By clicking here, one can download a simulation from PhET at the University of Colorado that can be used to create direct current circuits using wires, resistors, batteries, light bulbs and switches (pictured to the left). To measure different values, two types of ammeters and a voltmeter can be used. Thus, I decided to test how the number of resistors in a circuit with a 9 volt battery and a light bulb with a resistance of 10 ohms affects the voltage in the circuit (pictured to the right).

Before testing this, though, a few standards had to be placed. The key one is to find where and how to measure the voltage in the circuit, as this depends on the placement of the nodes of the voltmeter as well as that of the resistors and the battery. Before I go into this, it is important to understand what exactly voltage is. It is measured in volts, or joules per coulomb, and is calculated by finding the difference between the **electric potential energy** of two different points. When using the voltmeter, these two points are the nodes of the voltmeter. To make sure that the numbers are always proportional, the **positive node** of the voltmeter is always connected to the wire that connects the positive end of the battery to one of the nodes of the light bulb and the **negative node **is connected to the wire connecting the resistors to the light bulb (the resistors are connected directly to the negative end of the battery).

**Resistors** are a component of a circuit that apply electrical resistance. The resistance is calculated in ohms. Each of our resistors, including the light bulb apply **10 ohms of resistance**. **Ohm’s Law** states that the new current throughout the circuit is the **voltage across the conductor (the circuit between the two nodes of the battery) divided by the resistance of the resistor**.

This brings us to our *HYPOTHESIS*

By solving Ohm’s law for voltage, we get **Voltage=Current*Resistance**. Thus, if we add resistors to a circuit, the voltage across the light bulb will decrease proportionally to the current because if the resistors change the current of a circuit, they will equally change the electric potential of any given point on the circuit. Since these are both calculated by each other, it is impossible to calculate either without a given relationship.

**PROCEDURE**

- In the program listed at the top, make the circuit listed directly above.
- Record the voltage according to the resistance (remember that the light bulb has a resistance, too).
- Add a resistor to the negative node of the battery (between it and the wire that currently connects it to the light bulb). See graphic to the right. Attempt to maintain the length of each of the wires, so as to prevent a change in the voltage simply from the length of the wires (this simulation doesn’t work on a large enough scale to have an effect of the voltage by distance).
- Repeat step 2.
- Repeat step 3 followed by step 2 until sufficient data has been gathered.

**RESULTS**

Our test used five resistors. The data is shown in the table and graph below.

Note, in both the graph and the table, there are two result columns. To confirm my results, I tested the same circuit, but with two 9 volt batteries (18 volts total). The results turned out to be the same, proportionally.

**DISCUSSION**

After graphing the data, I used Microsoft Excel to find a trendline for the points. Interestingly enough, using a power trendline, the resulting equation was very straightforward. With a 9 volt battery and a set of 10 ohm resistors, the equation that gives any voltage based on the resistance in ohms is y (voltage) = 90x^-1 (resistance). With the battery expanded to 18 volts, the equation changes to y = 180x^-1. However, this doesn’t directly prove our claim that the voltage is affected proportionally to the **current**. Therefore, below are another table and graph detailing the current in amps in the same situation.

It is apparent that the the equation given to us for the current, y = 9x^-1 is directly proportional to the equation for voltage, which is just multiplied by a factor of 10 (this can be confirmed by the points acknowledged on the table.

Interestingly enough, the actual procedure of this experiment took me a while to understand. As mentioned earlier, the voltage depends greatly upon where the nodes of the voltmeter are placed. For example, if they are placed on the same wire of a circuit, the voltage will be 0, as there is no difference between the electric potentials of each. Similarly, if the nodes are placed on both ends of the battery, no matter how many resisters there are, the voltage will be 9, since the full voltage of a battery flows from one end to the other. This led me to my next question.

**Why does the voltage change due to resistors?**

To answer this, I first considered what was happening in the original circuit and consulted with the great physicist of Eastside Prep, Mr. Cross. Since voltage is a **difference **between electric potentials, it is indicated that one point has a potential 9 joules per coulomb greater than that of the other point. It is possible that this means one point has a potential of zero. In the battery, this is the + node, since its positive charge attracts electrons, preventing them from doing work, as the electrons from the – node of the battery do to power to light bulb. Therefore, the electric potential of the wire directly connected to the – node of a 9-volt battery **is always 9, and the electric potential of the positive node of the batter is always 0**. In the default circuit, this allows the full electric potential to go into powering the light bulb. However, when resistors are added, the electric potential is decreased, as, as Mr. Cross describes it, “a certain number of volts have to be ‘dropped off’ at each resistor.” Therefore, after each resistor, the voltage is one greater (assuming one of the nodes of the voltmeter remains on the negative side of the battery). **Remember, this sounds strange because voltage is a difference**. For example, imagine that there is a resistance of 30 ohms (1 light bulb, 2 resistors). The electric potential out of the negative node of the batter is 9; after the light bulb, it is 6; after the first resistor, it is 3; and finally, at the positive node of the battery (after the second resistor), it is 0. The voltage, in relation, is 0, 3, 6, and 9 volts respectively.

**In Conclusion,**

With the addition of resistors to a circuit, the voltage changes proportionally to the current, as stated by Ohm’s law, V=IR, where V is the voltage, I is the current in amps, and R is the resistance in ohm. In our situation, this equation is y = 90x^-1, as the voltage across the circuit is 9 volts, and the resistance of each resistor is 10 ohms. This change is because the resistors not only affect the current, but also the electric potential of any given point on the circuit, and thus the voltage, which is the difference between the electric potential of any two given points.

**RESOURCES**

Reid, Sam. “Circuit Construction Kit (DC Only).” *PhET*. University of Colorado, 2011. Web. 01 June 2012. <http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc>.

Wikipedia contributors. “Ampere.” *Wikipedia, The Free Encyclopedia*. Wikipedia, The Free Encyclopedia, 23 May. 2012. Web. 2 Jun. 2012.

Wikipedia contributors. “Resistor.” *Wikipedia, The Free Encyclopedia*. Wikipedia, The Free Encyclopedia, 1 Jun. 2012. Web. 2 Jun. 2012.

Wikipedia contributors. “Voltage.” *Wikipedia, The Free Encyclopedia*. Wikipedia, The Free Encyclopedia, 16 May. 2012. Web. 2 Jun. 2012.