It's helpful to know where you found this; it's possible it makes sense, but it's also possible the explanation there is in error.

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...

Regarding where I got this question from (@3DDave), this is a standalone question in a GAMSAT practise booklet. This exam and their practise booklets are, if I must say so myself, notorious for very badly worded questions and answers that are more about corrupting the wording of the question and/or answers in order to trick and confuse candidates, than providing a fair and clear objective question to test candidates, so there is no context to speak of or to base this question on, and there is no way for me to know if I am supposed to do something or not, to get to this answer.

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?

The more typical presentation of such a problem usually requires you to calculate the actual voltage somewhere in the loop, say across the right side battery.

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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The voltage is the directed sum of the voltages.
+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 law
Not just a good idea;
It's the LAW!

If EMF is calculating the terminal voltage of the battery then first calculate the current in the circuit then use that current to calculate the terminal voltage.

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...

EMF versus terminal voltage.
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 law
Not just a good idea;
It's the LAW!

By the way, I agree with Lionel's post on this.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

Thanks everyone, as I said, I don't know anything about electrical engineering so there are a lot of terms I don't know, like what terminal voltage is.

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 without a 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 IS a 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.

Wouldn't it be easiest to subtract 6 Volts from the circuit making it an 18V circuit with 3 resistors in series? Then just add the 6V back to whatever got calculated.

Your current calculation is incorrect. You ignored our admonitions to meticulously keep track of the sign of the current.

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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If the goal is to "get the number 18", then it is easier to subtract 6V from 24 to give 18V.

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.

It's algebra.

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.

@3DDave
Okay thank you, I understand this equation and thanks for telling me about the convention.

But my other question remains, which is why use 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 for EMF. 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.

EMF is the unloaded voltage and is taken as a constant. Normally that constant is used when a resistor is put in series with a battery to calculate the internal resistance of the battery, but that resistance is given.

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.

Thank you 3DDave for the explanations. Unfortunately my confusions have not been fully cleared yet so I am going to leave this for awhile and maybe come back to it later. But if you or anyone has nothing better to do... then here are a few more of my questions.

"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?

Just realised I didn't actually post actual wording of the question itself. Edited my first post to include the question now. So sorry everyone.

Quote:

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

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.

Quote (Question)

"What is the resultant EMF in the circuit?"

24V EMF minus 6V EMF = 18 Volts EMF.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

EMF stands for electro-motive force, and as such, is a arcane term, since voltage is not a force, i.e., kg*m/s^2. "Electro-motive" is an allusion to something that causes electricity to flow; EMF is something I might have seen in textbooks when I was in high school. I think more plausibly, the internal battery voltage would be called "open-circuit voltage" which is typically what you see on a datasheet for a battery

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)
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The formula is not used because the answer is given - the EMFs of the two cells are plainly stated and EMF is constant. With the cells in opposition the EMFs for the circuit subtract.

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

Dear Mr Total Novice
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 V and the RHS battery EMF is 24 V.
Note: (a)This is the EMF voltage across the battery when NOT connected 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 with the 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 A
Note: (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)

Quote:

2.3 the RHS battery is discharging 3 A

And the LHS battery is charging 3 A.
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 law
Not just a good idea;
It's the LAW!

Dear Mr Dear Mr Total Novice

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 + and LHS - )
Che Kuan Yau (Singapore)

Quote (IRStuff: 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.)

Firstly I apologise for "ignoring the signs". I don't study electrical circuitry and learnt the basics just for this question so I don't really know how to do that.

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

Quote (+24V - i*4.5 -6V -i*1 -i*.5 = 0
+24-6V -i*(4.5+1+.5) = 0
18V = i*6.)

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

Quote (Dear Mr Total Novice
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 V and the RHS battery EMF is 24 V.
Note: (a)This is the EMF voltage across the battery when NOT connected 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 with the 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 A
Note: (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))

@Che12345,
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?

This is the OP's circuit annotated. We arbitrarily assert that the conventional current is flowing CCW, although we pretty much know that has to be, since the larger battery will dictate the current flow.

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.

TTFN (ta ta for now)
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Quote:

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?

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?

TTFN (ta ta for now)
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"EMF is the voltage across a battery when it is NOT connected to a load"

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.

Dear Mr Total Novice (Electrical)(OP)
".....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 EMF of the LHS battery is 6 V and the RHS battery is 24 V ; when the circuit is Open without may current flowing through.
2. When the circuit is closed with a load of 4.5 Ohm, with current of 3 A flowing; the voltage ( 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 EMF and potential difference.
Che Kuan Yau (Singapore)

Okay I am going to make some observations here:
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:

Quote (2. When the circuit is closed with a load of 4.5 Ohm, with current of 3 A flowing; the voltage ( NOT EMF) or call it potential difference at each point)

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

but then at times it seems that EMF IS the same as voltage for example:
by Che12345:

Quote (The EMF of the LHS battery is 6 V and the RHS battery is 24 V)

(here EMF was used instead of voltage)

and by 3DDave:

Quote (EMF is a constant and is given in the diagram.)

(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 load on 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.

Quote:

These questions are not done in a classroom, it is from a practise booklet created by the examiners.

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

EMF is given in the problem statement with identical values for E1 and E2. Since they are identical they cannot be in-circuit voltages.

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.

Stuff like this goes right to getting the spoon out....

This problem is really old, and the apparent answer is supposed to the EMF of the simplified circuit, which is the difference of the battery voltages, which makes sense that the answer might be 18V

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

TTFN (ta ta for now)
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Forget Internal resistance for now.
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 law
Not just a good idea;
It's the LAW!

More info - from an test-prep site:

-----

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:

Quote:

Hello. The question is taken from physics and the question is The figure shows like it. In which the positive terminal of the two batteries are connected via a 4.5 register. The batteries are indicated by the dashed line. The M F of the battery are six for 10. 24 world and their respective internal resistance is r one moment 50.5. So what this stuff is R E M F Industry ticket. Okay So EMF is six whole days here. 24 World is here. So net IMF will be equal to 90 MF will be equal to 24 world minus six world. So that is equal to 18 world. It involved is an 80 M. F. Now what we have to evaluate what is the current flowing in the circuit. So in order to evaluate the current we have to use the Kv L O. Okay so I double the distances 4.5 and then six words and then there is a resistance Having a resistance value one then there is a battery. 24 world and then resistance it is further connected. It is 24 world and 0.5 words mm Okay so let us use the Gabriel ice to govern. So that is I into one. So one I -6 plus 4.5 into I Plus 24 and plus 0.5 in two I 0.5 Y is equal to zero. So the guarantee 0.5 into 4.5 is five and +16 I that is equal to minus of 80 involved. So I. easy. Sorry. And Bill, actually this is involved. I is equal to minus three mps. Okay? But so from here, the direction of covered must be opposed. So I will be equal to three mps and the direction of current either anti clockwise direction. So the required current flow into the psyche. It is three mps. So this is the required solution for the given question. So hope this clears your routing.

Best of luck.

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This is not done as part of a course in a classroom, the exam is also not part of any course. It's like IELTS - you can sit the exam if you want and there is no official curriculum for the exam content.

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

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But I still haven't gotten an understandable definition of EMF so no, I don't know what it is.

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.

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

I just wanted to provide some context on this exam for people who have been helping me.
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.

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However, it must be stressed that success in GAMSAT is unlikely without knowledge and ability in the biological and physical sciences.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

Quote (However, it must be stressed that success in GAMSAT is unlikely without knowledge and ability in the biological and physical sciences.)

Absolutely true. But I was pointing out (as per your previous comment) the fact that anyone can sit the exam (passing or not is irrelevant) to explain why this is not a question to test what you have learnt, simply because it isn't part of a class and whether you have learnt a particular topic or not, does not stop your eligibility from sitting the rest of the test. ie "I am trying to learn what this topic is about just to answer the question", rather than "I am doing this question to revise what I already learned about this topic".
Thanks and have a great day/night.