April 21 DC Circuits

Purpose: In todays class, we began our studies of direct circuits (DC). A direct current is the unidirectional flow of electric charge, which is usually produced by sources such as batteries. Direct current flows in a conductor such as a wire, and its current flows in a constant direction which is what makes it different from AC circuits. We discussed how DC circuits work in series and in parallel, and how resistors work in such circuits and how they are color coded in order to be read.

We began with the discussion of voltage and current in a DC circuit. We were asked to make some predictions of what would happen if the switch was closed instead of open in each scenario pictured below. 

Our prediction for this one was that all three light bulbs would light up. After conducting the experiment, we found our predictions to be incorrect. This is because the center of the circuit is negligible; the current flows on the outer part of the circuit. 

Our prediction here was that the upper light bulb would become dimmer and the lower light bulb would become brighter.
After experimentation, we found that there was no change. 

Final conclusion to experiment was that there was no change on the light bulbs in this DC circuit when the switch was closed. The reason for this is that voltage in a parallel circuit is the same throughout the whole circuit, it remain constant.



Our next activity we were asked the following:

Activity:  As a table have one half setup the two batteries and two bulbs from last week to make the bulbs as bright as possible.  Have the other half setup the bulbs to make them as dim as possible.  Observe the circuits and complete the following table:

Circuit Element Series / Parallel

Dim Bulbs Seeries
Bright Bulbs Parallel

Bright Bulbs Parallel

Dim Batteries Parallel

Bright Batteries series

Using a Multimeter

A digital multimeter is a device that can be used to measure either current, voltage, or resistance depending on how it is set up.  

SERIES CIRCUITS

We will start by making measurements of the current and voltage of your circuits

 VOLTAGE

 Measure the voltage across each of the batteries as shown.  First use both batteries in series, followed by just one battery.  

# Batteries	Voltage of source	VR1	VR2
1	1.26	0.62	0.62
2	2.55	1.26	1.26

# Batteries

Voltage of source

VR1

    VR2

1

1.26

0.62

     0.62

2

2.55

1.26

     1.26

Write a relationship between the voltage of the source (batteries) and the voltage across  the bulbs. 

The sum of the resistors is equal to the source of voltage.

2.  Use the same circuit as before EXCEPT put ammeters in positions 1, 2, and 3 in the circuit.

# Batteries

Voltage of source

I1 (A)

I2 (A)

I3 (A)

1			1.26            1.01     1.01	1.01
2			2.05              2.05    2.05	2.05

PARALLEL CIRCUITS.

1. Now measure the voltage across each of the bulbs and the batteries as shown.  

# Batteries	Voltage of source	VR1	VR2
1	1.26	1.26	1.26
2	2.55	2.55	2.55

2.  Use the same circuit as before EXCEPT put ammeters in positions 1, 2, and 3 in the circuit. 

#batteries	I1(A)	I2(A)	I3 (A)
1	2.0	2.0	2.0
2	4.0	4.0	8.1

Important rules in solving circuit problems:

components in a series circuit will have the same current throughout and voltage will divide.

Components connected in parallel will have the same voltage throughout and current will divide. 

Resistance and Its Measurement

For our next activity, we continue with the study of resistors. A resistor is a electrical component that induces electrical resistance as a circuit element. In our pervious activities and experiments, we created circuits which involved the use of light bulbs. Light bulbs are a perfect example of a type of resistor. Electrical energy is transferred to light and heat energy inside a light bulb when connected to a circuit, and even though all the current returns to the battery after flowing through the light bulb, potential energy has been lost. Potential energy has been lost because the light bulb puts resistance on the flow of electric current, thus, a light bulb is one kind of electrical resistance. However, a light bulb has a resistance that increases with temperature and current, so they do not made a good circuit element. To get a better understanding of resistors, we examine carbon resistors. Using Ohm's law, we have an equation for resistance in terms of potential difference and current, which is: R=V/I. We were told that carbon resistors will have a constant R for a range of different currents, which doesn't apply to light bulbs which is why they make poor ohmic resistors. Carbon resistors are measured in Ohms's. A carbon resistor usually has colored markings around it to signify its value in ohms

Here we have a detailed description of carbon resistors. The last band on a carbon resistor signifies its tolerance and the orientation of the other colors can tell us the resistance level which can be calculated by if equation: value in ohms = AB × 10^c ± D.   

Activity:  Decoding and Measuring Resistors

a. Decode the five resistors, determine the resistance of each.

Here we calculated the resistance of each resistor by hand using each one's color band orientation.

Here we have a resistor close up and as you can see the tolerance of this +/- 5, because it has a gold band at the end. We then measured the resistance of each resistor with a multimeter to find if our theoretical values match our experimental.

A blog question we were given was: Did the resistors match the colored coded value within the uncertainty.
The answer to this question was yes, the values we calculated by hand (pictured above) did in fact match the experimental.

Carbon Resistors in Parallel and Series

Activity:  Resistances for Series Wiring

a. If you have three different carbon resistors, what do you think the equivalent resistance to the flow of electrical current will be if the resistors are wired in series?  Explain the reasons for your prediction based on your previous observations with batteries and bulbs.

answer: we said that resistors would add together

b. Compare the calculated and measured values of equivalent resistance of the series network as follows:  

Write down the measured values of each of the three resistors:

R₁ =  ________________________  Ω

R₂ =  ________________________  Ω

R₃ =  ________________________  Ω

Our values for part b

Here we have our resistors in parallel and in series with our measured value noted right below each corresponding diagram. What we found was that when resistors are in series the total resistance is its sum of all the resistors. However, when resistors are in parallel, the total resistance is the sum of the inverse of each resistor. 

Note:the resistance of carbon resistors doesn’t change as current is increased

Kirchhoff’s Laws

WWe move on with our study of Kirchhoff's Law. In the real work circuits are not all simple, in fact they are usually set in a combination of series and parallel which may contain an X amount of resistors. In order to analyze a circuit we use something called Kirchhoff's Laws, Which are as follows:

Kirchhoff’s Laws

1. Junction (or node) Rule (based on charge conservation):  The sum of all the currents entering any node or branch point of a circuit (that is, where two or more wires merge)must equal the sum of all currents leaving the node.

2. Loop Rule (based on energy conservation):  Around any closed loop in a circuit, the sum of all emfs, voltage gains provided by batteries or other power sources, (ε = emf) and all the potential drops across resistors and other circuit elements must equal zero.

We were given as example of how to use Kirchhoff's Laws which is pictures below. We were given an outline of how to perform such calculation. First, we assign currents, next apply the loop rule -/+ through 1/r + V and finally indicate direction of current of loop with respect to V.

Python electric potential loop

For this assignment, we were asked to construct a 3D model of 3 point charges along with their observation points in Vpython. We were given two options of how to construct our 3D model, I chose the loop method. 

First, k constant and pi were defined. Then, using the coding we used in class I was able to re-write that code and construct my model. I created new locations and new charges, then wrote in the formulas in order to find their electric potential with respect to their corresponding observation points. My final result is pictured below along with its coding. 

4/16 The Potential Difference Due to Continuous Charge Distributions

Purpose: We will study the potential from a continuous charge distribution and find how to calculate it different ways. In the end each calculate will produce the same answer. The method we will use are integration and approximation by summing up several finite elements of charge q by using excel. Finally, use Gauss's law to find the electric field along with the defining equation for potential difference to set up a line integral. 

We started by defining an equation for the charge distribution on a ring on a point. The ring has radius a and the distance from the ring to the point is x. We found that V=sqrt(x^2 + a^2)

We were given that the ring had 20 charges on it so we divided them equal through the whole ring.

We started by calculating the potential difference by hand (pictured below), we then used excel to calculate the same problem, and finally we derived an integral which we used to calculate the same problem. All methods yield the same solution.

Here we have the our solution to the electric potential by hand.

Here is our excel solution 

Here we derived the electric potential of the ring on a point distance x, which we found to be equal to  kqcos(theta) divided by x.

Here we took that equation we derived above and integrated it to get the equation we needed to solve for the electric potential on a ring. After integrating we found that it was equal to kq divided by sqrt(x^2 + a^2).

This is our solution using integration which we found to be 4.99E5 V



We also found that we can calculate the change in potential by using the electric field.

Activity:  ΔV from a Ring using the E-field Method

Our final solution was that E is equal to kQx divided by (x^2 + a^2)^3/2

In the next part of class, we begin our study of the electric potential on a charged rod. We are given an observation point at (0.1, 0.15). Picture below is our solution.

Equipotential Surfaces

Sometimes it is possible to move along a surface without doing any work.  Thus, it is possible to remain at the same potential energy anywhere along such a surface.  If an electric charge can travel along a surface without doing any work, the surface is called an equipotential surface.

Here is our definition of an equipotential surface and the solution to the activity above.

Electric Potential Lab/Activity

In this lab, we measured the electric potential at various points on a conducting sheet using a power supply. We set up our experiment by laying out a conducting paper then getting a power supply and connecting our wire connector. We measured the potential difference in 1cm intervals, going to the left (negative) and then going to the right (positive). Our work is pictured below. What we concluded was that the potential energy goes in the direction of the electric field.

April 14 Electric Potential

Purpose: Today we will continue our studies with electric potential energy and electric power. We will see how work is related to the two and also see how it is similar to the potential energy in kinematics. The difference now is that we are dealing with voltage and electric charges.



Dimmest and Brightest Light Bulb Experiment



We began the day with a light bulb experiment. We were asked to create circuit setups in which one would allow the light bulb to light at its brightest and the other set up would make the light bulb be at its dimmest.

Here we have our light bulb setup in series which made it light at its brightest. It is lighting at its brightest in series because  here the voltage is at its highest. If there is higher voltage this means that there is a higher potential energy and thus more work being done.

Next setup, we have our light bulb at its dimmest and in parallel. When we have a parallel circuit, the current is at its highest but it doesn't mean voltage is also at its highest, this is a perfect example of that concept.

Professor Mason then showed us how to properly draw circuits, for example, a light bulb, battery, etc. We drew a picture of our newly created circuits then re-drew them using the drawing techniques Mason showed us.

Water Heater Experiment

In this next experiment, we were shown how voltage affects the temperature of water.

We began we heating the water at a constant temperature for two minutes, then doubling the voltage for two minutes and graphing that.

This is our graph of the heated water before doubling the voltage

This is our graph of heating the water after doubling the voltage.

After conducting both scenarios, we clearly see that the slope became steep when the voltage was doubled. A steep slope indicates that there was a rise in temperature. However, we found that there was a change in temperature by a factor of 7 experimentally, we were told that this number is too high so we were asked to calculate this factor theoretically.
We used P=VI and V=IR

2V=2IR

P=2V2I

P=4VI
We then find a more accurate factor, which is 4.

Here a discussed what variables we needed to change the temperature of the water.

In our next activity(pictured below), we were asked to calculate work in an electric field under different circumstances.

Activity: Work done on a charge traveling in a uniform electric field
1.) A charge q travels a distance d from point A to point B;  the path is parallel to a uniform electric field of magnitude E. 
 The work would be mgh.

2.) The charge q travels a distance d from point A to point B in a uniform electric field of magnitude E, but this time the path is perpendicular to the field lines. 

The work would be potential energy.

3.)The charge q travels a distance d from point A to point B in a uniform electric field of magnitude E.  The path lies at a 45° angle to the field lines

The work done would be zero.

Here we were asked to rank work from most to least, as shown above. We found that A produced the most work which is given by the equation W=Eqd, then it was C which is W= Eqdcos(theta), and finally B which is W=0.

Next we derived the equation for electric potential energy, which is shown in the picture below. We found that electric potential energy is equal to the integral of the vector of the electric field multiplied by the vector  ds. We also derived the electric potential energy equation for a equal potential line, which says that electric potential energy is equal to Kq divided by r times vector ds.

In the last past of class, we were asked to conduct an electric potential field in Vpython. We were given the code and began to construct (picture below). 

We were then asked to calculate the electric potential of each point with respect to every corresponding observation point(pictured below). We were then asked to re-write the given program and also add an addition charge and corresponding observation point and then calculating the electric potential for each( pictured below).

April 9 Current and Resistance

We began the class with the discussion of current flow, and to gain a better understanding we conducted our first experiment. 

Lighting a Bulb

You can begin to explore circuits and currents by lighting a bulb with a battery.  You will need:

• 1 #14 V bulb

• 1 D-cell battery, 1.5 V, alkaline

• A piece of wire

Use the materials listed above to find some arrangements in which the bulb lights and some in which it does not light.  For instance, does the bulb light up in the following arrangement?

Activity:  Arrangements that Cause Light

a. Sketch two different arrangements in which the bulb lights

b. Sketch two arrangements in which the bulb doesn’t light.

Our Answesr to the questions above. We drew two scenario that would produce light and two that wouldn't.

 c. Examine the bulb closely.  Use a magnifying glass, if available.  The image shows the parts of the bulb that are hidden from view.  Why is the filament of the bulb connected in this way? 

The filament of a light bulb is connected directly to the conducting metal so that a charge can be produced thus creating energy and producing light. 

      d. Describe as fully as possible what conditions are needed if the bulb is to light and how these conditions are not satisfied in the arrangements that fail to cause the bulb to light.

In order for a light bulb to light there must be a closed circuit and electric flow must pass through the filament. In both scenarios that failed to produce light, there was no closed circuit and the electric current was not passing through the filament. With no electric flow there is no energy thus there is no heat to light up the bulb.

e.  Now use a second battery and connect it so that the bulb lights twice a bright as previously. 

part E

We were asked to make the light bulb as dim as possible with a specific arrangement.

We then continued class with our next demonstration of an electroscope. An electroscope is an early scientific instrument that is used to detect the presence and magnitude of electric charge on a body. Electroscopes detect electric charge by the motion of a test object due to the Coulomb electrostatic force.

Professor Mason began the experiment by rubbing the electroscope with wool to build up electrostatic (negative charge), the two plates inside the box then gain the same charge thus causing them to repel each other. He then touches the rod of the electroscope with the positive end of a battery and nothing happens, this is because there is no closed circuit for the current to flow through.

Next part of class we conduct the modeling of a simple electric circuit

Here we answered the question: why do you need a wire to go back from the bulb to the battery?

Our answer: So that the electrons can flow in and out of the battery causing an equilibrium charge.

Next activity we are given the following question: 

our answer was that we needed to  know the height of the waterfall and the rate of the current flow. What we found is that electric circuit is much like the flow of water. For example, a pump does work on the water to raise it to the top (pump gives Potential energy to the water). It maintains a difference in height between the two side. The water falls and gives up that stored PE to the waterwheel; meaning it produces work. In an electric circuit, the battery serves as a "pump" to pump the changes to the top of an electrical "cliff." The battery gives potential energy to each charge it raises up to the "top". When the charges "fall down" they give up their energy to the bulb, their PE is then transformed into light or heat.

Here we discussed the relationship between voltage(which determines how much potential energy the battery gives to each charge it raises up to the "top", current (measures how many charges flow through the bulb each second, and Power (measures the amount of energy delivered to and radiated by the bulb as heat and light every second.

We concluded that voltage is equal to power over current. We also did an example, which is pictured above.

Next activity we did was measuring current with an ammeter.

Our answer to this experiment was 46 milliamps, and we found a positive answer. What we found is that current doesn't change when energy changes. This is because there is an equal current going in as there is going out.

Current in a Wire

In our next activity we discuss what four things we need to know to find the current in the wire.

we said: charge of q, density, volume, and drift velocity.

Current is a measure of the rate at which charge is flowing past point in a wire.

Definition: Instantaneous Current

In this next example, we are introduced to drift velocity.  

Relationship between current and drift velocity:

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.

In a vacuum the electric field would cause a charge to accelerate. In a wire, collisions of the conduction charges with impurities, imperfections, and vibrations of the atomic lattice causes the motion of the conduction charges to be slowed down. This represents a loss of energy which is dissipated as heat.

Over a wide range of conditions, the flow of the charges quickly achieves a steady state value and remains constant. The "average speed" at which the "free" charges are moving in the wire is called the drift velocity v_d.

The charge carriers in a wire are normally electrons. The number of conduction electrons "free" to participate in the current flow depends upon the atomic structure of material making up the wire. Conductors have many electrons that are able to participate, where as insulators have few free electron. Semiconductors are materials that lie somewhere between these two extremes. 

The current in a wire can be expressed as function of the number of charge carriers/volume, the magnitude of the charge carriers, the drift velocity of the charge carriers, and the cross-sectional are of the wire, 

Here we solved for drift velocity using the equation we derived in class for current.

Example: the 12-gauge copper wire in a home 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. Calculate the drift speed of the electrons.

We continued class with the study of resistance and Ohms Law

We found that when a potential difference is applied to the ends of a wire, current begins to flow. However, the current is different for every type of material even if the voltage is the same. 

Resistors are used in a circuit to control the current flow or the voltage at different locations in a circuit. With this idea we moved onto our next experiment in which we measured the current and potential difference of a wire. The following graphs are the results of the experiment.

Here we have our prediction of what we thought the potential difference and current graphs would look like.  We made our linear which was obviously incorrect. Next to our predictions we have our derivation for potential difference.

Here we found that resistance is proportional to the length of the wire. If we increase the length of a wire or make its cross-sectional area smaller, the wire's resistance will increase.