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

ericgibbs

Joined Jan 29, 2010
18,980
HI ES,
One possible interpretation of the question is to test your understanding of the circuit.
They say the EMF's of the batteries are 24V and 6V and then ask what is the resultant EMF of the circuit.

It is just possible the 4.5R and the internal resistance values have been added in order to 'confuse/mislead' the student into working out the PD, rather than realizing that the resultant EMF is in fact 24V-6V =18V

E.
BTW: Nice chatting with you.;)
 

Irving

Joined Jan 30, 2016
3,971
Another way to look at this is Kirchhoff's voltage law which states that "the sum of the voltage differences around any closed loop in a circuit must be zero. A loop in a circuit is any path which ends at the same point at which it starts."

Equating "voltage difference" to "EMF generation" gives the equation:
EMF(bat1) + EMF(bat2) + EMF(res) = 0, or, rearranging​
EMF(bat1) + EMF(bat2) = -EMF(res)

Traversing the circuit clockwise from top right gives:
EMF(bat1) = 24, EMF(bat2) = -6, therefore the resultant EMF in the circuit (irrespective of the fact its split over 3 separate resistors) is -18v, the sign just serving to show which way round another "EMF generator" would need to be to reduce loop current to zero.
 

Thread Starter

Electric Sheep

Joined Mar 8, 2023
19
HI ES,
One possible interpretation of the question is to test your understanding of the circuit.
They say the EMF's of the batteries are 24V and 6V and then ask what is the resultant EMF of the circuit.

It is just possible the 4.5R and the internal resistance values have been added in order to 'confuse/mislead' the student into working out the PD, rather than realizing that the resultant EMF is in fact 24V-6V =18V

E.
BTW: Nice chatting with you.;)
Ahh okay so you are saying it is more of a trick question?
Another way to look at this is Kirchhoff's voltage law which states that "the sum of the voltage differences around any closed loop in a circuit must be zero. A loop in a circuit is any path which ends at the same point at which it starts."

Equating "voltage difference" to "EMF generation" gives the equation:
EMF(bat1) + EMF(bat2) + EMF(res) = 0, or, rearranging​
EMF(bat1) + EMF(bat2) = -EMF(res)

Traversing the circuit clockwise from top right gives:
EMF(bat1) = 24, EMF(bat2) = -6, therefore the resultant EMF in the circuit (irrespective of the fact its split over 3 separate resistors) is -18v, the sign just serving to show which way round another "EMF generator" would need to be to reduce loop current to zero.
This explanation makes partial sense.
I see how rearranging the formula would give 24 - 6 if you define EFM(bat1) as 24V and EMF(bat2) as 6V. But what is -EMF(res)? Is it the resistance of something? But nothing has a resistance of -18 in this circuit. Even the sums of all the resistors is 6: 0.5 + 1 + 4.5) and nowhere near -18 or +18.

Only when using the rearranged formula do you realise that EMF(res), whatever it is, must be -18:
EMF(bat1) + EMF(bat2) = -EMF(res)
24V - 6V = 18
therefore
24V - 6V -18 = 0 and EMF(res) is -18.
 

WBahn

Joined Mar 31, 2012
30,230
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?
I don't think that the phrase "EMF of the circuit" has any unambiguous meaning. Imagine a more complex circuit with six batteries and twelve resistors randomly connected. What could possibly be meant by the EMF of such a circuit.

EMF is a voltage and a voltage is always the potential difference between two points. Unless the two points being used is so glaringly obvious that no one could mistake another pair of points as being the points in question, whoever asks for the EMF needs to specify which two points (and the polarity of the two points) before an answer can be given.
 
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WBahn

Joined Mar 31, 2012
30,230
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.
The term "electromotive force" refers to the voltage difference between two points created by a SOURCE that converts energy FROM some other form INTO electrical energy.

The term "potential difference" refers to the voltage difference between two points, irrespective of how that voltage came to be about.

Let's annotate your original diagram a bit:

1678462228535.png

In your circuit, if we agree that the two points in question are across the 4.5 Ω resistor, then our terminals are A and B. It is arbitrary which one we call the positive terminal and which one we call the negative terminal. I have chosen to call Point A the positive terminal, so my terminal voltage, V_T, is Vab -- the voltage at Point A relative to the voltage at Point B.

Since all of the components in this particular circuit are in series, they all have the same current, I_T. As with the voltage, I could have defined this current to be going in either direction. I chose to define it going clockwise.

I could flip a coin to decide the polarity of V_T and flip the coin again to decide the polarity of I_T. But I didn't quite do that. Instead, I chose one arbitrarily (the current direction, as it happens) and then I chose the polarity of the other in accordance with the "passive sign convention" which says to choose polarities so that, through passive components (such as resistors) the current goes from positive to negative. The result is that if the numerical value for both have the same sign (so that their product, which is the measure of electrical power, is positive), electrical energy is being delivered to the component. If they are opposite, so that the power is negative, then the component is delivering power to the circuit (by converting it from some other form).

If we now go around the circuit and use KVL and Ohm's Law to sum up the voltage gains, we get (starting at the bottom node and going around the circuit in the direction of I_T:

\(
-I_T * R_1 \; + \; E_1 \; - R_L * I_T \; - \; E_2 \; - \; I_T * R_2 \; = \; 0
\)

The terminal voltage, V_T, is just

\(
V_T \; = \; I_T*R_L
\)

Substituting this into the prior equation, we have:

\(
-I_T * R_1 \; + \; E_1 \; - V_T \; - \; E_2 \; - \; I_T * R_2 \; = \; 0
\)

Solving this for V_T, we have:

\(
V_T \; = \; -I_T * R_1 \; + \; E_1 \; - \; E_2 \; - \; I_T * R_2
\)

Rearranging this a bit, we get:

\(
V_T \; = \; \left( E_1 \; - \; E_2 \right) \; - \; I_T * \left( R_1 \; + \; R_2 \right)
\)

As you can see, the voltage that we would measure between A and B is a mixture of the voltage potentials produced by the sources (the batteries), and the potential differences that result due to the resistances consuming electrical energy. So this is NOT the EMF since that term ONLY considers the voltage potentials that are due to the sources converting energy from some other form to electrical energy.

But what happens if we remove R_L from the circuit? Then we don't have a complete circuit and I_T must go to zero. Under those conditions (known as the "open-circuit". or "open-terminal", or "no-load" conditions, the terminal voltage that we would measure would be:

\(
V_T \; = \; \left( E_1 \; - \; E_2 \right) \;\;\;\; \left( I_T \; = \; 0 \right)
\)

We can, somewhat reasonably, call this the "EMF" of the circuit. In this case, because I chose the terminal voltage to be Vab and not Vba, the EMF is -18 V. This underscores that it is important, ANY time you are talking about voltages or currents, to clearly define what the reference polarity is.

But this does NOT generalize to an arbitrary circuit. A more complex connection of components would make it impossible to separate the voltage observed between two points into that due to the EMFs of the batteries and that due to the consumption of the resistors and to then just remove a load resistance to get one but not the other. What we COULD still do, using this approach, would be to find the "equivalent" source voltage. This is known as the Thevenin voltage (and also an equivalent resistance that is in series with it) such that, if we replaced the actual components inside our black box (which, in this case, contains everything except R_L) with the Thevenin equivalent circuit, the world outside the box (namely R_L) could not tell the difference.

The equivalent circuit in this case is a -18 V source in series with a 1.5 Ω resistor (with the positive terminal connected to Point A and the negative terminal connected to Point B).

If you swap the polarity of the terminals (so that B is the positive terminal), the the Thevenin voltage becomes +18 V.
 

crutschow

Joined Mar 14, 2008
34,671
Fact: The answer is 18 V.
Fact: There a load is present.

So how can the answer be 24V - 6V =18?
The reason they say the EMF is for "no load" is that all real EMF sources have internal resistance (which here is apparently shown by the 1 ohm and 0.5 ohm resistors).
So, with a load, the external battery voltage will change, but it's internal EMF stays the same (at least until it discharges significantly).
Thus, with the load connected, you still say the battery internal EMF's are 24V and 6V.
 

Irving

Joined Jan 30, 2016
3,971
Ahh okay so you are saying it is more of a trick question?

This explanation makes partial sense.
I see how rearranging the formula would give 24 - 6 if you define EFM(bat1) as 24V and EMF(bat2) as 6V. But what is -EMF(res)? Is it the resistance of something? But nothing has a resistance of -18 in this circuit. Even the sums of all the resistors is 6: 0.5 + 1 + 4.5) and nowhere near -18 or +18.

Only when using the rearranged formula do you realise that EMF(res), whatever it is, must be -18:
EMF(bat1) + EMF(bat2) = -EMF(res)
24V - 6V = 18
therefore
24V - 6V -18 = 0 and EMF(res) is -18.
"res" is short for RESidual or RESult. My bad for assuming it would be obvious...
 

Thread Starter

Electric Sheep

Joined Mar 8, 2023
19
The reason they say the EMF is for "no load" is that all real EMF sources have internal resistance (which here is apparently shown by the 1 ohm and 0.5 ohm resistors).
So, with a load, the external battery voltage will change, but it's internal EMF stays the same (at least until it discharges significantly).
Thus, with the load connected, you still say the battery internal EMF's are 24V and 6V.
This seems like the closest thing to solving my confusion because it seems to be telling me why the definitions say what they say, but I am having trouble connecting the first sentence logically with the rest of this statement.

What does all real EMF sources having internal resistance have to do with a circuit with a load having external battery voltage changing and internal EMF not changing?
 

Thread Starter

Electric Sheep

Joined Mar 8, 2023
19
I would like to post something I found out today. Please feel free to correct or add to it if anything is wrong.

This same question was asked on this website.

I have the exact same issue and question as the poster in the link: we both have a formula for emf that we found, and the only way we could see how to get 18V was to "simply minus emf of battery 1 from battery2", and like the poster in that link, I also wanted to know "if this is correct rather than a coincidence, what is the rule behind it".

Based on my research, EMF is apparently not the same as voltage, so my question/confusion boiled down to: in the diagram, it is labelled as voltage while in the wording they called it EMF, so which is it if they are not the same? The interchangeable usage of the two terms is confusing me. As an analogy, to me it's like saying: Fact: bananas are not the same as apples, the bananas in the basket are $24 and $6 each. Here is a picture of some apples worth $24 and $6 each. Now find the difference in bananas from the picture. The two are not the same, you can't do it.

BUT I think I may have a solution now:

The answer in the above link says:
"i've no idea why they tell you all about the resistances. the emf in a circuit (a loop) is simply the sum of the individual emfs (and you can see from the long and short lines that the batteries are "facing" opposite ways, so one of them will be minus the other)"

Firstly, the answerer is saying that the resistance is not needed to make any calculations and is irrelevant to the question and answer, since the answer is just the sum of +24V and -6V.

It just occurred to me that EMF has units. Up until now, I thought that EMF was its own unit, like how the units for Voltage is itself, Volts. I looked up the units for EMF and apparently the units for EMF is also volts. Perhaps that is why the text says the "EMFs of the batteries are 6 VOLTS and 24 VOLTS" and the diagram also labels them as 24V and 6V. In this case, the answer would be very simple if you followed the instructions, that "the emf in a circuit (a loop) is simply the sum of the individual emfs" (I do understand that the batteries are facing opposite ways so you subtract). This method is the simplest rule I have seen so far and does not require me to know anything else or use any formulas. The only problem is, this method does not use the formula emf = V + IR and every time I search for "emf in a circuit" online, the results are all some variation of "emf is voltage across the cell terminals when there is no current or load in the circuit." This is very different to "the emf in a circuit (a loop) is simply the sum of the individual emfs". The 2 rules/methods bear no resemblance with each other, the first has a condition (that there is no load or current), the second one just says emf in a circuit is just the sum of individual emfs. Also, until today, I have never seen this method/rule in any reliable source anywhere else before except in that link. Without a source to confirm that it is a real method, I don't know if its reliable.

Would someone be able to confirm its usability and validity? If this method works, I will accept this method as the answer to my question. But then, why even bother with a formula that looks nothing like it?
 
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Thread Starter

Electric Sheep

Joined Mar 8, 2023
19
The term "electromotive force" refers to the voltage difference between two points created by a SOURCE that converts energy FROM some other form INTO electrical energy.

The term "potential difference" refers to the voltage difference between two points, irrespective of how that voltage came to be about.

Let's annotate your original diagram a bit:

View attachment 289422

In your circuit, if we agree that the two points in question are across the 4.5 Ω resistor, then our terminals are A and B. It is arbitrary which one we call the positive terminal and which one we call the negative terminal. I have chosen to call Point A the positive terminal, so my terminal voltage, V_T, is Vab -- the voltage at Point A relative to the voltage at Point B.

Since all of the components in this particular circuit are in series, they all have the same current, I_T. As with the voltage, I could have defined this current to be going in either direction. I chose to define it going clockwise.

I could flip a coin to decide the polarity of V_T and flip the coin again to decide the polarity of I_T. But I didn't quite do that. Instead, I chose one arbitrarily (the current direction, as it happens) and then I chose the polarity of the other in accordance with the "passive sign convention" which says to choose polarities so that, through passive components (such as resistors) the current goes from positive to negative. The result is that if the numerical value for both have the same sign (so that their product, which is the measure of electrical power, is positive), electrical energy is being delivered to the component. If they are opposite, so that the power is negative, then the component is delivering power to the circuit (by converting it from some other form).

If we now go around the circuit and use KVL and Ohm's Law to sum up the voltage gains, we get (starting at the bottom node and going around the circuit in the direction of I_T:

\(
-I_T * R_1 \; + \; E_1 \; - R_L * I_T \; - \; E_2 \; - \; I_T * R_2 \; = \; 0
\)

The terminal voltage, V_T, is just

\(
V_T \; = \; I_T*R_L
\)

Substituting this into the prior equation, we have:

\(
-I_T * R_1 \; + \; E_1 \; - V_T \; - \; E_2 \; - \; I_T * R_2 \; = \; 0
\)

Solving this for V_T, we have:

\(
V_T \; = \; -I_T * R_1 \; + \; E_1 \; - \; E_2 \; - \; I_T * R_2
\)

Rearranging this a bit, we get:

\(
V_T \; = \; \left( E_1 \; - \; E_2 \right) \; - \; I_T * \left( R_1 \; + \; R_2 \right)
\)

As you can see, the voltage that we would measure between A and B is a mixture of the voltage potentials produced by the sources (the batteries), and the potential differences that result due to the resistances consuming electrical energy. So this is NOT the EMF since that term ONLY considers the voltage potentials that are due to the sources converting energy from some other form to electrical energy.

But what happens if we remove R_L from the circuit? Then we don't have a complete circuit and I_T must go to zero. Under those conditions (known as the "open-circuit". or "open-terminal", or "no-load" conditions, the terminal voltage that we would measure would be:

\(
V_T \; = \; \left( E_1 \; - \; E_2 \right) \;\;\;\; \left( I_T \; = \; 0 \right)
\)

We can, somewhat reasonably, call this the "EMF" of the circuit. In this case, because I chose the terminal voltage to be Vab and not Vba, the EMF is -18 V. This underscores that it is important, ANY time you are talking about voltages or currents, to clearly define what the reference polarity is.

But this does NOT generalize to an arbitrary circuit. A more complex connection of components would make it impossible to separate the voltage observed between two points into that due to the EMFs of the batteries and that due to the consumption of the resistors and to then just remove a load resistance to get one but not the other. What we COULD still do, using this approach, would be to find the "equivalent" source voltage. This is known as the Thevenin voltage (and also an equivalent resistance that is in series with it) such that, if we replaced the actual components inside our black box (which, in this case, contains everything except R_L) with the Thevenin equivalent circuit, the world outside the box (namely R_L) could not tell the difference.

The equivalent circuit in this case is a -18 V source in series with a 1.5 Ω resistor (with the positive terminal connected to Point A and the negative terminal connected to Point B).

If you swap the polarity of the terminals (so that B is the positive terminal), the the Thevenin voltage becomes +18 V.
I think I have an answer in my previous post (number 29). But please let me ask a question here. In my previous post, I said "If this method (the 'emf is the sum of all individual emfs in a circuit' method) works, why even bother with a formula that looks nothing like it?"

When I say it "looks nothing like it" I mean "e = V + Ir" looks nothing like "emf is the sum of all individual emfs in a circuit" (this statement is from an answer someone else gave in this link, also mentioned in post #29) which is simply "+24V + -6V.

Now, I can see why your working would turn out to be in the form of 'e = V + Ir', so, your working would fulfil my desire and expectation for using the formula to find EMF and the solution to this question. But, ultimately, it does not give me 18V, which is the correct answer.

I understand that I*R is V, but as the person in the link above, and also in another forum pointed out, the values for resistance and current and ohms etc are not relevant to the calculation of this question at all and is only there to bother you on purpose, you have no use for them... so why the −IT∗R1+E1−VT−E2−IT∗R2=0 and VT=−IT∗R1+E1−E2−IT∗R2?

Let's accept your final simplified equation which was VT=(E1−E2)−IT∗(R1+R2).
If I substitute the known values for R1 = 1 and R2 = 0.5 and E1 = +6 and E2 = +24:
+24 - +6 - I(1+0.5) = VT
(we don't know the current I but we do know the answer can't be 18V).
 
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WBahn

Joined Mar 31, 2012
30,230
Where does this "emf = V + IR" rule that you keep talking about come from?

You keep throwing it out there and want to use it, but you don't seem to want to understand what it means or where it comes from.

Without that understanding, it becomes just another equation that you want to throw at a problem whether or not it applies.

That equation is for a very specific situation -- it's not just any emf, any V, any I, and any R. If you don't use exactly the right values for each, you will end up with nonsense.

Consider the following circuit:

1678551267378.png

The gold box represents a real voltage source, such as a battery. Part of that battery converts chemical energy into electrical energy. The voltage that is produced is the battery's EMF and is represented by the voltage source Vs within that box. But a real battery can't actually deliver all of that energy to the circuit that it is powering; some of it will be converted to heat within the battery itself. This is modeled by including some internal resistance.

Because both of these things are part and parcel of the same thing -- the battery's internal construction -- we can't measure them separately. We can only make voltage and current measurements external to the battery, namely Vo and Io above, and then analyze them to figure out what the value of Vs is and what the value of Rs is.

In the circuit above, which is a simple series circuit, we have:

\(
I_0 \; = \; \frac{V_S}{R_S \; + \; R_L}
\)

We also have

\(
V_0 \; = \; V_S \; - \; I_0 * R_S
\)

Solve that last equation for V_S:

\(
V_S \; = \; V_0 \; + \; I_0 * R_S
\)

Does that last equation look familiar?

It says that the EMF of the battery is equal to the terminal voltage of the battery plus the voltage drop across the internal resistance of the battery.

To use it, Vo MUST be the voltage measured across the battery's terminals AT THE SAME TIME that there is a current Io flowing through the battery.

You also need to know the internal resistance of the battery, which is often one of the things that you are trying to find out.

But what happens if we make two measurements?

And what if one of those measurements is with Io being so small that it is basically zero?

In that case, the Io*Rs term goes away and we are left with the result that the terminal voltage of the battery when there is no current flowing through it is equal to the battery's EMF voltage.

Now Vs is a known quantity and we can put in a load resistor that draws some appreciable current and measure Vo and Io under those conditions and use that same equation to solve for Rs.
 
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WBahn

Joined Mar 31, 2012
30,230
I think I have an answer in my previous post (number 29). But please let me ask a question here. In my previous post, I said "If this method (the 'emf is the sum of all individual emfs in a circuit' method) works, why even bother with a formula that looks nothing like it?"

When I say it "looks nothing like it" I mean "e = V + Ir" looks nothing like "emf is the sum of all individual emfs in a circuit" (this statement is from an answer someone else gave in this link, also mentioned in post #29) which is simply "+24V + -6V.

Now, I can see why your working would turn out to be in the form of 'e = V + Ir', so, your working would fulfil my desire and expectation for using the formula to find EMF and the solution to this question. But, ultimately, it does not give me 18V, which is the correct answer.

I understand that I*R is V, but as the person in the link above, and also in another forum pointed out, the values for resistance and current and ohms etc are not relevant to the calculation of this question at all and is only there to bother you on purpose, you have no use for them... so why the −IT∗R1+E1−VT−E2−IT∗R2=0 and VT=−IT∗R1+E1−E2−IT∗R2?

Let's accept your final simplified equation which was VT=(E1−E2)−IT∗(R1+R2).
If I substitute the known values for R1 = 1 and R2 = 0.5 and E1 = -6 and E2 = +24:
-6 - +24 - I(1+0.5) = VT
(we don't know the current I but we do know the answer can't be 18V).
But E1 doesn't equal -6 V (you REALLY need to start using units properly).

E1 = 6 V and E2 = 24 V -- look a the diagram that explicitly defines what E1 and E2 are.

So you have

VT = (6 V - 24 V) - IT(1 Ω + 0.5 Ω) = (-18 V) - IT(1.5 Ω)

If a load resistor is chosen to be so large that IT is negligibly small, then VT = -18 V.

The minus sign is merely indicative of the arbitrary choice of VT being Vab rather than Vba. This underscores the fact that just asking for something like "the EMF of the circuit" is inherently ambiguous. If nothing else, it fails to specify the polarity of the desired result -- +18 V and -18 V are equally correct answers, each for a different choice of the polarity of the terminal voltage and the question (as you've presented) didn't indicate which polarity it wants.

As for throwing everything away and just "summing up the individual EMFs in a circuit", I've already addressed that more than once. It works IN THIS CIRCUIT because everything is in series. But what if the circuit were move complex, with a combination of series and parallel connected batteries?
 

Thread Starter

Electric Sheep

Joined Mar 8, 2023
19
Thank you so much, I think I actually understand this now, at least for this series circuit question.

Where does this "emf = V + IR" rule that you keep talking about come from?

You keep throwing it out there and want to use it, but you don't seem to want to understand what it means or where it comes from.

Without that understanding, it becomes just another equation that you want to throw at a problem whether or not it applies.

That equation is for a very specific situation -- it's not just any emf, any V, any I, and any R. If you don't use exactly the right values for each, you will end up with nonsense.
I always look for a formula when there is a calculation question that involves something I don't know> This formula is on virtually every website that discusses this topic, mostly online textbooks or tutorial sites, Here is one of them. I understand what I V and R are, but you are right in saying that I want to use it without understanding what it means. Like any physics or maths question, given a formula you can simply substitute in the values that are given and you get an answer, so that was what I was doing. I had no idea it is for a specific situation.

The gold box represents a real voltage source, such as a battery. Part of that battery converts chemical energy into electrical energy. The voltage that is produced is the battery's EMF and is represented by the voltage source Vs within that box. But a real battery can't actually deliver all of that energy to the circuit that it is powering; some of it will be converted to heat within the battery itself. This is modeled by including some internal resistance.

Because both of these things are part and parcel of the same thing -- the battery's internal construction -- we can't measure them separately. We can only make voltage and current measurements external to the battery, namely Vo and Io above, and then analyze them to figure out what the value of Vs is and what the value of Rs is.
Okay, this is totally clear and I understand this. And this actually defines EMF much clearer than what I was accepting as the definition.

But what happens if we make two measurements?

And what if one of those measurements is with Io being so small that it is basically zero?

In that case, the Io*Rs term goes away and we are left with the result that the terminal voltage of the battery when there is no current flowing through it is equal to the battery's EMF voltage.
I completely understand this now. This was the largest part of my confusion and I could not get over this until now and it hindered my understanding of everything else in this question. Until I saw this line I thought the official definition of EMF is "whatever the battery voltage is when there is no current". So I thought, in order to define or calculate EMF, you always have to remove the current (or load) (and, since there is a formula for it, I thought you should then use that formula). I now understand (as a person who otherwise has no understanding of electrical circuitry) that this statement is merely a consequence of making the Io (current) zero but not a compulsory requirement to define or to calculate EMF. My confusion was I had always thought the current or load MUST be zero for you to know and calculate the EMF, but load and current is clearly in the diagram.

Now it is very clear, thank you very very much for your help!:)
 
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Thread Starter

Electric Sheep

Joined Mar 8, 2023
19
But E1 doesn't equal -6 V (you REALLY need to start using units properly).

E1 = 6 V and E2 = 24 V -- look a the diagram that explicitly defines what E1 and E2 are.

So you have

VT = (6 V - 24 V) - IT(1 Ω + 0.5 Ω) = (-18 V) - IT(1.5 Ω)

If a load resistor is chosen to be so large that IT is negligibly small, then VT = -18 V.

The minus sign is merely indicative of the arbitrary choice of VT being Vab rather than Vba. This underscores the fact that just asking for something like "the EMF of the circuit" is inherently ambiguous. If nothing else, it fails to specify the polarity of the desired result -- +18 V and -18 V are equally correct answers, each for a different choice of the polarity of the terminal voltage and the question (as you've presented) didn't indicate which polarity it wants.

As for throwing everything away and just "summing up the individual EMFs in a circuit", I've already addressed that more than once. It works IN THIS CIRCUIT because everything is in series. But what if the circuit were move complex, with a combination of series and parallel connected batteries?
I apologise for the incorrect sign on the 6V value, I have changed it. In my original calculation I went counterclockwise and started with 24V being positive, and since the 2 voltages oppose each other and 24V will overpower 6V, I designated + to 24V and - to 6V. Since I already knew 6V was going to be a negative value, and the "shortcut method" said to just add up the EMFs, so instead of putting (+24) + (-6V), I put -6 first and +24 second so -6V + 24V to match the clockwise direction in your annotated diagram. The extra - in the middle was a typo.

Yes it is abundantly clear now that "summing up all the EMFs" only works here because it is in series.

The minus sign is merely indicative of the arbitrary choice of VT being Vab rather than Vba. This underscores the fact that just asking for something like "the EMF of the circuit" is inherently ambiguous. If nothing else, it fails to specify the polarity of the desired result -- +18 V and -18 V are equally correct answers, each for a different choice of the polarity of the terminal voltage and the question (as you've presented) didn't indicate which polarity it wants.
Thank you for mentioning this because it is important. The answer choices were:
A 6 V
B 18 V
C 24 V
D 30 V

but now we know what the answer is if they decide to change the signs to negative in future editions.

Many thanks again!
 

Thread Starter

Electric Sheep

Joined Mar 8, 2023
19
I just wanted to say thank you again to everyone for helping me with this question, there was no way I could have understood the reasoning behind it without human help from you all. Thanks and have a great day!:D
 

Thread Starter

Electric Sheep

Joined Mar 8, 2023
19
HI ES,
One possible interpretation of the question is to test your understanding of the circuit.
They say the EMF's of the batteries are 24V and 6V and then ask what is the resultant EMF of the circuit.

It is just possible the 4.5R and the internal resistance values have been added in order to 'confuse/mislead' the student into working out the PD, rather than realizing that the resultant EMF is in fact 24V-6V =18V
Hi E,
Thank you for pointing this out, I understand this now. For this question, yes you are absolutely right, I found other sources that also say this and I understand why now.
Have a nice day.
 
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