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Hypothetically, suppose our resistance is 200 ohms. Quantitatively calculate the impact of a 1 Ohm ammeter resistance and a 1 megaohm voltmeter resistance (rather than ideal circumstances). Based on the uncertainties in the experiment, are these systemic errors likely to be relevant?
The experiment is measuring the voltage and current of a single resistor. (green, blue, brown, gold) with uncertainties of 0.01 V and 0.0001A
The systematic error based on 0.01V uncertainty is relevant for the Voltmeter measurements on both experiments.
The systematic error based on 0.0001A uncertainty is not relevant for the Ammeter measurements on both experiments.
Explanation:
CONNECTING AMMETER AND VOLTMETER TO A CIRCUIT FOR MEASUREMENT
When using a voltmeter or ammeter, you are connecting another resistor to the existing circuit and, thus, altering the circuit. Ideally, voltmeters and ammeters have a negligible effect on the circuit, but it is necessary to examine the circumstances under which they do or do not interfere.
THE VOLTMETER
The voltmeter, which is always connected in parallel with the device being measured. If the resistance of the voltmeter is appreciably greater than that of the device being measure, very little current flows through the voltmeter, and so no appreciable effect on the circuit. Because a large resistance in parallel with a small one has a combined resistance essentially equal to the small one. However, if the voltmeter’s resistance is comparable to that of the device being measured, this will result in a smaller resistance, this will appreciably affect the circuit. In this case, the voltage across the device is not the same as when the voltmeter is not part of the circuit.
THE AMMETER
An ammeter is usually connected in series to measure current in the branch of the circuit being measured so that its resistance adds to that branch. Often the Ammeter is designed with a resistance of that is very small compared to the resistances of the devices in the circuit, such that the extra resistance is negligible. In a situation when a very small load resistances are involved, or if the ammeter resistance is not as low in resistance as it ought to be, then the total series resistance is significantly greater, resulting in a reduction in the current of the branch being measured.
The experiment was designed to measure the resistance of 560Ω +/- 5% (588 – 532). To know if the systematic errors are likely to be relevant, we calculate the effect of the voltmeter and ammeter on the measurement of the 560Ω and compare it with that of measuring 200 Ω resistor. Â
TAKING MEASUREMENTS WITH 1MΩ VOLTMETER ACROSS 560Ω
The equivalent parallel resistance RE = RV*R1/RV+R1
Note R1 is the resistance been measure and RV is the voltmeter resistance
RE= 1,000,000*560/(1,000,000+569) = 560,000,000/1,000,560 =559.6865Ω
Difference from the actual value of resistor =560 – 559.6865 = 0.3135
TAKING MEASUREMENTS WITH 1MΩ VOLTMETER ACROSS 200Ω
The equivalent parallel resistance RE = RV*R2/RV+R2
Note R2 is the 200Ω resistance been measured
RE =1,000,000*200/(1,000,000+200)  =200,000,000/1,000,200 = 199.9600Ω
Difference from the actual value of resistor being measure = 200-199.96 = 0.04
The difference between the first and second measurements = 0.3135 -0.04 =0.2735 Â
This shows that the voltmeter reading is more accurate in the second measurement, the systematic error can be considered as not relevant when compared to the measured resistance value. But when the uncertain of 0.01V is taken into consideration, we can say that the error is relevant, as the measured values both exceeded the 0.01V uncertainty.
TAKING MEASUREMENTS WITH 1Ω AMMETER AT 9V
Assuming we connect a 6V battery to power the circuit such that we will have to measure the current flowing through the circuit. The first instance will be when we connect the 560Ω and the second instance is when we connect the 200Ω resistor.
MEASURING CIRCUIT CURRENT WITH 560Ω CONNECTED
The total series resistance RS = RA+R1 = 1+560 =561Ω
Current through the circuit I = V/RS =9/561 = 0.016042A
Current through the circuit without the Ammeter resistance I =V/R1 =9/560 = 0.01607
Difference in current =0.01607-0.01604 = 0.00003
MEASURING CIRCUIT CURRENT WITH 200Ω CONNECTED
The total series resistance RS = RA+R2 = 1+200 =201Ω
Current through the circuit I =V/RS = 9/201= 0.04477A
Current through the circuit without the Ammeter resistance I =V/R2 =9/200 = 0.045A
Difference in current =0.045-0.04477 = 0.00023
This shows that the accuracy of the Ammeter is better in the first experiment set up with the 560Ω than in the second experiment set up with 200Ω. When compare to the uncertainty requirement of 0.0001A the systematic error is not relevant.
