Potential Dividers and Voltmeters

[Peter G.]

Hi,

I have a simple arrangement: Two 500 kΩ resistors connected in series. The teacher illustrates what it would be like to measure the p.d across one of these 500 kΩ resistors with a voltmeter with 50 kΩ resistance. To do so, he calculated the resistance in parallel of the voltmeter and one of the resistors. He got 45.5 kΩ. So far so good, but what he did next however is what I don't understand. He divides 45.5 kΩ by the total 545 kΩ and multiplies it by the e.m.f. Why would the voltmeter read (45.5 kΩ / 545 kΩ) x 12 V? I thought it would measure 6V out of the 12 of the source since both resistors had the same resistance. The voltmeter accounts for itself?

Thanks,
Peter G.

 

[PeterO]

Hi,

I have a simple arrangement: Two 500 kΩ resistors connected in series. The teacher illustrates what it would be like to measure the p.d across one of these 500 kΩ resistors with a voltmeter with 50 kΩ resistance. To do so, he calculated the resistance in parallel of the voltmeter and one of the resistors. He got 45.5 kΩ. So far so good, but what he did next however is what I don't understand. He divides 45.5 kΩ by the total 545 kΩ and multiplies it by the e.m.f. Why would the voltmeter read (45.5 kΩ / 545 kΩ) x 12 V? I thought it would measure 6V out of the 12 of the source since both resistors had the same resistance. The voltmeter accounts for itself?

Thanks,
Peter G.

What he did is correct. Voltmeters give erroneous readings if you use them to measure the PD across a resistor with resistance of a similar , or larger, value to the meter itself.
Actually they give accurate readings, but the inclusion of the meter distorts the circuit.
To measure the PD in the circuit mentioned you would need a Voltmeter of much higher resistance. A CRO would be better.

 

[Peter G.]

But then why when we set up the same arrangement, but, instead of a voltmeter we use a resistor the p.d across a resistor would be 6 V in this example?

 

[PeterO]

But then why when we set up the same arrangement, but, instead of a voltmeter we use a resistor the p.d across a resistor would be 6 V in this example?

If you set up a voltage divider where the two parts were:
#1 a 500 000 Ohm resistor
#2 a 500 000 Ohm and a 50 000 Ohm resistor in parallel

Then the voltage would not be divided 6 - 6

it would be divided 1 - 11, since one part of the divider has 11 times the resistance of the other.

 

[rwisz]

Hello Peter,

Thank you for sharing your question about potential dividers and voltmeters. I can understand why you may be confused about the calculation your teacher did. Let me explain it in more detail.

Firstly, the reason why the voltmeter reads (45.5 kΩ / 545 kΩ) x 12 V is because of the way potential dividers work. In this simple arrangement with two 500 kΩ resistors in series, the total resistance is 1000 kΩ. This means that the voltage across each resistor will be half of the source voltage, which is 6 V. However, when we add the voltmeter with a resistance of 50 kΩ in parallel with one of the resistors, it changes the effective resistance in that branch. This is where the calculation of 45.5 kΩ comes from - it is the parallel resistance of the voltmeter and one of the 500 kΩ resistors.

Now, why do we divide this by the total resistance of 545 kΩ? This is because the voltage across each component in a series circuit is proportional to its resistance. So, if we have two components with different resistances in series, the voltage drop across each component will also be different. In this case, the voltmeter has a much lower resistance compared to the 500 kΩ resistor, so it will have a larger voltage drop across it. By dividing the resistance of the voltmeter and one resistor by the total resistance, we are essentially calculating the fraction of the total voltage that is dropped across the voltmeter.

Finally, we multiply this fraction by the source voltage to get the voltage reading on the voltmeter. This is because the voltmeter is measuring the voltage across the 500 kΩ resistor, not the total voltage of the circuit. So, in your example, the voltmeter would read 6 V because it is measuring the voltage across the 500 kΩ resistor, which is half of the total voltage of 12 V.

I hope this helps to clarify your doubts. Let me know if you have any further questions.

Best,