Circuits Pre-Lab 7 - Physics Pre Lab PDF

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Victor Sahagun Circuits Pre-Lab Introduction Circuitry—the sub-field of physics that focusses on the flow of electrons through networks of metallic wires and the functions that different circuits perform—is important to many areas of engineering from designing circuit boards in computers and calculators to designing the electrical wiring in a house. Furthermore, we can extend the ideas of circuits and electricity to the flow of charged ions in fluids which can lead us to many insights about biological systems. In this week’s lab, your task will be to use the ideas of circuits and electricity to model the behavior of a biological neuron. This pre-lab is designed to help introduce you to the ideas of simple circuits in the context of typical electrical circuits if you have not yet encountered these in class and to prepare you to extend these ideas to a more complex system. Plan on spending roughly 45 minutes on this. You will have an opportunity in lab to check your results with your group members and your TA.

Getting Set Up This pre-lab makes use of the PhET circuit construction simulator. • Visit the site https://phet.colorado.edu/en/simulation/circuit-construction-kit-dc • This simulation is in HTML5 and can be run directly in your browser, tablet, or smart phone and you won’t need to download anything. • Click the “play” icon and select the lab option.

Current in Simple Circuits Begin by setting up the following circuit, which includes a battery, a resistor, and two wires:

You can use the “non-contact ammeter” to measure the electrical current at various parts of the circuit. You can click on the resistor to change its resistance. You can also click on the battery to change its voltage (not pictured). 1

1. Describe in words what you see happen when you make the connection. Describe in words what

happens when the circuit is unconnected and does not make a complete loop. When the connection was made in the circuit in a loop, the electrons flow directly through the wires, the battery, and the resistor. When the circuit does not make a complete loop and is unconnected the flow odf the electrons in the circuit comes to a stop. The electrons in the circuit move in the opposite direction of the conventional current.

2. Move the non-contact ammeter to different parts of your circuit. Describe in words what happens

to the value of the current being measured. When the non-contact ammeter is moved to different parts of the curvuit the values of the current that is measured remain the same, they do not change. The current stays at 0.90 Amps at all of the regions located in the circuit.

The current in a circuit is really just many billions of electrons moving around the loop. These electrons are never “used up” in the circuit, which is reflected by the fact that you measured the same current at every part of your circuit. This is entirely analogous to water flowing through a pipe; you should expect all of the water that enters one end to leave the other end. This idea is known as Kirchoff ’s Junction Rule 3. Consider the following circuit: Use the idea of kirchoff’s junction rule to determine

• I3: 40 = (30) +I3 = 10 mA • I4: 30 mA • I5: 40 mA 4.Write an algebraic equation relating I1, I2, and I3.

I1= I2+I3

Equations like this relating together currents in different branches are a “junction rule equations.” Kirchoff’s Junction rule is usually applied at junctions in a circuit—places where current has two or more paths to follow—however the junction rule can also be applied to circuit “branches.” 5. Write an algebraic equation relating the two currents in the rightmost branch, I2 and I4 . Write an

algebraic equation relating the two currents in the leftmost branch, I1 and I5.

I2= I4= 30 mA I1= I5= 40 mA I5= I3+ I4 = 40 mA

Current, Resistance, and Ohm’s Law 1. Going back to the circuit you have set up in the PhET simulation, change the resistance of the

resistor. Describe in words what happens to the current measured by the non-contact ammeter. Once the resistance of the resistor was changed in the PhET simulation the current in the circuit decreased when the resistor is increased. When the resistance in the resistor is decreased it does the opposite, the current in the circuit increases. The resistance and current, therefore, have an inverse relationship with each other since the opposite thing happen when it is decreased and vice versa. In regard to the electron flow, the flow of electrons decreases when the resistance in the resistor is increased. The opposite happened when the resistance is decreased.

2. Change the voltage of the battery. Describe in words what happens to the current measured by the

non-contact ammeter. When the voltage of the battery is increased the current and the electron flow increases. When the voltage is decreased the current and the electron flow decreases. This means that this also has an inverse relationship. The current is affected by the amount of voltage a battery carries.

Electrons are pushed around the circuit by the voltage of the power sources (the battery here, but other things can provide a so-called “motive force” like capacitors, generators, solar panels, and yes, even the diffusion across cell membranes.). Meanwhile the wires themselves are “crowded”; electrons moving through them “bang” into the structural atoms that make up the wire as well as other elec- trons. This prevents them from fully accelerating. At the macroscopic scale, we measure this as electrical resistance.

The amount of current is proportional to the voltage supplied by the battery and inversely propor- tional to the resistance of the circuit. This relationship between the voltage across a section of circuit (∆V ), that portion’s resistance R, and the current through the portion, I , is given by “Ohm’s Law:” ∆V = IR

(1)

3. Confirm that this relation holds for different values of battery voltage and resistor resistance in

your circuit. You can use the voltmeter tool for this—make sure that one lead is connected on one side of the circuit element and the other lead is connected to the other side. Voltage measures in volts (v) And current measures in Amperes (A) 2

4.If you set your battery to 9 V and your resistor to 20 Ω (Ohms), what will be the resulting current?

Do this calculation and then test it in the PhET simulation. ΔV=IR 9V/ 20Ω= I (20Ω/20Ω) I = 0.45 A

Resistance The resistance of an object is not just due to the details of the material from which it is made (i.e. how “crowded” the material is at an atomic scale), but it is also related to the object’s geometry. In particular, the two things that matter are the length of a resistor (the dimension along the flow of current), and the cross-sectional-area of a resistor (the area perpendicular to the flow of current). Consider some comparisons: 1. Which of these two resistors do you expect will have a larger resistance? (Assume that they are

made from the same material)

R= ρL/ A  R=ρ2L/ A

A larger length with make the resistance be larger, therefore, the resistor with the length of 2L will have the later resistance.

2. Which of these two resistors have a larger resistance (again, assume that they are made from the

same material)? (Hint: thinking about roads and traffic flow, what has more ‘resistance’ a 4-lane highway or a 1-lane road? ) R= pL/ 2A

The resistor that has the smaller cross-sectional area will have the larger resistance. This is the resistor that has 1A.

Resistance can be calculated according to the following formula: R=

ρL A

(2)

,

where ρ is the resistivity of a material (i.e. how “crowded” it is from a molecular standpoint). Resis- tivity is a material property, so, for example, all copper everywhere has a resistivity of 1.68×10−8 Ωm Check to make sure that this relationship makes sense to you. 3. Let’s practice calculating resistances. Suppose that the length of this circular resistor is 1 cm and

the radius is 2 mm. ρL

R= =

(3)

A ρL

(4)

2 πr (1.68 × 10−8 Ωm)(0.01 m)

= 10

π(0.002 m)2

−6

= 1.34 ×

Ω

(5)

Now, you find the resistance of a square block of copper with dimensions L = 0.01 m, h = 0.003 m, and b = 0.004 m.

A= (0.004)(0.003) A=1.2 *10^-5 m Ρ= 1.68*10^-8 Ωm R= ((1.68*10^-8)(0.01))/ 1.2*10^-5 = 1.4*10^-5 Ω

Circuits and Voltage Set up the following circuit, which includes a 9 V battery, 10 Ω resistor, a 20 Ω resistor, and three wires:

You can use the “voltmeter” to measure voltages across all of the elements (resistors, wires, etc.). 1. Describe in words what happens when you reverse the order of the black and red voltmeter probes.

Do you think that this reflects anything about the circuit? When the order of the black and red voltmeter probes are reverses it produces an equal, but negative value of the voltage. The black probe is negative and the red one is positive, when the red probe is connected upstream is will show a negative value of the voltage.

2. Measure the voltage across all elements; How does the battery voltage compare to the voltages of

all of the other elements? The battery voltage which is 9 is larger than the voltages of the rest of the elements. The individual resistors were 6 volts and 3 volts respectively.

You may have noticed that the voltages of all of the wires and resistors add up to the battery voltage. Energy is conserved in a circuit so all of the energy provided by the battery (voltage is energy per charge) is dissipated in the wires and resistors (as thermal energy, light energy (in lightbulbs), motional energy (in motors), etc). This is known as Kirchoff ’s Loop Rule: ∆Vsource =

Σ all other elements on the loop

∆V

(6)

3. Use the loop rule to calculate the voltage in the remaining bulb ΔVsource= ∑elements

12V= 9V +x -9 -9 X= 3Volts For more complex circuits we need to be careful.

Luckily, the loop rule is valid for any loop. For example, we could pick the leftmost loop:

(12 V) = (9 V) + ∆VA ∆VA = (12 V) − (9 V) = 3 V

(7) (8)

Notice that we can completely ignore the bulbs that aren’t on the loop! Why can we do that? Voltage is like electrical “height:” you can figure out how far up or down you’ve climbed between your lab room and your house, but you wouldn’t care about any staircases or hills that aren’t on your path. For circuits, the voltage along each path must independently add up to zero. 4.Write down an expression for the voltage across bulb B in terms of the other voltages on the loop.

Calculate the voltage in bulb B. 12 v= (9v)+(1v) +ΔVB

12 v = 10 v +vB VB= 12 V- 10 V = 2 V

5. Build this circuit in the PhET simulation and compare the measured results to what you calculated.

My measured results match the calculated ones as shows above in the calculations.

Read the Lab Introduction Now that you have worked through the pre-lab, read over the lab introduction prior to coming to class....