1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Current
(mA)
0.258
0.525
0.850
1.150
1.380
1.700
Data Set 2
Voltage
(V)
1.52
3.13
4.53
6.06
7.48
9.05
Current vs Voltage for batteries and resistor
Column
1.5
Linear Fit for: Data Set 2 | Current
x = my+b
m (Slope): 0.1928 mA/V
b (Y-Intercept): -0.04397 mA
Correlation: 0.9991
RMSE: 0.02588
1.0
Report by Ben Coburn
Page down for analysis and additional
data.
0.5
1
3
5
Voltage (V)
7
9
Goal: Get a relationship between current and voltage in a simple circuit.
Method: Attach two battery holders to a current meter and a resistor, using wires with alligator clips. The system is setup such that both battery holders can either be
connected in series or not connected at all, to vary the number of connected batteries from 1 to 6. A second meter is used to measure voltage across the resistor.
Equipment: 1 2 battery holder, 1 4 battery holder, 1 5.59k ohm resistor, 2 meters, one digital, one analog.
Data: See previous page.
Analysis: The fit equation from the graph is Current(I)= mVoltage(V)+b. The accepted equation for current with resistance is I = V*(1/R), so I and V correspond to themselves,
and m = 1/R. b shouldn't exist, and is probably a result of some experimental error, possibly resistance in the wires or measurement inaccuracies.
m from the equation = .1928mA/V = .0001928A/V = 0.001928ohms
1/5.59k ohms = 1.789*10^-4ohms, which is reaonably close to the calculated value.
Possible sources of error include the meters, possible errors reading the needle on my analog current meter, and maybe the resistance of the wires. After the second half of this
lab gave me a more off result, I check all my resistors with a better meter, and got a value of 5.71kohms for the resistor I used in this part, which is a probable source of error,
and causes me to wonder about the accuracy of the meter I used for this one in general, not just for resistance.
If I had the equipment and were doing this more seriously, I'd check the meters with a voltage and current standard, then start wondering about the accuracy of Ohm's Law if the
difference remained.
This is an excellent discussion that demonstrates an understanding of possible errors. The format of including method,
analysis, and discussion of error on the same page is an improvement over that given in the lab instructions.
Possible improvements:
Percentage difference should be calculated between the calculated and measured resistances.
The units on 0.001928 should be 1/ohm.
1
2
3
4
5
6
7
8
9
Current
(mA)
15.100
10.400
5.725
2.620
1.845
1.550
0.860
Data Set
Resistance
(kohms)
0.559
0.817
1.513
3.350
4.710
5.710
10.060
Current vs Resistance for batteries and resistor
Co
15
Auto Fit for Data Set | Current
Y = A/X
A: 8.552
RMSE: 0.06698
10
Teacher's comments:
It's generally good practice to include
the possibility of an intercept with the
fit (in the same way as was done in
Part A). Thus, one would fit Y = A/X +
B instead of Y = A/X. By seeing if the
intercept is close to the expected value
(0 in this case), one has additional
confirmation of the data and method.
For this fit, the intercept represents the
current for infinite resistance (1/R = 0).
I tried a fit and found the intercept to
be 0.08 ma, which is much smaller
than any current measured in the
experiment.
5
0
0
5
Resistance (kohms)
10
Goal: Examine the relationship between current and resistance in a simple circuit
Equipment: 1 2 battery holder, 1 4 battery holder, 2 meters, both capable of reading voltage and resistance, 6 resistors, from .560 to 10k ohms.
Method: First, we measured all the resistors with the good meter. Then we measured the voltage across a 6 batteries in 2 seperate holders, arranged in series, again using the good meter. We
tested the current for each resistor by attaching a 1 ohm resistor to the precise meter, then attaching leads to that and attaching them to opposite sides of the resistor we were testing. We used the
second meter to determine which resistor we were testing, and got the resistance from the voltage using V=IR, with R = 1
data: see previous page
Analysis: The fit equation from the graph is Current(I) = A/Resistance(R). The accepted equation for current with resistance is I = V/R, so I and R correspond to themselves and A = Voltage.
The measured voltage at the end of the experiment was 8.84V, which is off by .29V from the calculated number taken from the graph coefficient. I have absolutely no idea where this difference came
from, because voltage was measured at both the beginning and end of the experiment and was about the same. All the measurements where taken from a digital meter that should be accurate to at
least 3 significant figures, possibly as many as 4 or 5. I also measured the resistance in the leads to be .25ohms, which should be negligable. The only possibility I see is that maybe the resistors I
have don't actually obey Ohm's Law, but that seems unlikely. It's really more likely that my meter is just wrong and I don't know it.
I tried this lab once before, using a different set of meters, one of which actually displayed current, and ended up with a measured voltage of 9.02V and calculated voltage of 9.456V. The graph for
this is also on the previous page.
I don't have the tools for it, but the meters should probably be checked with voltage and resistance standards, and it might be interesting to somehow check if my resistors really do obey Ohm's Law,
probably by doing an experiment like the first part of this lab.
This is an excellent example of error investigation that goes beyond
discussion and includes the results of additional experimentation. The
experiment was repeated with a better meter and outstanding agreement
between the measured and calculated voltage was obtained.