logic" related questions. You may come from an entirely non-science background, eg you could be an actor for all they care, and have never learnt maths or any form of science so it is entirely possible and 100% normal to have not learnt or studied anything about electrical circuitry (or biology or chemistry), so there is no "learning" to speak of.

Now I would like to just say thank you to everyone who took the time to help me with this, I did a of research before hand but it is not the same as getting answers from humans. Thanks again and have a nice day.

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

## RE: How to calculate the EMF in a circuit with 2 opposing batteries?

Thanks and have a great day/night.