# Potential Divider AQA A level Question

#1
I need help with this 1 mark. I don't understand how to calculate the voltage between 2 points in a circuit, can anyone explain me how to work through these types of question please?

This is an example:

In this resistor network, the emf of the supply is 12 V and it has negligible internal resistance.

What is the reading on a voltmeter connected between points X and Y?
0
2 years ago
#2
(Original post by I-ZAAA)
I need help with this 1 mark. I don't understand how to calculate the voltage between 2 points in a circuit, can anyone explain me how to work through these types of question please?

This is an example:

In this resistor network, the emf of the supply is 12 V and it has negligible internal resistance.

What is the reading on a voltmeter connected between points X and Y?
Before we get to that bit - if you were asked to find the potential differences across the 2 ohm and 3 ohm resistors, could you do that? This is an important step to getting the answer you want, and should be a little easier.
0
#3
(Original post by Pangol)
Before we get to that bit - if you were asked to find the potential differences across the 2 ohm and 3 ohm resistors, could you do that? This is an important step to getting the answer you want, and should be a little easier.
If I've done the right calculations I get the p.d. to be 8 V across the 2 ohm resistor and 9 V across the 3 ohm resistor.
Last edited by I-ZAAA; 2 years ago
0
2 years ago
#4
(Original post by I-ZAAA)
If I've done the right calculations I get the p.d. to be 8 V across the 2 ohm resistor and 9 V across the 3 ohm resistor.
All good so far. The best way to proceed is to think about this in a way that is not often described in text books. We need to think about the electrical potential at various points.

The question doesn't say if this is a dc or ac supply, but for simplicity, let's say it's a dc supply with the positive terminal at the top. All of this still works whatever the orientation of a dc supply or if it is an ac supply. Since this is a 12V supply, that means that each coulomb of charge has 12 J of energy when it leaves the supply. As there is a pd of 8 V across the 2 ohm resistor, every coulomb of charge that passes through this resistor now only has 4 J of energy. So, the electrical potential at X is 4 V. Similarly, every coulomb of charge that passes through the 3 ohm resistor now only has 3 J of energy. The electrical potential at Y is therefore 3 V.

What you want is the potential difference between X and Y, which is the difference in the potentials at X and Y. Easy!
Last edited by Pangol; 2 years ago
1
#5
(Original post by Pangol)
All good so far. The best way to proceed is to think about this in a way that is not often described in text books. We need to think about the electrical potential at various points.

The question doesn't say if this is a dc or ac supply, but for simplicity, let's say it's a dc supply with the positive terminal at the top. All of this still works whatever the orientation of a dc supply or if it is an ac supply. Since this is a 12V supply, that means that each coulomb of charge has 12 J of energy when it leaves the supply. As there is a pd of 8 V across the 2 ohm resistor, every coulomb of charge that passes through this resistor now only has 4 J of energy. So, the electrical potential at X is 4 V. Similarly, every coulomb of charge that passes through the 3 ohm resistor now only has 3 J of energy. The electrical potential at Y is therefore 3 V.

What you want is the potential difference between X and Y, which is the difference in the potentials at X and Y. Easy!
OMG!! Your explanation was perfect. I completely understood it straightaway. Thanks!
0
2 years ago
#6
(Original post by Pangol)
…..The question doesn't say if this is a dc or ac supply, but for simplicity, let's say it's a dc supply with the positive terminal at the top. All of this still works whatever the orientation of a dc supply or if it is an ac supply. Since this is a 12V supply, that means that each coulomb of charge has 12 J of energy when it leaves the supply. As there is a pd of 8 V across the 2 ohm resistor, every coulomb of charge that passes through this resistor now only has 4 J of energy. So, the electrical potential at X is 4 V. Similarly, every coulomb of charge that passes through the 3 ohm resistor now only has 3 J of energy. The electrical potential at Y is therefore 3 V. ….
I cannot really agree with this “dangerous” way (or perhaps wrong way) of explaining the electric potential. When you mention that “…each coulomb of charge has 12 J of energy when it leaves the supply. As there is a pd of 8 V across the 2 ohm resistor, every coulomb of charge that passes through this resistor now only has 4 J of energy.”, you seem to imply that when one coulomb of charge exit from 12 V supply “carries” 12 J of [electric potential] energy and “deposits” 8 J of energy to the 2-ohm resistor. As a result, the coulomb of charge has 4 J of energy after passing through the 2-ohm resistor. If I interpret your writing correctly, I would write more about it after your reply.
0
2 years ago
#7
(Original post by Eimmanuel)
I cannot really agree with this “dangerous” way (or perhaps wrong way) of explaining the electric potential. When you mention that “…each coulomb of charge has 12 J of energy when it leaves the supply. As there is a pd of 8 V across the 2 ohm resistor, every coulomb of charge that passes through this resistor now only has 4 J of energy.”, you seem to imply that when one coulomb of charge exit from 12 V supply “carries” 12 J of [electric potential] energy and “deposits” 8 J of energy to the 2-ohm resistor. As a result, the coulomb of charge has 4 J of energy after passing through the 2-ohm resistor. If I interpret your writing correctly, I would write more about it after your reply.
It is of course not actually true that charges only receive energy when they pass through a supply and then use it up when the pass through components. Like a lot of concepts that are introduced at GCSE or A Level, the reality is quite different. The "power station" model of electric circuits is, as you allude to, essentially a fiction.

But it is a perfectly adequate analogy for this level, and gives the right answers to the questions.

I do often wonder though why more students don't see it's problems and ask how the charges "know", when passing through components, how much energy they will have to "save" to get through the rest. But there you go!
0
2 years ago
#8
(Original post by Pangol)
It is of course not actually true that charges only receive energy when they pass through a supply and then use it up when the pass through components. Like a lot of concepts that are introduced at GCSE or A Level, the reality is quite different. The "power station" model of electric circuits is, as you allude to, essentially a fiction.

But it is a perfectly adequate analogy for this level, and gives the right answers to the questions.

I do often wonder though why more students don't see it's problems and ask how the charges "know", when passing through components, how much energy they will have to "save" to get through the rest. But there you go!
I find that you have an ingenious way of avoiding the answer and “claiming” of me suggesting a “power station” model which I have no idea what is the model.

It seems that you know there are flaws in the description in post #4 but you still choose to write in such way and expecting students not to misinterpret them.

IMO, there is NO perfect analogy! Analogy does not aid in understanding, in fact it imperils understanding.

…I do often wonder though why more students don't see it's problems and ask how the charges "know", when passing through components, how much energy they will have to "save" to get through the rest. But there you go!
From the way, you are writing post #4, I believe I know an answer NOT the answer.
0
2 years ago
#9
(Original post by Eimmanuel)
I find that you have an ingenious way of avoiding the answer and “claiming” of me suggesting a “power station” model which I have no idea what is the model.

It seems that you know there are flaws in the description in post #4 but you still choose to write in such way and expecting students not to misinterpret them.

IMO, there is NO perfect analogy! Analogy does not aid in understanding, in fact it imperils understanding.

From the way, you are writing post #4, I believe I know an answer NOT the answer.
Well this is needlessly aggressive and confrontational, especially from a study forum helper.

You seemed to be criticising my explanation that charges receive energy when passing though supplies which they then use up when they pass through components. This is what is often referred to as the "power station" model, where things move through a power station and are given energy which they then use up when they go on a journey, arriving back at the power station depleted to need another injection of energy.

Is this what happens in a circuit? No. Does it provide a useful way for students to think about and understand problems at this level? I would say so.

As you say, pretty much every analogy used in physics is not perfect. For example, when students are introduced to electric circuits, the flow of electrons is often compared to the flow of water through pipes. This is actually not a bad analogy at all, and gives students a way to imagine what emf, current and resistance "feel" like. But of course it is not perfect. Break open a pipe and the water will leak out, whereas the electrons do not keep flowing out of a cut wire!

If any UK school teacher teaches physics without the use of analogies, I would be very surprised.

So I'm not really sure what your point is, or why you need to say it in the way that you have.
1
2 years ago
#10
(Original post by I-ZAAA)
I need help with this 1 mark. I don't understand how to calculate the voltage between 2 points in a circuit, can anyone explain me how to work through these types of question please?

This is an example:

In this resistor network, the emf of the supply is 12 V and it has negligible internal resistance.

What is the reading on a voltmeter connected between points X and Y?
The issue that I have with this question is that it doesn't label the positive and negative terminals of the battery. That prevents us from getting the sign of the PD.

You can either jump to seeing each two series resistors as a potential divider, or calculate the current (and hence voltage drop on each resistor) for each parallel path.

I'd jump to the voltage divider, so X has a PD of 1/3 of the battery's, and Y 1/4 (assuming the top terminal is positive). If the bottom terminal is positive, it'll just change the sign of the difference between X and Y.

1/3 - 1/4 = 1/12, so you quickly get the answer (multiply by the battery voltage).
0
2 years ago
#11
(Original post by Pangol)
Well this is needlessly aggressive and confrontational, especially from a study forum helper.

You seemed to be criticising my explanation that charges receive energy when passing though supplies which they then use up when they pass through components. This is what is often referred to as the "power station" model, where things move through a power station and are given energy which they then use up when they go on a journey, arriving back at the power station depleted to need another injection of energy.

Is this what happens in a circuit? No. Does it provide a useful way for students to think about and understand problems at this level? I would say so.

As you say, pretty much every analogy used in physics is not perfect. For example, when students are introduced to electric circuits, the flow of electrons is often compared to the flow of water through pipes. This is actually not a bad analogy at all, and gives students a way to imagine what emf, current and resistance "feel" like. But of course it is not perfect. Break open a pipe and the water will leak out, whereas the electrons do not keep flowing out of a cut wire!

If any UK school teacher teaches physics without the use of analogies, I would be very surprised.

So I'm not really sure what your point is, or why you need to say it in the way that you have.

My point is really simple:
Analogies (may or may not) help students to solve problems successfully but the students tend to arrive the answer with a false understanding of the concepts.

I am good with the correct explanation of the physics concepts followed by the usage of analogy to give students the “intuition” of the abstract concepts with the warnings of the caveats of the given analogy. The correct explanation of physics concepts does not mean the correct explanation of the reality.

Perhaps, you can ask OP or anyone at GCSE or A level who have read your explanation in post#4 to explain the incorrect physics concepts that were mentioned if they were not told or informed already. If students are NOT taught of the correct physics concepts, how would they know the caveat(s) of the analogy?

Needless to say, you seem to “soft-pedal” correct thinking in physics concepts and emphasize on students’ intuition. Again, I may interpret your writing wrongly.

If any UK school teacher teaches physics without the use of analogies, I would be very surprised.
I hope this is not really true. It seems that you like generalize thing which I doubt that you have really asked every UK school teacher.

I remember reading an article from a physics professor from a university in regard to a job applicant who was applying a teaching position to the university and wrote the following (unfortunately I cannot remember most of the writing):
The use of analogies helps students to understand concepts better…

The physics professor and his colleagues did not choose this applicant and write the danger of using analogies to do physics teaching which imperils the students’ understanding and thinking.

I believe this is the last post that I would reply to you in this thread as you tend to over-interpret my writing emotionally and it does not really help the OP to understand the question.
0
1 year ago
#12
(Original post by I-ZAAA)
If I've done the right calculations I get the p.d. to be 8 V across the 2 ohm resistor and 9 V across the 3 ohm resistor.
wait how did you get 8 and 9 ?
0
1 year ago
#13
(Original post by In a Nutshell.21)
wait how did you get 8 and 9 ?
Please start a new thread. This one is now over 4 months old.

Thanks.
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