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Physics A level potential divider question
Hi, I was wondering if someone could help me with the following potential divider question:
Resistors X and Y are connected in series with a 6.0 V battery of negligible internal resistance. X has resistance R and Y has resistance R/2 . A voltmeter of resistance R is connected across Y.
What is the reading on the voltmeter?
I thought that the voltage would be split in a 2:1 ratio so 4:2 so the difference would be 2V but the actual answer is 1.5V. I can't really see how this is the case so was wondering if someone could help me. Thanks a lot!!
I've attached the question paper here if anyone would like to see the diagram, it is question 32: https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2019/june/AQA-74072-QP-JUN19.PDF
Resistors X and Y are connected in series with a 6.0 V battery of negligible internal resistance. X has resistance R and Y has resistance R/2 . A voltmeter of resistance R is connected across Y.
What is the reading on the voltmeter?
I thought that the voltage would be split in a 2:1 ratio so 4:2 so the difference would be 2V but the actual answer is 1.5V. I can't really see how this is the case so was wondering if someone could help me. Thanks a lot!!
I've attached the question paper here if anyone would like to see the diagram, it is question 32: https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2019/june/AQA-74072-QP-JUN19.PDF
Original post by Felix'sfreckles
Hi, I was wondering if someone could help me with the following potential divider question:
Resistors X and Y are connected in series with a 6.0 V battery of negligible internal resistance. X has resistance R and Y has resistance R/2 . A voltmeter of resistance R is connected across Y.
What is the reading on the voltmeter?
I thought that the voltage would be split in a 2:1 ratio so 4:2 so the difference would be 2V but the actual answer is 1.5V. I can't really see how this is the case so was wondering if someone could help me. Thanks a lot!!
I've attached the question paper here if anyone would like to see the diagram, it is question 32: https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2019/june/AQA-74072-QP-JUN19.PDF
Resistors X and Y are connected in series with a 6.0 V battery of negligible internal resistance. X has resistance R and Y has resistance R/2 . A voltmeter of resistance R is connected across Y.
What is the reading on the voltmeter?
I thought that the voltage would be split in a 2:1 ratio so 4:2 so the difference would be 2V but the actual answer is 1.5V. I can't really see how this is the case so was wondering if someone could help me. Thanks a lot!!
I've attached the question paper here if anyone would like to see the diagram, it is question 32: https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2019/june/AQA-74072-QP-JUN19.PDF
Looks like a mistake in the question.
If the resistors are in the ratio 2:1 then so will be the pds across them.
That is; 4V and 2V
The answer given, 1.5V, would be the case if the resistors were in the ratio 3:1
That is 4.5V and 1.5V
But that is not the case.
Possibly a misprint and '2' should have been a 3 in R/2?
Original post by Felix'sfreckles
Hi, I was wondering if someone could help me with the following potential divider question:
Resistors X and Y are connected in series with a 6.0 V battery of negligible internal resistance. X has resistance R and Y has resistance R/2 . A voltmeter of resistance R is connected across Y.
What is the reading on the voltmeter?
I thought that the voltage would be split in a 2:1 ratio so 4:2 so the difference would be 2V but the actual answer is 1.5V. I can't really see how this is the case so was wondering if someone could help me. Thanks a lot!!
I've attached the question paper here if anyone would like to see the diagram, it is question 32: https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2019/june/AQA-74072-QP-JUN19.PDF
Resistors X and Y are connected in series with a 6.0 V battery of negligible internal resistance. X has resistance R and Y has resistance R/2 . A voltmeter of resistance R is connected across Y.
What is the reading on the voltmeter?
I thought that the voltage would be split in a 2:1 ratio so 4:2 so the difference would be 2V but the actual answer is 1.5V. I can't really see how this is the case so was wondering if someone could help me. Thanks a lot!!
I've attached the question paper here if anyone would like to see the diagram, it is question 32: https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2019/june/AQA-74072-QP-JUN19.PDF
Original post by Stonebridge
Looks like a mistake in the question.
If the resistors are in the ratio 2:1 then so will be the pds across them.
That is; 4V and 2V
The answer given, 1.5V, would be the case if the resistors were in the ratio 3:1
That is 4.5V and 1.5V
But that is not the case.
Possibly a misprint and '2' should have been a 3 in R/2?
If the resistors are in the ratio 2:1 then so will be the pds across them.
That is; 4V and 2V
The answer given, 1.5V, would be the case if the resistors were in the ratio 3:1
That is 4.5V and 1.5V
But that is not the case.
Possibly a misprint and '2' should have been a 3 in R/2?
I think you guys misread the question. The voltmeter has resistance R (not infinite resistance), so it has to be 1.5V.
GGG
Original post by Driving_Mad
I think you guys misread the question. The voltmeter has resistance R (not infinite resistance), so it has to be 1.5V.
GGG
GGG
You are absolutely right. Apologies for misreading the question.
(edited 1 year ago)
Original post by Stonebridge
You are absolutely right. Apologies for misreading the question.
but how is it still 1.5V if it is in parallel should it not be 3V? as voltage is not shared in parallel
Original post by shiv2k5
but how is it still 1.5V if it is in parallel should it not be 3V? as voltage is not shared in parallel
The resistance of the voltmeter in parallel with Y reduces the resistance in the bottom half of the divider to less than the resistance of Y.
Its an unusual set up cos normally you'd use a voltmeter that was very much higher resistance that the components in the circuit under test, but in this case the voltmeter is close to the values of X and Y... so it's having a significant effect on the PD it's trying to measure.
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