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# Proving Ohm's Law

Proving Ohm’s Law

Virgil Nunan
April 14, 2013
Denver School of the Arts, Physics Department Period 3, Professor Christopher Lyke

1.
Motivation and Main Objective
1.1
Purpose
The purpose of this lab was to learn how to construct circuits and measure or calculate basic circuit properties while proving Ohm’s law.
1.2.
Materials
- 10 Wires with alligator clips.
- Table to hold the apparatus.
- Ammeter to measure current.
- Voltmeter to measure voltage.
- Pencil & paper to record data.
- Two batteries for energy.
- Two light bulbs to provide resistance in the circuit.

2.
Equations
2.1.
Letters and Symbols
- ? (omega) is the unit for resistance, measured in Ohms (volts/ampere).
- “I” is the letter for current, measured in amperes (volts/?).
- “V” is the letter for voltage, measured in volts (joule/coulomb).
- “R” is the letter for resistance as represented in equations (? = unit for resistance).
2.2.
Equations**

? = V/I
V = I • ?
?V = I • R

2.3.
What is Ohm’s Law?
Ohms quantify the current resistance at a point on a circuit. Taking volts over current at a certain point (the point of resistance–a light bulb, for example) will tell you how many volts are present per ampere. Comparing this number to a point that isn't on the resistor will show that the current is restricted at the resistor, and unrestricted elsewhere on the circuit.

To obtain "R," one must simply divide voltage by ?. (if you have the voltage and current of the circuit, use one of the equations in §2.2 to solve for ?).

Ohm's law states that:

?V = R • I

This statement will be the main idea in the proof of Ohm's Law throughout this paper.

3.
Testing Ohm's Law
3.1.
Procedure For a Circuit with One Battery and One Resistor

3.1.1.
Determining Locations of
Various Apparatuses. We determined that the ammeter must be in series with the circuit because it is measuring the current, so the current must flow through the resistor and ammeter. The voltmeter can be in parallel to the circuit because the voltmeter is measuring the voltage before and after the resistor.

3.1.2.
Testing the Circuit. To make sure the battery had chemical energy, we connected the battery to the voltmeter, and it measured at 1.5 Volts (it was a 1.5 volt battery).

3.1.3.
Putting Everything Together in the Circuit. We connected the ammeter, voltmeter, lightbulb, and battery with alligator clips (ammeter in series, voltmeter in parallel as explained in §3.1.1). The battery was connected to the ammeter, then the lightbulb, followed by the other terminal of the battery. Then, another two wires ran from each of the terminals of the battery to the ports of the voltmeter. This is illustrated in §3.1.4.

3.1.5.
Recording Results. After the connections are made, we read the ammeter and voltmeter, and recorded those results. We then used the equations from §2.2 to calculate the resistance created by the resistor (lightbulb).
3.2.
Procedure for For a Circuit with x
Batteries and n Resistors
This process has steps similar to those in §3.1. The only difference between these two is the look of the circuit and the results from the test. For example, a circuit with an extra battery will have more voltage than the circuit with one battery. This will increase the resistance because voltage has a direct relationship with ?.

3.2.1.
Variations We Used in our Lab Exploration.
- One battery, two resistors.
- Two batteries, one resistor.
- Two batteries, two resistors.

4.
Interpreting Results
4.1.
Understanding the Meters
Once we run our circuit, our meters give us results. The ammeter gives us the current of the circuit in amperes, also known in equations such as the ones in §2.2 as “I.” The voltmeter gives us the voltage of the battery in volts, also known in equations such as the ones in §2.2 as “V.”
4.2.
Determining ?
Now that we have the elements needed to solve for ?, we can take one of two of the equations from §2.2 to solve for ?:

? = V/I
Or
V = I • ?

5.
Results and Data From our Experiment
5.1.
One Battery and One Resistor Circuit

5.1.1.
The Meter Readings. Because we used one, 1.5 V battery, our voltmeter read 1.5 volts. We had one lightbulb resistor, and our ammeter read 2.6 amperes.

5.1.2.
Determining ?. Now that we have our “V” and our “I” values, we can plug those into the equation:

? = V/I
?? = 1.5V/2.6A
.58? = 1.5V/2.6A
From the equation above, we now know that the resistance is equal to .58 Ohms.

5.2.
Two Batteries and One Resistor Circuit

5.2.1.
The Meter Readings. Because we used two, 1.5 V battery, our voltmeter read 3.0 volts. Because we kept only one resistor, our ? value stayed the same, because there will be the same amount of resistance (we are not altering the resistor in any way that it would change to more or less resistance). Therefore, our ? = .58, and V = 3.

5.2.2.
Determining “I.” Now that we have our “V” and our ? values, we can plug those into the equation:

V = I • ?
?3V = I • .58?
3V/.58? = I
5.17A = I

From the equation above, we now know that the current is equal to 5.17 amperes.
5.3.
One Battery and Two Resistor Circuit

5.3.1.
The Meter Readings. One, 1.5 V batter gives us a 1.5 voltmeter reading. Our ammeter read 1.3 amperes.

5.3.2.
Determining ?. We have our “V” and our “I” values, we can plug those into the equation:

? = V/I
?? = 1.5V/1.3A
1.15? = 1.5V/1.3A
5.4.
Two Batteries and Two Resistor Circuit.

5.4.1.
The Meter Readings. Because we used two, 1.5 V batteries, our voltmeter read 3.0 volts. Because the resistance is doubled (and the resistor we added is the same as the other resistor), we can say that the resistance is 2 • .58? (.58? is the resistance we found that one lightbulb had in a circuit), or 1.16.

5.4.2.
Determining “I.” Now that we have our “V” and our ? values, we can plug those into the equation:

V = I • ?
?3V = I • 1.16?
3V/1.16? = I
2.61A = I

From this equation, we now now that doubling resistance and batteries will give us 2.61 amperes of current.

6.
Data-Based Conclusions
Was Georg Ohm correct? Yes, he was (yes, we achieved our purpose). This is why:
6.1.
Doubling Resistors will Double ?
In §5.1.2, we found that one resistor and one battery created .58? of resistivity. In §5.3.2, we found that with the same number of batteries (one) and doubling the resistors (two), our resistance became 1.15?:

1.15? ? 2 • .58?

The resistance doubled when the number of resistors doubled.
6.2.
Doubling the Resistance Will Halve
the Current
In §5.1.2, we found that one resistor and one battery created .58? of resistivity and 2.6A of current. In §5.3.2, we found that with the same number of batteries (one) and doubling the resistors (two), our resistance became 1.15? and our current was 1.3A:

(1/2) • 2.6A = 1.3A

The current was halved when the resistance was doubled.
6.3.
Doubling the Voltage Will Double Current, but Leave Resistance

Unchanged

6.3.1.
Proving Algebraically. According to the equation:

V = I • ?

It can be rearranged to solve for "I" by doing:

I = V/?

Because resistivity only changed when you changed the number of resistors in the circuit, let’s set ? as a constant n because we are keeping the number of resistors constant, so that:

I = V/n

Now, when we use 2 for “V,” “I” will equal 2, meaning it’s doubled.

6.3.2.
Proving in our Experiment. In §5.2.1, we had one battery and one resistor, and the current was 2.6A. In §5.2.2, we had two batteries (doubling the voltage) and one resistor, and the current was 1.3A.
6.4
Data Table

7.
Final Conclusion
Ohm’s law works because when you decrease the amount of resistance in a circuit, you are decreasing the amount of “stuff” blocking the way of current. Increasing the amount of voltage in a circuit is like increasing the amount of “pressure” you are putting on the current, so the current will increase.
7.1.
River Metaphor
A river may flow at a certain speed until beavers start building a dam (that’s the resistance). The dam causes the water flow to slow down. All of a sudden, massive snowmelt upstream causes an increase in the pressure (that’s the voltage) of the water stream, causing the water to flow much faster.

8.
Future Improvements
In the future, if I were to conduct this lab again, I would be more organized. Wires and meters and lightbulbs and batteries–it gets a little crazy sometimes when trying to arrange them in a way that works.