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# Why does the output voltage remain the same potential divider component changed? watch

1. Say you have a potential divider circuit. There are 2 resistors both of 20 x 10^3 ohms in series. Across one of the resistors a voltmeter is connected also with resistance 20 x 10^3 ohms. If the voltmeter is removed and replaced with another voltmeter of resistance 10 x 10^6 ohms, why does the output voltage remain the same?

The mark scheme says exactly "High resistance >> R1" and "No change in circuit/no current flows through 10 MΩ" for 2 marks. Is this first point trying to say the resistance of the voltmeter is greater than that of R1, if so then I don't understand how this means the output voltage is the same.

My guess would have been that both the resistors remain the same, but then wouldn't the voltage split in a different ratio since the the voltmeter has changed?
2. (Original post by TiernanW)
Say you have a potential divider circuit. There are 2 resistors both of 20 x 10^3 ohms in series. Across one of the resistors a voltmeter is connected also with resistance 20 x 10^3 ohms. If the voltmeter is removed and replaced with another voltmeter of resistance 10 x 10^6 ohms, why does the output voltage remain the same?

The mark scheme says exactly "High resistance >> R1" and "No change in circuit/no current flows through 10 MΩ" for 2 marks. Is this first point trying to say the resistance of the voltmeter is greater than that of R1, if so then I don't understand how this means the output voltage is the same.

My guess would have been that both the resistors remain the same, but then wouldn't the voltage split in a different ratio since the the voltmeter has changed?
Unless your description of the circuit and voltmeters used are incorrect, the voltage reading between the two voltmeters cannot be the same.

The series resistance across the supply must change when the voltmeter is replaced which changes the series current in all resistances. if the supply voltage remains constant, the voltmeter reading MUST change.

Case 1:

Series resistance = 20K + (20K || 20K) = 20K + 10K = 30K.
Current = V/30K

Case 2:

Series resistance = 20K + (20K || 10M) = 20K + 19.96K = 39.96K
Current = V/39.96K

3. (Original post by uberteknik)
Unless your description of the circuit and voltmeters used are incorrect, the voltage reading between the two voltmeters cannot be the same.

The series resistance across the supply must change when the voltmeter is replaced which changes the series current in all resistances. if the supply voltage remains constant, the voltmeter reading MUST change.

Case 1:

Series resistance = 20K + (20K || 20K) = 20K + 10K = 30K.
Current = V/30K

Case 2:

Series resistance = 20K + (20K || 10M) = 20K + 19.96K = 39.96K
Current = V/39.96K

I apologize. Don't know how I misread it, because I kept re-reading the question to check I had read it right and even after I hadn't. It asked why the voltage when there is no component connected to the circuit would be the same for when a 10 x 10^6 resistance component (voltmeter) is connected. Doing calculations the total resistance of the parallel bit (excluding the single series resistor) is about 1999Ohms with it connected and 2000 without it, so a negligible difference.

But still how can no current flow through the 10 x 10^6 resistance component? Is it so small that is negligible basically?
4. (Original post by TiernanW)
I apologize. Don't know how I misread it, because I kept re-reading the question to check I had read it right and even after I hadn't. It asked why the voltage when there is no component connected to the circuit would be the same for when a 10 x 10^6 resistance component (voltmeter) is connected. Doing calculations the total resistance of the parallel bit (excluding the single series resistor) is about 1999Ohms with it connected and 2000 without it, so a negligible difference.

But still how can no current flow through the 10 x 10^6 resistance component? Is it so small that is negligible basically?
Don't forget mark schemes are prompts for examiners and not meant as model answers for students.

Strictly speaking, the prompt is incorrect since there will still be a small (uA) current flowing in the 10M voltmeter resistance.

If there were no voltmeter in circuit, the true potential divider is in the ratio 1:1 (20K + 20K).

With a 20K Ohms voltmeter in parallel with one of the 20K resistors, the effective potential divider changes to a ratio of 2:1 (20K + 10K). Hence the voltmeter will produce a substantially erroneous reading.

With a 10M Ohms voltmeter in parallel, the effective potential divider remains close to the true 1:1 ratio (20K + 19.96K). i.e. the high resistance voltmeter has not 'loaded' the circuit and the reading will be close to the true p.d.

This is the reason why real world voltmeters are designed to have a very high internal resistance. As such, they can be 'ignored' when measuring p.d.'s across parallel resistances 3 or more orders of magnitude less than the voltmeter internal resistance.

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Updated: May 23, 2016
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