# To find which of the circuits, shown below, are most suitable to measure a range of resistances, which the meters (the voltmeter and the ammeter) could be used to measure.

Nigel Evans

Physics Coursework – “How Do I Connect This Voltmeter”

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How do I connect this Voltmeter?

## Aim

To find which of the circuits, shown below, are most suitable to measure a range of resistances, which the meters (the voltmeter and the ammeter) could be used to measure.

## Prediction

I did not know which circuit would be the most accurate, to start with so I did a preliminary investigation, which consisted of setting up the two circuits above and then just putting two resistors in each and working out the required resistance. I did not see at the time that different value of resistance would make much of a difference until I commenced with my calculations. I discovered by using resistance values of 2200Ω and 4700Ω, that Circuit Two was better.

Circuit One

2200Ω = 2105Ω by meter readings

4700Ω = 4364Ω by meter readings

Circuit Two

2200Ω = 2222Ω by meter readings

4700Ω = 4600Ω by meter readings

As you can see from these initial findings Circuit Two is the better circuit for measuring resistance values according to the labelled resistance.

## Hypothesis

The manufacturers’ specifications as given on the Voltmeter and Ammeter are as follows:

Ammeter – Maximum Current 2mA, resistance 40Ω

Voltmeter – Maximum Pd 5V, Maximum current 100μA

After inspecting the above apparatus I have decided that the smallest value of the Current that I can accurately measure is 1 x 10-4 A and the maximum is 2 x 10-3 A. Any higher than 2 x 10-3A will be too high a reading for the ammeter. Also the minimum voltage that I think that I can accurately measure is 0.1V and the maximum is 5V (any higher will be to high for the voltmeter).

From the above values I have established the biggest and smallest resistance that these two meters can be used to measure:

From Ohms law it follows that

For the biggest resistance =

5 x 104 Ω

For the smallest resistance =

50Ω

Therefore the range of resistances that these meters can be used to measure is from 50Ω to 50000Ω.

I will be assuming that the Lab Pack has a negligible internal resistance.

The resistance of the Voltmeter calculated from Ohms Law, using the values specified on the device itself is 50000Ω:

= 5 x 104 Ω

The resistance of the ammeter is as specified on the ammeter itself which is 40Ω.

Prediction for Circuit One:

Circuit One as shown in the diagram above as the voltmeter connected in parallel around the resistor only. The ammeter is connected in series with the resistor and is not included in the parallel connection.

In these predictions I will be picking different resistances for the variable resistor, and then trying to calculate the resistance using the meter readings.

For 50Ω:

Specified Current in Resistor (I1) = 1 x 10-4

Set Resistance of R1 = 50Ω

Pd across Resistor from Ohms Law

V = IR

V = 50Ω x 1 x 10-4A

V = 0.005V

Current in Voltmeter (I2) – Using the Pd for the circuit calculated above and the resistance reading specified form the voltmeter readings.

= 1 x 10-7 A

The Total Current as derived from Kirchhoff’s first law:

I = I1 + I2

I = (1 x 10-4) + (1 x 10-7)

I = 1.001 x 10-4A

Therefore the resistance that should be obtained from the meter readings is as follows:

R = 49.95Ω

At the start the resistance I specified on the variable resistor was 50Ω. The reading calculated from the theoretical work above is 0.05Ω off this value. 0.05 expressed as a percentage of the original 50Ω is only 0.1%. So I derived from this that Circuit One is accurate for low resistances.

For 50000Ω:

Current in Resistor (I1) = 1 x 10-4

Resistance of R1 = 50000Ω

Pd across Resistor from Ohms Law

V = IR

V = 50000Ω x (1 x 10-4)

V = 5V

Current in V (I2) – Using the Pd for the circuit calculated above and the resistance reading specified form the voltmeter readings.

= 1 x 10-4 A

From Kirchhoff’s first law, the total current is:

I = I1 + I2

I = (1 x 10-4) + (1 x 10-4)

I = 2 x 10-4A

Therefore the resistance that should be obtained from the meter readings is as follows:

R = 25000Ω

At the start the resistance I specified on the variable resistor was 50000Ω. The reading above that I calculated from the theoretical work is 25000Ω off this value. 25000Ω expressed as a percentage of the original 50000Ω is 50%. This is a vast reading off the labelled resistance, so I can assume that Circuit One is not very accurate for high resistances.

For 25000Ω:

Current in Resistor (I1) = 1 x 10-4

Resistance of R1 = 25000Ω

Pd across Resistor from Ohms Law:

V = IR

V = 25000Ω x (1 x 10-4)

V = 2.5V

Current in V (I2) – Using the Pd for the circuit calculated above and the resistance reading specified form the voltmeter readings.

= 5 x 10-5 A

From Kirchhoff’s first law, the total current is:

I = I1 + I2

I = (1 x 10-4) + (5 x 10-5)

I = 1.5 x 10-4A

Therefore the resistance that should be obtained from the meter readings is as follows:

R = 16666.67Ω

At the start the resistance I specified on the variable resistor was 25000Ω. The reading that I calculated above is 8333.33Ω off this value. 8333.33Ω expressed as a percentage of the original 25000Ω is 33.3%. This is still quite a high value off the labelled resistance so I reflect that Circuit One is only accurate for relatively low resistances, but I concur that further calculations are needed, to investigate the matter further.

As I continued on with my calculations in exactly the same manner, I obtained the following readings:

As is shown above, and in the calculations I have made, this circuit set-up is better for lower resistances than it is for higher ones. The results range from only a 0.1% margin difference to a vast 50% difference.

I have actually plotted a graph of labelled resistance against the resistance obtained by the calculations above (graph on next page) and this is a curve. This shows that as the resistances become bigger, the value that Circuit One obtains due to the meter readings will become even further from the labelled resistance values.

Prediction for Circuit Two:

In these predictions I will be picking different resistances for the variable resistor, and then trying to calculate the resistance using the meter readings.

In these calculation the resistance of the ammeter will be R2 for the resistor this will be R1.

For 50Ω:

Current through Resistor (and therefore the ammeter also) = 1 x 10-4A

Total Resistance for the resistor and the ammeter in series (Rs) =

Rs = R1 + R2

Rs = 50 + 40

Rs = 90 Ω

Pd across resistor (and ammeter), from Ohms Law =

V = IR

V = (1 x 10-4) x 90

V = 0.009V

Total resistance calculated from meter readings, from Ohms law =

= 90Ω

For 50000Ω:

Current through Resistor (and therefore the ammeter also) = 1 x 10-4

Total Resistance for the resistor and the ammeter in series (Rs) =

Rs = R1 + R2

Rs = 50000 + 40

Rs = 50040 Ω

Pd across resistor (and ammeter), from Ohms Law =

V = IR

V = (1 x 10-4) x 50040