# as level really confused by physics question - aqa unit 1 jan 2013Watch

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Thread starter 3 years ago
#1
Cant post link but can someone look at question 6b on this paper? It says c-d is 2 volts but can someone explain this to me
0
3 years ago
#2
(Original post by jakestevens70)
Cant post link but can someone look at question 6b on this paper? It says c-d is 2 volts but can someone explain this to me
Why can't you post a link? Take a screenshot?
0
Thread starter 3 years ago
#3
My internet connection isnt working so its difficult doing it on my phone
0
3 years ago
#4
I assume you know how to work out the potential difference across each of the branches AE and BF. This is 12V for both branches.

Now, before I carry on I would like to show you a method of thinking about potential difference.

Say you have some branch in a circuit called XY. (This is like AE and BF in the circuit but it starts at X and ends at Y.) Now, consider a resistor in this branch, with a potential difference of 4V, and another one with potential difference of 6V. This means the potential difference across XY is 4+6=10V. Now, imagine that the arrangement is: X node, resistor with potential difference 4V, resistor with potential 6V, Y node. Now imagine moving along XY, starting from point X. At point X you are on a platform that is 10V from the ground. Then at the 4V potential difference, there is a drop of 4V. Then you drop down this and move along, and when you reach the 6V potential difference there is another drop, this time it is a drop of 6V. When you drop down this one, you are at 0V, because you started at point X at 10V, then you dropped 4+6=10V, so you are at 0V, i.e. ground level. Then if you move forward a bit more you reach the Y node, i.e. point Y.

This is an intuitive way of thinking about potential difference.

How do we apply this to the problem?

Well, if you think about it, what the voltmeter is doing is measuring the "height difference" (potential difference) between the platform on the AE branch and the platform on the BF branch. Furthermore, it is measuring the height difference BETWEEN each pair of resistors. You should have worked these out as 6V across the 20k resistors, 4V across the thermistor, and 8V across the 10k resistor.

Consider the AE branch: It is easy to calculate that the potential difference across this whole branch is 12V, so when you start at point A you are on a platform 12V from the ground. Then when you reach the first 20k resistor, you drop to a platform of 12-6=6V from the ground. Finally, moving along this platform you reach the second 20k resistor, where there is a further drop of 6V. Then you are at ground level, and moving along you reach point E. So if you think about it, the point where the wire from the voltmeter is connected to the branch AE, is in between the two 20k resistors, and is therefore 6V from the ground.

Similarly, for the BF branch we start at 12V again at point B. THen moving along we reach the 10k resistor where there is a drop of 8V, so we are now on a platform of height 4V. Then moving further we reach the thermistor, which drops us to a platform of 0V. Finally, if we keep on moving we reach point F. Now in between the thermistor and the 10k resistor, we were on a platform of height 4V.

Therefore, the two platforms across which the voltmeter is taking a reading have heights 6V (on the AE branch) and 4V (on the BF branch). The difference of these gives the potential difference across these branches at those points, which is 6-4=2V.
7
Thread starter 3 years ago
#5
Amazing thank you
0
3 years ago
#6
(Original post by PotterPhysics)
I assume you know how to work out the potential difference across each of the branches AE and BF. This is 12V for both branches.

Now, before I carry on I would like to show you a method of thinking about potential difference.

Say you have some branch in a circuit called XY. (This is like AE and BF in the circuit but it starts at X and ends at Y.) Now, consider a resistor in this branch, with a potential difference of 4V, and another one with potential difference of 6V. This means the potential difference across XY is 4+6=10V. Now, imagine that the arrangement is: X node, resistor with potential difference 4V, resistor with potential 6V, Y node. Now imagine moving along XY, starting from point X. At point X you are on a platform that is 10V from the ground. Then at the 4V potential difference, there is a drop of 4V. Then you drop down this and move along, and when you reach the 6V potential difference there is another drop, this time it is a drop of 6V. When you drop down this one, you are at 0V, because you started at point X at 10V, then you dropped 4+6=10V, so you are at 0V, i.e. ground level. Then if you move forward a bit more you reach the Y node, i.e. point Y.

This is an intuitive way of thinking about potential difference.

How do we apply this to the problem?

Well, if you think about it, what the voltmeter is doing is measuring the "height difference" (potential difference) between the platform on the AE branch and the platform on the BF branch. Furthermore, it is measuring the height difference BETWEEN each pair of resistors. You should have worked these out as 6V across the 20k resistors, 4V across the thermistor, and 8V across the 10k resistor.

Consider the AE branch: It is easy to calculate that the potential difference across this whole branch is 12V, so when you start at point A you are on a platform 12V from the ground. Then when you reach the first 20k resistor, you drop to a platform of 12-6=6V from the ground. Finally, moving along this platform you reach the second 20k resistor, where there is a further drop of 6V. Then you are at ground level, and moving along you reach point E. So if you think about it, the point where the wire from the voltmeter is connected to the branch AE, is in between the two 20k resistors, and is therefore 6V from the ground.

Similarly, for the BF branch we start at 12V again at point B. THen moving along we reach the 10k resistor where there is a drop of 8V, so we are now on a platform of height 4V. Then moving further we reach the thermistor, which drops us to a platform of 0V. Finally, if we keep on moving we reach point F. Now in between the thermistor and the 10k resistor, we were on a platform of height 4V.

Therefore, the two platforms across which the voltmeter is taking a reading have heights 6V (on the AE branch) and 4V (on the BF branch). The difference of these gives the potential difference across these branches at those points, which is 6-4=2V.
I.. holy crap. That was amazing. Thank you!
0
3 years ago
#7
(Original post by Mattematics)
I.. holy crap. That was amazing. Thank you!
Haha no problem, glad it helped!
0
3 years ago
#8
(Original post by PotterPhysics)
I assume you know how to work out the potential difference across each of the branches AE and BF. This is 12V for both branches.

Now, before I carry on I would like to show you a method of thinking about potential difference.

Say you have some branch in a circuit called XY. (This is like AE and BF in the circuit but it starts at X and ends at Y.) Now, consider a resistor in this branch, with a potential difference of 4V, and another one with potential difference of 6V. This means the potential difference across XY is 4+6=10V. Now, imagine that the arrangement is: X node, resistor with potential difference 4V, resistor with potential 6V, Y node. Now imagine moving along XY, starting from point X. At point X you are on a platform that is 10V from the ground. Then at the 4V potential difference, there is a drop of 4V. Then you drop down this and move along, and when you reach the 6V potential difference there is another drop, this time it is a drop of 6V. When you drop down this one, you are at 0V, because you started at point X at 10V, then you dropped 4+6=10V, so you are at 0V, i.e. ground level. Then if you move forward a bit more you reach the Y node, i.e. point Y.

This is an intuitive way of thinking about potential difference.

How do we apply this to the problem?

Well, if you think about it, what the voltmeter is doing is measuring the "height difference" (potential difference) between the platform on the AE branch and the platform on the BF branch. Furthermore, it is measuring the height difference BETWEEN each pair of resistors. You should have worked these out as 6V across the 20k resistors, 4V across the thermistor, and 8V across the 10k resistor.

Consider the AE branch: It is easy to calculate that the potential difference across this whole branch is 12V, so when you start at point A you are on a platform 12V from the ground. Then when you reach the first 20k resistor, you drop to a platform of 12-6=6V from the ground. Finally, moving along this platform you reach the second 20k resistor, where there is a further drop of 6V. Then you are at ground level, and moving along you reach point E. So if you think about it, the point where the wire from the voltmeter is connected to the branch AE, is in between the two 20k resistors, and is therefore 6V from the ground.

Similarly, for the BF branch we start at 12V again at point B. THen moving along we reach the 10k resistor where there is a drop of 8V, so we are now on a platform of height 4V. Then moving further we reach the thermistor, which drops us to a platform of 0V. Finally, if we keep on moving we reach point F. Now in between the thermistor and the 10k resistor, we were on a platform of height 4V.

Therefore, the two platforms across which the voltmeter is taking a reading have heights 6V (on the AE branch) and 4V (on the BF branch). The difference of these gives the potential difference across these branches at those points, which is 6-4=2V.
Wow thank you so much. You explained it so well. 0
3 years ago
#9
(Original post by PotterPhysics)

Consider the AE branch: It is easy to calculate that the potential difference across this whole branch is 12V,
I'm not sure what you mean here? 0
3 years ago
#10
(Original post by Stallzy)
I'm not sure what you mean here? EMF travels in closed loops - that is, every parallel branch of a circuit has the same total EMF (in this case, anyway). As the EMF of the battery is 12 volts, the potential difference across that branch is also 12 volts.
0
3 years ago
#11
PotterPhysics thank you so much, you've saved my physics AS
0
3 years ago
#12
(Original post by PotterPhysics)
I assume you know how to work out the potential difference across each of the branches AE and BF. This is 12V for both branches.

Now, before I carry on I would like to show you a method of thinking about potential difference.

Say you have some branch in a circuit called XY. (This is like AE and BF in the circuit but it starts at X and ends at Y.) Now, consider a resistor in this branch, with a potential difference of 4V, and another one with potential difference of 6V. This means the potential difference across XY is 4+6=10V. Now, imagine that the arrangement is: X node, resistor with potential difference 4V, resistor with potential 6V, Y node. Now imagine moving along XY, starting from point X. At point X you are on a platform that is 10V from the ground. Then at the 4V potential difference, there is a drop of 4V. Then you drop down this and move along, and when you reach the 6V potential difference there is another drop, this time it is a drop of 6V. When you drop down this one, you are at 0V, because you started at point X at 10V, then you dropped 4+6=10V, so you are at 0V, i.e. ground level. Then if you move forward a bit more you reach the Y node, i.e. point Y.

This is an intuitive way of thinking about potential difference.

How do we apply this to the problem?

Well, if you think about it, what the voltmeter is doing is measuring the "height difference" (potential difference) between the platform on the AE branch and the platform on the BF branch. Furthermore, it is measuring the height difference BETWEEN each pair of resistors. You should have worked these out as 6V across the 20k resistors, 4V across the thermistor, and 8V across the 10k resistor.

Consider the AE branch: It is easy to calculate that the potential difference across this whole branch is 12V, so when you start at point A you are on a platform 12V from the ground. Then when you reach the first 20k resistor, you drop to a platform of 12-6=6V from the ground. Finally, moving along this platform you reach the second 20k resistor, where there is a further drop of 6V. Then you are at ground level, and moving along you reach point E. So if you think about it, the point where the wire from the voltmeter is connected to the branch AE, is in between the two 20k resistors, and is therefore 6V from the ground.

Similarly, for the BF branch we start at 12V again at point B. THen moving along we reach the 10k resistor where there is a drop of 8V, so we are now on a platform of height 4V. Then moving further we reach the thermistor, which drops us to a platform of 0V. Finally, if we keep on moving we reach point F. Now in between the thermistor and the 10k resistor, we were on a platform of height 4V.

Therefore, the two platforms across which the voltmeter is taking a reading have heights 6V (on the AE branch) and 4V (on the BF branch). The difference of these gives the potential difference across these branches at those points, which is 6-4=2V.
Cannot thank you enough for this answer! Amazing!!!
0
3 years ago
#13
(Original post by Mattematics)
EMF travels in closed loops - that is, every parallel branch of a circuit has the same total EMF (in this case, anyway). As the EMF of the battery is 12 volts, the potential difference across that branch is also 12 volts.
Thanks! 0
3 years ago
#14
(Original post by jakestevens70)
Cant post link but can someone look at question 6b on this paper? It says c-d is 2 volts but can someone explain this to me
because the overall voltage is 12 and youve worked out 6V between A-C and 4V between D-F so 12V-6V-4V=2V
0
3 years ago
#15
(Original post by PotterPhysics)
I assume you know how to work out the potential difference across each of the branches AE and BF. This is 12V for both branches.

Now, before I carry on I would like to show you a method of thinking about potential difference.

Say you have some branch in a circuit called XY. (This is like AE and BF in the circuit but it starts at X and ends at Y.) Now, consider a resistor in this branch, with a potential difference of 4V, and another one with potential difference of 6V. This means the potential difference across XY is 4+6=10V. Now, imagine that the arrangement is: X node, resistor with potential difference 4V, resistor with potential 6V, Y node. Now imagine moving along XY, starting from point X. At point X you are on a platform that is 10V from the ground. Then at the 4V potential difference, there is a drop of 4V. Then you drop down this and move along, and when you reach the 6V potential difference there is another drop, this time it is a drop of 6V. When you drop down this one, you are at 0V, because you started at point X at 10V, then you dropped 4+6=10V, so you are at 0V, i.e. ground level. Then if you move forward a bit more you reach the Y node, i.e. point Y.

This is an intuitive way of thinking about potential difference.

How do we apply this to the problem?

Well, if you think about it, what the voltmeter is doing is measuring the "height difference" (potential difference) between the platform on the AE branch and the platform on the BF branch. Furthermore, it is measuring the height difference BETWEEN each pair of resistors. You should have worked these out as 6V across the 20k resistors, 4V across the thermistor, and 8V across the 10k resistor.

Consider the AE branch: It is easy to calculate that the potential difference across this whole branch is 12V, so when you start at point A you are on a platform 12V from the ground. Then when you reach the first 20k resistor, you drop to a platform of 12-6=6V from the ground. Finally, moving along this platform you reach the second 20k resistor, where there is a further drop of 6V. Then you are at ground level, and moving along you reach point E. So if you think about it, the point where the wire from the voltmeter is connected to the branch AE, is in between the two 20k resistors, and is therefore 6V from the ground.

Similarly, for the BF branch we start at 12V again at point B. THen moving along we reach the 10k resistor where there is a drop of 8V, so we are now on a platform of height 4V. Then moving further we reach the thermistor, which drops us to a platform of 0V. Finally, if we keep on moving we reach point F. Now in between the thermistor and the 10k resistor, we were on a platform of height 4V.

Therefore, the two platforms across which the voltmeter is taking a reading have heights 6V (on the AE branch) and 4V (on the BF branch). The difference of these gives the potential difference across these branches at those points, which is 6-4=2V.
That was awesome. Thank **** we have people this clever, its people like you who gave us electricity.
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