Kirchhoffs Rules Lab Online
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Electrical Engineering
Date
Apr 3, 2024
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1 Kirchhoff’s Rules
Lab Online
Purpose
In this activity we will be examining the means to analyze the voltages and currents passing over circuit components via Kirchhoff’s Rules.
Theory
Whenever a circuit is complicated (sometimes known as a network), analyzing it by application of the formulas for resistors in parallel or in series becomes impractical (or impossible). In such cases, the analysis of the circuits can be done with the help of Kirchhoff’s Rules
. The rules are direct consequences of two fundamental principles: The Conservation of Energy –
The total amount of energy in a closed system is conserved, and The Conservation of Charge –
The net charge in a closed system is conserved. Kirchho
ff’s Rules
: The Loop Rule –
The net change in electric potential difference around any closed path in a circuit sums to zero. ∆𝑉
?𝑒𝑡
= ∑ ∆𝑉
?
= 0
?
?
The Junction Rule –
The net current entering a junction must be equal to the net current leaving that same junction. ∑ 𝑖
?
= ∑ 𝑖
?
?
?
?
?
The loop Rule
is a consequence of the Conservation of Energy. Since the electric potential energy a charged particle has is a function of its location in an electrical field, if a charged particle starts at a random location in a circuit, then travels around that circuit only to return to the exact same location that it started from, then it must have the exact same electric potential energy at that location as it did when it started its journey. This results in the net change in electric potential difference being zero along whichever path it took. The Junction Rule
is a consequence of the Conservation of Charge. A Junction is any location in a circuit where three or more branches of a circuit meet, and therefore give the current multiple paths to travel along
. Since the amount of charge can’t change
, any bit of charge that enters a
2 junction must also leave that same junction, and this results in the amount of current entering a junction being equal to the amount of current leaving that same junction. How to apply The Loop Rule
: Let us start off with a diagram of a circuit. To the right there is a diagram of a circuit consisting of an electric potential source, and four resistors. 1.
First we need to pick a location in the circuit, then pick a path (loop) to travel around the circuit returning to the original location. It is common, but not required, to pick the potential difference source as the original location. (A valid loop is not even required to have an electric potential source in it) 2.
Then Let us pick the path (loop) starting from the potential difference source, then moving over resistor 1, then resistor 2, and then finally returning the potential difference source. 3.
Now we need to label the ends of the circuit components that are in our loop either positive or negative based on their relationships to our potential difference source. The side of a resistor that is directly connected to the positive side of the potential difference source will be positive, its other side will be negative, and vice versa. (If there are multiple resistors in series then all the resistors in series will have the same sides labeled as positive, and the same sides labeled as negative) 4.
From Ohm’s Law The potential difference over a resistor will have the magnitude of 𝛥𝑉
?
= 𝑖
?
𝑅
?
5.
We will write our Loop Rule equation as we travel around our loop. When you move over a circuit component traveling from the negative side to the positive side (also known as moving from a low potential to a high potential) the potential difference of that component will have a positive value. However, when you move over a circuit component traveling from the positive side to the negative side (also known as moving
3 from a high potential to a low potential) the potential difference of that component will have a negative value. Finally, the equation always equals zero. •
In our example, where we start at our potential difference source moving from its negative side to its positive side, then over resistor 1, moving from its positive side to its negative side, then over resistor 2, moving from its positive side to its negative side, then finally returning to our potential difference source we get the following equation; ∆𝑉 − 𝑖
1
𝑅
1
− 𝑖
2
𝑅
2
= 0
How to apply the Junction Rule: Let us start off with the same circuit diagram as before. This diagram has two junctions in it. The Junction at the top, where the current coming from the potential difference source that can either enter the branch that will make it pass through resistor 1, or it can enter the branch that will make it pass through resistor 3. This gives us the junction equation; 𝑖 = 𝑖
1
+ 𝑖
3
It also has the junction at the bottom, where the currents that passed over resistor 2, and resistor 4 enter the junction, and then combine to become the current that is leaving the junction to entering the potential difference source. This gives us the junction equation; 𝑖
2
+ 𝑖
4
= 𝑖
There is something that should be noted about these two junction equations: they are the same equation
! The first one is the current coming from the potential difference source entering a junction, and then splitting off into two different branches of the circuit. The second equation is the current from those same two branches entering a junction, and combining to be the current entering the potential difference source. Remember, by conservation of charge we know that all the currents passing over resistors in series must be the same. (In this example that means that
𝑖
1
= 𝑖
2
, 𝑎𝑛𝑑 𝑖
3
= 𝑖
4
) When applying the Junction Rule to a circuit there will always be at least one pair of junction equations that are mathematically redundant, and therefore you can only use one of those equations to help you solve the circuit. It is said you ‘solve the circuit’ by applying these two rules to a circuit as many times as needed till you have constructed as many mathematically independent equations as there are
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4 unknowns (if there are 5 unknowns, then you need 5 mathematically independent equations). You solve those equations for the unknowns using either the substitution or matrix method. Setup Circuit 1 1.
Go to the following website: https://phet.colorado.edu/en/simulation/circuit-construction-kit-dc-virtual-lab 2.
You should now see the following: 3.
Click the down arrow to download the software. Once finished, you should see the following.
5 4.
Near the bottom left of the screen, click on the Potential Difference symbol. 5.
Near the top right of the screen, click on the Green/White Plus “Advanced” icon. Make sure the Wire Resistivity and Battery Resistance are both set to ‘tiny’
, then click the Red Minus to close it. 6.
In the White Box at the top right of your screen make sure “Show Current” is checked, select “Conventional”, make sure “Labels” is checked, and make sure that “Values” is checked.
Procedure Circuit 1 1.
On the Left side of your screen you will see a white box with the symbols of various basic circuit components. You will ‘Click and Drag” the various components to build simple circuit boards. 2.
On the right side of your screen you will see a white box with a voltmeter and an ammeter. You will ‘insert’ the ammeter
into the circuit you build to measure the current. 3.
Build the following circuit: (You can use the plus and minus at the bottom of the left side of your screen to give yourself more room to work with if you need it.)
6 4.
Click on the Potential Difference source (Battery) and set it to 100 V. 5.
Click on each of the resistors to set its resistance. Set R
1
= 50.0 Ohms, R
2 = 10.0 Ohms, and R
3 = 30.0 Ohms. 6.
Read off the currents for each ammeter in front of each resistor and record the currents for each resistor in Table 1. Setup Circuit 2 1.
Use the orange button in the bottom right of your screen to reset the simulator. a.
In the top right box, select “conventional” for the current and recheck “values”.
b.
Your screen should no look like the following: Procedure Circuit 2 1.
Build the following circuit: (You can use the plus and minus at the bottom of the left side of your screen to give yourself more room to work with if you need it.)
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7 2.
Set the Potential Difference (Battery) to 50.0 V, and the resistors as follows, R
1
= 20 Ohms, R
2
= 50.0 Ohms, R
3
= 75.0 Ohms, and R
4
= 25.0 Ohms. 3.
Record the currents for each resistor in Table 2. Setup Circuit 3
1.
Use the orange button in the bottom right of your screen to reset the simulator. a.
In the top right box, select “conventional” for the current and recheck “values”
. b.
Your screen should now look like the following:
8 Procedure Circuit 3 1.
Build the following circuit: (You can use the plus and minus at the bottom of the left side of your screen to give yourself more room to work with if you need it.) 2.
Set the Potential Difference (Battery) to 50.0 V, and the resistors as follows, R
1
= 10 Ohms, R
2
= 25.0 Ohms, R
3
= 40.0 Ohms, R
4
= 100.0 Ohms, and R
5
= 50 Ohms. 3.
Record the currents for each resistor in Table 3. Setup Circuit 4
1.
Use the orange button in the bottom right of your screen to reset the simulator. a.
In the top right box, select “conventional” for the current and recheck “values”.
b.
Your screen should now look like the following:
9 Procedure Circuit 4
1.
Build the following circuit: (You can use the plus and minus at the bottom of the left side of your screen to give yourself more room to work with if you need it.)
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10 2.
Set the Potential Difference 1 (Battery 1) to V1 = 50.0 V, Potential Difference 2 (Battery 2) to V2 = 20 V, and the resistors as follows, R
1
= 10.0 Ohms, R
2
= 25.0 Ohms, and R
3
= 50.0 Ohms 3.
Record the currents for each resistor in Table 4.
11 Analysis of Kirchhoff’s Rules
Lab Online Name______________________________________________ Course/Section______________________________________ Instructor___________________________________________ NOTE: You must use Kirchhoff’s Rules, and show your work, to receive any credit for questions 1, 2, 3, and 4. If you use any other means to solve questions 1, 2, 3, and 4, even if you show your work and it is correct, you will receive ZERO credit for question 1, 2, 3, and 4. Table 1 (5 points) ΔV____
___100__
______
R(Ω)
i
m
(A) i
t
(A) % Error R1 50 0.43 0.43 0% R2 10 2.17 2.17 0% R3 30 2.61 2.61 0% 1.
Using Kirchhoff’s Rules
, construct enough mathematically independent equations to solve for the current of each resistor. Then calculate the % error between your measured and theoretical values for the current of each resistor. You MUST use Kirchhoff’s Rules and show work to receive any credit.
(15 points) Percent error= 0.43-0.43/0.43 *100 = 0% I1= 21.74/50=0.434 i2=100/30= I2(1/50) +(1/10) +(1/30) [23 i2 /150 = 3.33] 21.74/10 I2= 2.174
12 Table 2 (5 points) ΔV1___
50___________
R(Ω)
i
m
(A) i
t
(A) % Error R1 20 1.49 1.49 0% R2 50 0.41 0.41 0% R3 75 0.27 0.27 0% R4 25 0.81 0.81 0% 2.
Using Kirchhoff’s Rules, construct enough mathematically independent equations to solve for the current of each resistor. Then calculate the % error between your measured and theoretical values for the current of each resistor. You MUST use Kirchhoff’s Rules and show work to receive any credit
. (15 points) (1/50) +(1/75) +(1/25) = 0.07333 I2 = 1/50 = 0.02 1.49[0.02/0.07333]= 0.4064 I3 = 1/75 = 0.01333 1.49[0.01333/0.07333]= 0.2708 I4 = 1/25 = 0.04 1.49[0.04/0.07333
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13 Table 3 (5 points) ΔV
1____50__________ R(Ω)
i
m
(A) i
t
(A) % Error R1 10 1.63 1.63 0% R2 25 0.45 0.45 0% R3 40 0.84 0.84 0% R4 100 0.34 0.34 0% R5 50 0.45 0.67 33% 3.
Using Kirchhoff’s Rules, construct enough mathematically independent equations to solve for the current of each resistor. Then calculate the % error between your measured and theoretical values for the current of each resistor. You MUST use Kirchhoff’s Rules and show work to receive any credit. (15 points) Percent error= 0.67-0.45/0.45 *100= 33% 1/10 + 1/100 + 1/75 = 1/0.0483 = 20.7 10 + 20.70 = 30.70
14 V=IR I1=V/R =50/30.70 = 1.6286 I2 = 1.63(0.0133/0.0483) I2=0.4488 I3= 1.63[(1/40) / (1/40 + 1/100 + 1/75)] 1.63[0.025/0.0483] I3=0.8436 I4= (1/100) 1.63[0.01/0.0483] = 0.3374 I5= (1/50) 1.63[0.02/0.0483] = 0.6749 Table 4 (5 points) ΔV1___
_50__________ ΔV2___
20__________ R(Ω)
i
m
(A) i
t
(A) % Error R1 10 0.57 0.5714 0.25% R2 25 0.57 0.5714 0.25% R3 50 0.60 0.60 0% 4.
Using Kirchhoff’s Rules, construct enough mathematically independent equations to solve for the current of each resistor. Then calculate the % error between your
15 measured and theoretical values for the current of each resistor. You MUST use Kirchhoff’s Rules and show work to receive any credit. (15 points) 20-10i-25i=0 20=35i 0.5714=i % error= 0.5714-0.57/0.57 *100 =0.25% 50-i*50-20=0 30 = i*50 0.6 = i 5.
In the theory section of this worksheet, detailed explanations are given as to how Kirchhoff’s Laws are connected to conservation laws. Briefly, and in your own words, summarize each of these explanations. (10 points) Kirchoffs current law states that current flowing into a node must be equal to the current flowing out of it which can be called law of conservation of charge, thats one way they are connected and another is. Kirchoff's laws are connected to conservation laws because the loop rules state the algebraic sum of all the voltages around any closed loop in a circuit is equal to zero, the loop will be closed meaning that there is no energy lost.
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16 6.
Did our experiments support the theory? Justify your answer. (10 points) Yes, the experiments support the theory because there are times when the current flow is equal to where it is flowing out, as well as times when the loop rule and it being showed that no energy is lost.
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