physics circuits help
Hi guys, I'm confused with the question in the pic attached. Does this mean the emf across cell X is 0.5V or 2V? I was thinking that since cell Y flows in the opposite direction, the total emf across cell X would be cell X-Y, but it also says there's no current across cell Y..... 
Actually what I don't get is, why do they take the voltage as 2V in the circuit if both cells flow in opposite directions? Shouldn't it be 0.5V? and how can current be zero? It states later in the mark scheme that length L is 94cm, so shouldn't there definitely be some sort of current if a cell is connected across it?
I think you can just use K's 2nd law for the loop of 2V cell and wire.
In that loop, sum emfs = 2V = sum(IR) = I (1 + 4)
this works as the 1.5V cell supplies no current - if it did I would vary round the circuit, making it much more complex.
In that loop, sum emfs = 2V = sum(IR) = I (1 + 4)
this works as the 1.5V cell supplies no current - if it did I would vary round the circuit, making it much more complex.
BTW this is an old circuit called a 'potentiometer'*. If you were doing your A levels before the 1980s, you'd have to know lots of applications of this circuit. The question setter must be as old as me!
*potentiometer is a term with several uses - some which are still current!
*potentiometer is a term with several uses - some which are still current!
Thanks for the reply! Yeah, there are still quite a few questions with potentiometers around them but I don't think they teach the applications anymore! I understand how Kirchoff's 2nd Law can be used but I still don't understand why 2V is the net voltage around the loop- and I still don't get all the other qs I posted in the first reply to my post....
Original post by old_teach
BTW this is an old circuit called a 'potentiometer'*. If you were doing your A levels before the 1980s, you'd have to know lots of applications of this circuit. The question setter must be as old as me!
*potentiometer is a term with several uses - some which are still current!
*potentiometer is a term with several uses - some which are still current!
By K's 2, you can go around any closed loop*, so choose the one not including the other cell!
If you chose the loop through both cells, it would be 0.5V, but we don't have enough info (ie the length) to do it this way.
The current is zero in the 1.5V cell, as the pd across that length of wire is 1.5V too, so there is no net pd to drive a current through that cell.
*this makes sense, as any individual electron would do this.
If you chose the loop through both cells, it would be 0.5V, but we don't have enough info (ie the length) to do it this way.
The current is zero in the 1.5V cell, as the pd across that length of wire is 1.5V too, so there is no net pd to drive a current through that cell.
*this makes sense, as any individual electron would do this.
1. Ahhh!! I get what you mean about the loops! However, since 2V is the emf, shouldn't the terminal pd for the circuit used to calculate the current in cell X be calculated from emf- voltage dissipated via internal resistance ?
2. Sorry- I'm really weak with circuits... Would you mind explaining why there could be a pd across a wire but still no electrons flowing in the wire? To be honest, I get that voltage is the amount of "energy" given to a unit charge while flowing past two connected points, but I don't understand how there could be a voltage present but still no electrons flowing. My idea is that it's a given that electrons flow past as long as the power source is connected into the circuit. I think I might have largely misunderstood the whole concept behind this.
2. Sorry- I'm really weak with circuits... Would you mind explaining why there could be a pd across a wire but still no electrons flowing in the wire? To be honest, I get that voltage is the amount of "energy" given to a unit charge while flowing past two connected points, but I don't understand how there could be a voltage present but still no electrons flowing. My idea is that it's a given that electrons flow past as long as the power source is connected into the circuit. I think I might have largely misunderstood the whole concept behind this.
Original post by old_teach
By K's 2, you can go around any closed loop*, so choose the one not including the other cell!
If you chose the loop through both cells, it would be 0.5V, but we don't have enough info (ie the length) to do it this way.
The current is zero in the 1.5V cell, as the pd across that length of wire is 1.5V too, so there is no net pd to drive a current through that cell.
*this makes sense, as any individual electron would do this.
If you chose the loop through both cells, it would be 0.5V, but we don't have enough info (ie the length) to do it this way.
The current is zero in the 1.5V cell, as the pd across that length of wire is 1.5V too, so there is no net pd to drive a current through that cell.
*this makes sense, as any individual electron would do this.
1) I do K2 as sum of emfs = sum of IRs
this is actually something straightforward. If you are an electron going round that closed loop, the emfs give you energy, and you lose it going through the IRs.
You could say terminal pd = pd across wire.
E - I r = IR
or E = I(r + R)
which is the same as K2 (Physics works!).
2) There's a pd across the wire as cell X is pushing a current through the wire.
As the pd across the length of wire is same as Y's emf, the electrons in Y's circuit don't have a potential difference so they don't flow.
These circuits were always tricky, and I'm embarrassed to say when I did my A levels, I just learnt all the circuits. I only understood what was happening when I taught it.
this is actually something straightforward. If you are an electron going round that closed loop, the emfs give you energy, and you lose it going through the IRs.
You could say terminal pd = pd across wire.
E - I r = IR
or E = I(r + R)
which is the same as K2 (Physics works!).
2) There's a pd across the wire as cell X is pushing a current through the wire.
As the pd across the length of wire is same as Y's emf, the electrons in Y's circuit don't have a potential difference so they don't flow.
These circuits were always tricky, and I'm embarrassed to say when I did my A levels, I just learnt all the circuits. I only understood what was happening when I taught it.
This is the easiest way imo, we know that e=I(R+r) where R is the load, and r is internal. The question says the circuit is connected to the wire in part (a) so I= 2/4+1 = 0.4 A
1. I completely get 1 now! Thank you so much for writing it in such simple terms!
2. So does that mean if cell Y wasn't there, the electrons would flow across BD? I have to imagine that cell X wants to push it across with a force, but cell Y is opposing with the same exact force, so across BC there isn't any current at all? If there isn't any current across BC how is there current in cell X? Wouldn't the electrons get "stuck" there?
2. So does that mean if cell Y wasn't there, the electrons would flow across BD? I have to imagine that cell X wants to push it across with a force, but cell Y is opposing with the same exact force, so across BC there isn't any current at all? If there isn't any current across BC how is there current in cell X? Wouldn't the electrons get "stuck" there?
Original post by old_teach
1) I do K2 as sum of emfs = sum of IRs
this is actually something straightforward. If you are an electron going round that closed loop, the emfs give you energy, and you lose it going through the IRs.
You could say terminal pd = pd across wire.
E - I r = IR
or E = I(r + R)
which is the same as K2 (Physics works!).
2) There's a pd across the wire as cell X is pushing a current through the wire.
As the pd across the length of wire is same as Y's emf, the electrons in Y's circuit don't have a potential difference so they don't flow.
These circuits were always tricky, and I'm embarrassed to say when I did my A levels, I just learnt all the circuits. I only understood what was happening when I taught it.
this is actually something straightforward. If you are an electron going round that closed loop, the emfs give you energy, and you lose it going through the IRs.
You could say terminal pd = pd across wire.
E - I r = IR
or E = I(r + R)
which is the same as K2 (Physics works!).
2) There's a pd across the wire as cell X is pushing a current through the wire.
As the pd across the length of wire is same as Y's emf, the electrons in Y's circuit don't have a potential difference so they don't flow.
These circuits were always tricky, and I'm embarrassed to say when I did my A levels, I just learnt all the circuits. I only understood what was happening when I taught it.
The picture I have in mind is that the opposing force of cell Y is like having an open circuit- if electrons can't flow there, then wouldn't it stop the whole circuit's electrons from flowing and thus there would be zero current in the circuit as a whole?
Original post by ruomin02
1. I completely get 1 now! Thank you so much for writing it in such simple terms!
2. So does that mean if cell Y wasn't there, the electrons would flow across BD? I have to imagine that cell X wants to push it across with a force, but cell Y is opposing with the same exact force, so across BC there isn't any current at all? If there isn't any current across BC how is there current in cell X? Wouldn't the electrons get "stuck" there?
2. So does that mean if cell Y wasn't there, the electrons would flow across BD? I have to imagine that cell X wants to push it across with a force, but cell Y is opposing with the same exact force, so across BC there isn't any current at all? If there isn't any current across BC how is there current in cell X? Wouldn't the electrons get "stuck" there?
Thanks for your reply too! I was being really silly hahah
Original post by homemadeclock
This is the easiest way imo, we know that e=I(R+r) where R is the load, and r is internal. The question says the circuit is connected to the wire in part (a) so I= 2/4+1 = 0.4 A
its cool bro, tip is not to panic when you see a different style of question
Original post by ruomin02
Thanks for your reply too! I was being really silly hahah
Electrons are flowing through B - D, simplest to say they've come from cell X and slowly making their way around the circuit.
If you are an electron at C you would not 'choose' to go round the loop to cell Y as there is no 'potential difference' to make you go that way, so you would carry on round the circuit to cell X.
I hope that helps you visualise the circuit.
If you are an electron at C you would not 'choose' to go round the loop to cell Y as there is no 'potential difference' to make you go that way, so you would carry on round the circuit to cell X.
I hope that helps you visualise the circuit.
One sec- just checking, in this part, if the length of the wire is 94cm and the total BD is 100cm (given in markscheme), so 0.94* 2V= 1.88V, should the pd across the length of wire BC be 1.88V instead of 1.5V?
Original post by old_teach
By K's 2, you can go around any closed loop*, so choose the one not including the other cell!
If you chose the loop through both cells, it would be 0.5V, but we don't have enough info (ie the length) to do it this way.
The current is zero in the 1.5V cell, as the pd across that length of wire is 1.5V too, so there is no net pd to drive a current through that cell.
*this makes sense, as any individual electron would do this.
If you chose the loop through both cells, it would be 0.5V, but we don't have enough info (ie the length) to do it this way.
The current is zero in the 1.5V cell, as the pd across that length of wire is 1.5V too, so there is no net pd to drive a current through that cell.
*this makes sense, as any individual electron would do this.
pd across wire is not 2V, some is lost in the internal resistance of the 2V cell.
Quick Reply
Related discussions
- physics alevel edexcel circuit question
- Physics GCSE question
- need help please!
- Applied science unit 3
- Isaac physics C4a.2 - struggling
- Btec applied science unit 3 2024 exam
- Physics Question help
- A level physics Electricity question (2)
- Isaac Physics.
- Physics A-Level AQA
- Use a simulator to plot the frequency response of the circuit
- Question about parallel circuit- OCR AS physics
- Really stuck on this electricity physics question. Please help!
- In what circumstance would a variable resistor be unaccepted to vary voltage?
- Physics Aqa GCSE help
- LDR's in GCSE physics
- GCSE physics question circuits
- A level physics electricity
- A level physics question Capacitance (2)
- Physics
