Batteries, Current Flow, and Ohm's Law (4/9/2015)

Current Flow: Lighting a Bulb

 

Pictured above are two examples of how we can light a bulb using only a wire and battery. A picture of a method that doesn't work is also shown. A second failure is depicted on the whiteboard below, but not pictured above. The reason two of our methods work, while the others do not, is because in order for the bulb to light, the current through the wire must flow from one end of the battery to the other, while passing through a conductive surface on the bulb. The reasoning for this is because the tungsten filament inside the bulb is responsible for lighting it. The filament rests against the conductive metal portion of the bulb, so for the filament to light up the current must be carried through the metal to the tungsten.

 

An example of how adding a second battery to the circuit can make the light brighter.

An example of how the current is disrupted when the batteries are placed at similar ends. Energy is not allowed to flow through the circuit due to opposite charges.

Charge Demo (What was the device called?)

Caption explaining what the device is

In the photo above-left, the device is holding a charge transferred to it by a metal rod which gathered static (by way of Mason rubbing it against an unknown animal fur). Each of the foil flaps gathered identical charges which result in them repelling each other as shown. In the above-right photo, the charge has been discharged into Professor Mason, which cause the foil flaps to fall back into starting position. 

Our incorrect prediction to what will happen when one end of a battery is placed on the orb.

The picture below proves our prediction wrong, which referred to what will happen when one end of the battery touches the orb. When the battery was placed on the orb, nothing happened and the foil stayed in place. The reason for this is because there is no current flowing through this system. There would need to be some sort of connection between the opposite ends of the battery to cause a charge to flow through the machine.

Modeling a Simple Electric Circuit

In the photo above, Prof Mason is explaining why you need a wire to go back from the bulb to the battery. The reason for this is shown in the picture, where positive charges are flowing into the bulb, but if there is no exit path for them, then they build up in the bulb and cause it to stop working. There must be an equilibrium of charges, where charges must be allowed to exit as more charges are entering, in order to prevent a buildup and provide a constant flow of charge and energy.

The Concept of Electric Potential Difference (Voltage)

For this section of the lab, there is a waterfall that is going to power a hydroelectric plant. The water at high elevation will travel through a pipe (penstock) to a turbine at lower elevation. This causes the water to strike a series of buckets on the perimeter of a wheel mounted of the generator, cause it to turn. We were asked to figure out how big the generator should be, but ultimately we needed to figure out two things that we must know about the waterfall in order to answer the question. These two things happened to be the height at which the water is falling from (related to potential energy), and the flow rate of the water (related to current). 

Model for What a Battery Does

A calculation of what units are being used in the Power = Voltage * Current equation.

In this problem, a battery with a voltage of 1.5 V is connected to a bulb, which is the connected back to the opposite end of the battery. A current of 0.12 A is running through the circuit and we were asked to calculate the amount of power produced.  Using the equation Power = Voltage * Current, we can easily calculate the answer, which is 0.18 J/s or 0.18 W.

Developing a Model for Current Flow

Shown above is the correct answer to which model of current flow is correct regarding circuits. For this model, the direction of the current flowed only one way from a battery, to a bulb, and back to the battery along a circuit. Also, the magnitude of the current was the same throughout the entire circuit. It makes sense that the current flows in only one direction, because if the current was not flowing this way, then the bulb would not work because there would not be a constant charge running through it. There must also be an equal current through the circuit because if not, then the bulb would not work due to a build up of charge at a certain point. This relates to the equilibrium of charges needed to power a bulb.

Measuring Current with an Ammeter

Pictured above is an ammeter which has now been introduced into our battery-bulb-battery circuit. The purpose of the ammeter is to measure the value of current flowing through the circuit we have been using. The reading shows a value of approximately 50 mA in the positive direction.

Pictured above is the same circuit, but this time the leads going into and out of the ammeter have been reversed. The result is another reading of approximately 50 mA.

The significance of this result is that it proves that the D model mentioned in the previous activity is indeed the correct. The current has not changed and remains the same throughout the circuit, regardless of which portion of the circuit is being measured.

Miscellaneous Information from the Lecture

We were asked what four things would we need to know to determine the charge in a volume. Our answers above are correct, which results in the formula Q=q r_n A v_d. 

When a Voltage is applied across the ends of a wire, an Electric Field is created inside the wire, E = V/L where L is the length of the wire.

Above is our correct calculation of the drift velocity in a 12-gauge copper wire in a home that has a cross-sectional area of 3.31x10^-6 m² and carries a current of 10 A. The conduction electron density in copper is 8.49x10²⁸ electrons/m³.

Ohm's Law

Pictured above are our two predictions for Current vs Voltage and Voltage vs Current. Our prediction is that there will be a direct proportionality resulting in a linear graph. So when voltage (or current) increases, current (or voltage) will increase as well. This makes sense from looking at the equation used earlier where Power = Voltage * Current. Logically, if either the value of voltage or current increases/decreases, then it will have a proportional effect on the opposite value.

A circuit consisting of a voltage source (power supply), a resistor, and an ammeter (logger pro in this case, in order to graph the results).

After Prof Mason fidgeted with the circuit shown above, the result were these graphs of Electric Potential (Voltage) vs Time and Current vs Time. As you can see, the graphs are proportional, which means they look the same, but have different values. So there is direct proportionality between electric potential and current, which accurately corresponds to our prediction graph of current vs voltage.

The second graph depicts current vs voltage, which further proves our prediction correct. 

In order to calculate the value of the slope in the graph above, one must take the derivative of Voltage over Current. This is also the value of R (resistance).

Here, we have calculated the units of resistance (R) in order to better understand what we are working with. Since R is voltage over current, the units become joules/second over coulombs/second, which results in J*s/C^2. We can further simplify this by breaking down joules and ending up with the equation: kg*m^2/s*C^2. Alternatively, this results in 1 volt per amp, which is equal to 1 Ohm.

Resistivity

We were told the lengths of two coils made of identical material, 200m and 160m, and asked to predict which would have more resistance, the longer or the shorter. We correctly predicted that the longer would have more resistance, which turned out to be 21.4 Ohms. The shorter coil had a resistance of 17.4 Ohms. This makes sense, because the longer coil has more ions that the charges flowing through it are running into. This will slow down the electrons flowing through the coil.

As seen above, when comparing the ratios (in red) for each of the wire's lengths and Ohms, the values were approximately the same. This signifies a direct proportionality between resistance and length. As stated in the previous paragraph, it makes sense that a longer coil will have more resistance due to more ions for the charges to collide with.

The values in black shown above are ratio calculations between a coil from the first example, and a new coil with a length of 80 m and 8.7 Ohms. The resulting ratios were 2 and 2, proving proportionality yet again.

The equation symbolizing the relationship between resistance and resistivity is R=(rho)*(L/A). In one example, we were dealing with two coils of equal length (200m) and equal resistivity (rho). We also had the diameter of each coil (D1=0.025m and D2=0.032m), and R1=21.4 Ohms. Since L and rho are constant, we can cancel them out and write the two equations as A1*R1=A2*R2 in order to solve for R2. The result is 13.1 Ohms for the second coil. The calculation in green seen above shows how values cancel out in this problem. In this case, the area is that of a circle, and after several variables cancel out you're left with D1^2*R1=D2^2*R2.

As you see, the resistance of the fatter coil was lower than that of the thinner one. This makes sense not only because of the equation for R (where R is directly proportional to L and rho, and inversely proportional to A), but also because if you think of how a water flows through a hose it becomes obvious that a hose with a larger diameter meets lower water flow resistance.

Summary

In summation, we have looked at current flow and its various properties, such as direction (positive) and what affects the rate of flow (resistance and resistivity). Based on these analysis, we have a much better understanding of how a circuit works, and how to insure a current will flow through it. We also looked into Ohm's Law, which allowed us to relate various properties of eletrical circuits, such as voltage, power, and current, and see how they affect one another.