Copy of Lab 2 - Ohm’s Law
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University of Massachusetts, Amherst *
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152
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Electrical Engineering
Date
Jan 9, 2024
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Lab 2 - Ohm’s Law
Overview
In this lab you will learn about electrical resistance and its connection to
Ohm’s Law
. Ohm’s Law
simply states when a
voltage
is applied to a circuit, a
current
will flow through the circuit.
The amount of current is directly proportional to the applied voltage.
(1)
The portionally between the voltage and current is the
resistance
of the circuit. If the resistance
is a constant (it does not change if the applied voltage changes), then resistance obeys
Ohm’s Law:
(2)
The units of resistance are measured (not surprisingly) in ohms (
). In this lab you will use
resistors, circuit elements with constant resistance, to construct different circuits and test Ohm’s
law. Since a resistor’s resistance is constant, it is said to be
ohmic
. But not all circuit elements
are ohmic as you will see in later labs.
Part 1 - The Voltage Divider
A voltage divider is one the simplest circuit you can construct. I basically consist of two resistors
connected in
series
with each other. You will construct a voltage divider using the breadboard
and two 10,000
or 10k
. Remember 1k
= 1,000
.
From the Electricity and Magnetism Accessory Pack take out two 10k
resistors from the
pack.You can check if you have the right resistors using the color bands on them. The color
bands represent a numerical code for the resistance of the resistor. For at 4-band resistor the
color code is:
A helpful calculator for resistance from color code
The 10k
resistors in your E&M pack have the color bands brown (
1
), black (
0
), orange (
1000
),
and gold (
5%
). The color corresponds to either a digit, a multiplier, or a tolerance. So for the
case of a 10k
, the resistance is:
(
1
10 +
0
1)
1000
5%
= 10,000
5%
The tolerance means that the manufacturer rates the value of resistance within a percentage of
the nominal value. For example, the actual value of the resistance of the 10k
with a tolerance
of 5% may be between 9.5k
and 10.5k
.
Now construct a voltage divider. Here is a circuit diagram of what the voltage divider looks like:
You will also need to connect wires to the iOLab. The iOLab will function as both a voltage
source and a voltmeter and ammeter.
●
Connect the top end of the first resistor (R1) to the 3.3V voltage source on the iOLab.
●
Connect the bottom end of R1 to the A7 input. A7 will function as a voltmeter.
●
Connect the bottom end of the second resistor (R2) to the GND on the iOLab.
When you are done you voltage divider may look like this:
●
Red alligator clip is connected to 3.3V
●
Blue alligator clip is connected to A7.
●
White alligator clip is connected to GND.
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Now let’s measure the voltage of the voltage divider. Setup the iOLab and iOLab app:
●
Connect the dongle to the USB port on your computer.
●
Turn on the iOLab.
●
Launch the iOLab app and select Analog 7 (A7) from the list of sensors.
●
Click Record and record the voltage measured by Analog 7. Allow the recording to run
for 5 to 10 seconds.
●
Click the Analog Mode button (bar chart button) on the toolbar in the app. Click and hold
the mouse arrow and drag the arrow across a segment of the data in the graph.
1.
Take a screenshot of the iOLab app with the data of the voltage divider and paste it here:
2.
Record the average voltage (
) +/- and the uncertainty (
) measured by A7 in the table
below:
Average Voltage (V)
1.6413 V +/- .0034 V
3.
How does the value of the average voltage compare to the voltage supplied by the
iOLab? Explain how the voltage divider “divides” the voltage in terms of the resistance of
the resistors. The voltage divider divides the voltage basically in half as the current has
to run through the resistors before reaching the output, resulting in the resistances
halving the input voltage by the time the current reaches the output. The resistors are in
series, thus the resistance is doubled and the voltage is halved.
Part 2 - Ohm’s Law
Now let’s test Ohm’s Law. You will construct a circuit and measure both the voltage across and
current through the circuit using the iOLab. First you will construct a circuit that looks like this:
●
Replace the second resistor R2 with a 1
resistor.
●
A 1
resistor has band colors brown(
1
), black(
0
), gold(
0.1
), and gold(
5%
).
Your circuit should look like this:
●
Red alligator clip is connected to DAC.
●
Blue alligator clip is connected to G+.
●
Yellow alligator clip is connected to G-.
●
White alligator clip is connected to GND.
Now you need to setup the iOlab app:
●
Uncheck Analog 7 and check High Gain sensor.
●
On the toolbar click the settings button (cog) > Expert Mode > Output configuration.
●
At the bottom of the iOLab app window the DAC Output should appear.
●
In the DAC Output menu select 0.0V then click On.
Now it is time to take some data:
●
Click the Record button and record the High Gain voltage for about 5 seconds.
●
If you cannot see any line at all, it is because the line is too high off the graph, so we
can’t see it. To fix this, click the settings wheel in the bottom left corner of the graph and
increase the maximum y value.
●
While still recording, change the DAC Output voltage menu to 0.5V. Record for 5 more
seconds.
●
Change the DAC Output voltage in 0.5V steps up to 3.0V. With each change in voltage
record the High Gain voltage for about 5 seconds.
When you are done your data should looks like this:
Each jump in the DAC Output voltage means the voltage across the 1
resistor jumps. But
what about the current flowing through the circuit? How can we measure that? It turns out the 1
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IS the ammeter! Because we measure the voltage across the 1
, and we know the resistor is
, then using Ohm’s Law the current through the resistor is simply voltage divided by
.
Now you will need to measure the current for each 0.5V step in your data. Open the Google
Sheet Ohm’s Law - Student Data to record your measurements:
Ohm's Law - Student Data
Make sure you make a copy of the spreadsheet for yourself. Click the menu File > Make a copy.
Your copy will be in your UMass Google Drive. You can edit your copy by entering data, making
graphs and performing calculations.
4.
Measure the current for each 0.5V step in your data.
a.
Use the Analysis Mode to measure the current for each flat segment of data.
b.
Record the values of current in the Google Sheet.
5.
Make a scatter plot of Voltage vs. Current:
a.
Click on the top of the column with the values of Current (mA) (Column A).
b.
Press and hold the Ctrl key and click on the top of the column with the Voltage
(V) values (Column B).
c.
Release Ctrl key and open the Insert menu and select Chart.
d.
The Charter Editor will open. In Setup choose Chart Type > Scatter Chart,
e.
Make sure the X-axis indicates Current (mA).
f.
Make sure the Series indicates Voltage (V). If Current (mA) is in the Series,
remove it.
g.
In Customize > Chart & axis titles add axes titles with units to your graph.
h.
Copy and paste your plot into this document below here:
6.
Look closely at your graph of Voltage vs. Current and you will see it should be linear.
What does the slope of Voltage vs. Current
physically
represent? Hint: look back at
Ohm’s Law Equation 2.
V/I = R so the slope of the graph represents resistance.
In the spreadsheet there is a built-in function called LINEST. LINEST estimates the best fit line
(the slope
and intercept
) for a set of data. LINEST will return a table with
results of the slope and intercept and the uncertainties of slope and intercept of a line that best
fits the relationship between Voltage and Current. The table of results is organized like this:
LINEST
Best Slope
Best Intercept
Uncertainty of Slope
Uncertainty of Intercept
Correlation Coefficient R2
Standard Error
F Statistics
Degrees of Freedom
Sum of Square Regression
Sum of Square Residuals
7.
Use the built in function LINEST to determine the best slope of Voltage vs. Current.
a.
In an empty cell E2 enter =LINEST(range of y,range of x,1,1).
b.
The range of y should be the range of cells with the values of Voltage in Column
B (B2:B8).
c.
The range of x should be the range of cells with the values of Current in Column
A (A2:A8).
Part 3 - Resistors in Series and Parallel
Real electrical circuits have complex networks of all kinds of circuit elements. Let’s consider
some basic combinations of circuit elements, namely
resistors in series and parallel
.
Parallel Resistors
Let’s start by analyzing the two resistors connected in parallel. Construct a circuit with two 10k
in parallel. Two resistors are in parallel with each other if the ends of the resistor are connected
to the same potentials. That means the same voltage is across the two resistors.
When two resistors are in parallel, there are two paths for the electric current to travel through.
The two resistors combined act as a single resistor with resistance
.
8.
Do you think the combined resistance
of the two resistors in parallel will be greater
than or less than the individual resistance of each resistor? Explain your answer.
The combined resistance of two resistors in parallel will be less than the individual
resistances, because two resistors in parallel of the same magnitude halves the total
resistance.
9.
Based on your answer to question 9, will more or less current flow through the two
resistors in parallel than an individual resistor? Explain your answer.
Since I = V/R, and since the resistance
should
be less, more current will flow through in
parallel than through an individual resistor.
Your circuit should look like this:
●
Red alligator clip is connected to 3.3V.
●
Blue alligator clip is connected to G+.
●
Yellow alligator clip is connected to G-.
●
White alligator clip is connected to GND.
When two or more resistors are connected in parallel, the combined resistance is:
(3)
10. Using the resistance of two 10k
resistor, calculate the combined resistance
if the
resistors are connected in parallel.
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(
)
1/10k +1/10k = 1/5000 = 1/R
R = 5000 ohms
11. If 3.3V is applied to the two resistors in parallel, calculate the amount of current flowing
through the circuit.
Current (A)
.00066 Amps
Make sure the iOlab app is setup and ready to record data:
●
Check High Gain sensor.
●
Click the Record button and record the High Gain voltage for about 5 seconds.
●
As the data is recording,
carefully
remove the second 10k
resistor.
●
Continue recording for about 5 more seconds then click Stop.
12. Take a screenshot of the iOLab app with the data of the two resistors in parallel and
paste it here. If you cannot see any line at all, it is because the line is too high off the
graph, so we can’t see it. To fix this, click the settings wheel in the bottom left corner of
the graph and increase the maximum y value.
13. When you removed the second 10k
resistor, did the current change (increase,
decrease or stay the same? Is this consistent with your answer to question 10? Explain
your answer. It decreased. The current halves, which is what it’s supposed to do
because I removed the resistor that was in parallel with the first resistor, increasing the
resistance, so all that was left were 2 resistors in series.
Series Resistors
Now let’s observe what happens when two resistors are connected in
series
. Two or more
resistors are in series if there is only one path the current can flow through the circuit. The
current flows sequentially through each resistor. Construct a circuit with a 10k
resistor and a
4.7k
resistor in series. The 4.7k
resistor in your E&M pack is in the packet. The color bands
on the 4.7k
resistor are yellow(
4
), purple(
7
), red(
100
), and gold(
5%
). Based on the bands the
nominal resistance is:
(
4
10+
7
1)
100
5%
= 4,700
5%
Your circuit should look like this:
●
Red alligator clip is connected to 3.3V.
●
Blue alligator clip is connected to G+.
●
Yellow alligator clip is connected to G-.
●
White alligator clip is connected to GND.
When two resistors are in series, there is only one path for the electric current to travel through.
The two resistors combined act as a single resistor with resistance
.
14. Do you think the combined resistance
of the two resistors in series will be greater
than or less than the individual resistance of each resistor? Explain your answer.
The combined resistance of the two resistors in series will be more than the individual
resistance, because the current has to flow through both resistors, so both resistances
act with each other against the current.
15. Based on your answer to question 14, will more or less current flow through the circuit
with two resistors in series, compared to the circuit with only one resistor? Explain your
answer.
Less current will flow with two resistors in series in comparison to the circuit with only
one resistor because I=V/R, so when R is greater, I decreases.
When two or more resistors are connected in series, the combined resistance is:
(3)
16. Using the resistance of two resistors, calculate the combined resistance
if the
resistors are connected in series.
(
)
14700 ohms
17. If 3.3V is applied to the two resistors in series, calculate the amount of current flowing
through the circuit.
Current (A)
2.24*10^-4
Now record data:
●
Check High Gain sensor.
●
Click the Record button and record the High Gain voltage for about 5 seconds then click
Stop.
●
Click the Analog Mode button on the toolbar in the app. Click and hold the mouse arrow
and drag the arrow across a segment of the data in the graph.
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18. Take a screenshot of the iOLab app with the data of the two resistors in series and paste
it here:
19. Record the average current (
) +/- the uncertainty (
) measured by High Gain in the
table below:
Current (A)
.196 +/- .00030
20. How does the value of the average current compare to the current calculated in question
17?
This value is way higher than the current calculated, and if anything, the uncertainty is
closer to the calculated current in comparison to the actual average current. I’m not sure why it
is so high, because my setup is exactly like that of the example.
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mewith Question
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no calculation
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