annotated-Ohm%27s%20Law%20Kirchoff%27s%20Law%20Online
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Subject
Electrical Engineering
Date
Dec 6, 2023
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17
Uploaded by Alberto325
Physics Lab (Online Simulation)
1
OHM’S LAW
-
KIRCHOFF’S LAW
Electricity and Light
TA name: Tej Raj Karki
Due Date:
10/5/2023
Student Name: Alberto Canales
Theory:
A D.C. circuit consists of sources of direct current (EMFs), connected to a network of
elements.
If the ele
ments are ohmic (obey Ohm’s Law) then the currents through the
elements are directly proportional to the voltages applied across the elements.
For ohmic
elements, the relation between the current through an element (I), in amps, and the
voltage across the element (V), in volts, is
V = IR
[1]
where R is the resistance of the element in ohms.
Resistors, which are commonly used in
electronic devices, are ohmic.
If two or more resistors are connected in series with a voltage source, as illustrated in
Figure 18-1, they collectively behave as an equivalent single resistor with resistance, R
ser
,
where
R
ser
= R
1
+ R
2
+ R
3
+
+ R
n
[2]
Figure 18-1
the resistors are connected in parallel to the voltage source, as illustrated in Figure 18-2, then
they collectively behave as an equivalent single resistor with a resistance, R
par
, where
R
1
R
2
R
3
R
n
V
Physics Lab (Online Simulation)
2
Figure 18-2
n
3
2
1
par
R
1
R
1
R
1
R
1
R
1
+
+
+
+
=
[3]
The circuits illustrated in Figures 18-1 and 18-2 can be reduced to a single loop
containing a single voltage source and a single resistor.
And Ohms Law can be used to
determine the currents in the circuit.
If however the circuit contains more than one voltage source in a network of resistors, the circuit usually
cannot be reduced to a single loop.
Kirchhoff’s rules are useful in a
nalyzing a multi-loop circuit.
This Prelab is worth 15 points
Type all your answers in Blue
(1) Write down an equation and solve for the total resistance of three resistors R
1
=100
Ω
,
R
2
=150 Ω, R
3
=350Ω in the resistors were arranged in the following combinat
ion:
a) All three in series
(1 point)
R = R1 + R2 + R3
100Ω + 150Ω + 350Ω = 600Ω
b) All three in parallel
(1 point)
R = (R1*R2*R3)/(R1 + R2 + R3)
R = ((100Ω
)(
150Ω
)(
350Ω
))
/(100Ω + 150Ω + 350Ω)
= 8750Ω
c)
How would you arrange these three resistors to get a
net resistance of 410 Ω. (1 point)
R = ((R1*R2)/(R1 + R2 )) + R3
(((100Ω)(150Ω))/(100Ω + 150Ω)) + 350 =410Ω
(2) If you have an experimental setup which has three unknowns, how many linearly
independent equations do you need to determine the unknowns to find one unique
solution? (1 point)
You need three equations
(3) What is Kir
choff’s law of current? What does it conserve? (1 point)
The Kirchoff’s law of current is the total sum of all currents
entering a junction and
conserves charge
V
R
n
R
3
R
2
R
1
Physics Lab (Online Simulation)
3
(4)
What is Kirchoff’s law of vol
tage? What does it conserve?
(1 point)
The Kirchoff’s law of vol
tage is the total sum of voltages in a closed loop and conserves
potential energy
(5)
Electrical circuits have two main problems: “Short” and “Open”. Define these two
conditions with diagram and an example showing the consequence of each of these
faults. (Use back of the page if necessary). (2+2 = 4 points)
Open:
The circuit stops which breaks due to the circuits loop being open
Short:
A short circuit is when the current goes through a different path on the circuit and has no
flow on that path
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Physics Lab (Online Simulation)
4
(6)
Apply Kirchoff’s law of current and voltage in loop 1 and loop 2 of the above circuit and
write down the corresponding equation. (2.5 + 2.5 = 5 points)
Loop I:
Current equation
: I3 = I2 + I1
Voltage equation:
0 = ((I3)(R3)) - V1 + ((I1)(R1))
Loop 2:
Current equation:
I3 = I2 + I1
Voltage equation:
0 = ((I3)(R3)) - V2 +((I2)(R2))
Part A: Ohm’s Law
This lab uses the
Ohm’s Law
and
Circuit Construction Kit DC
simulation from PhET
Interactive Simulations at University of Colorado Boulder, under the CC-BY 4.0 license.
https://phet.colorado.edu/sims/html/ohms-law/latest/ohms-law_en.html
https://phet.colorado.edu/sims/html/circuit-construction-kit-dc/latest/circuit-construction-kit-dc_en.html
Develop your understanding:
Open
Ohm’s Law
, then explore to develop your own ideas about how resistance, current, and
battery voltage are related.
R
1
R
2
R
3
I
1
I
2
I
3
V
1
V
2
I
2
I
1
Loop 1
Loop 2
Physics Lab (Online Simulation)
5
1.
As you change the value of the battery voltage, how does this change the current through the
circuit and the resistance of the resistor? If the current or resistance remains constant, why do you
think it is?
→
As you change the voltage, the current increases too and resitance stays constant. If current stays
constant then resistance will increase with the voltage. I think this happens because as V
increases, one variable has to increase in change to offset the other variable.
2.
As you change the value of the resistance of the resistor, how does this change the current
through the circuit and the battery voltage? If the current or voltage remains constant, why do you
think?
→
If the voltage stays constant, resistance and current are inversly porportional so when resistance
increases, current decreases. If the current stays constant, resistance and voltage are directly
porportional so when resistance increases, voltage increases.
3.
Use understanding to make predictions about a circuit with lights and batteries.
→
The more lights you add to the series, the less bright it will be
→
The more batteries you add to the series, the more bright it will be
Demonstrate your understanding:
Directions: As you answer the questions,
explain in your own words why your answer makes
sense and provide evidence from your #1 experiments. Add more experiments to #1 if you need
to get better evidence.
Physics Lab (Online Simulation)
6
2. If you change the value of the battery voltage:
a.
How does the current through the circuit change? (answer, explain)
When the resistance is constant, the current is directly porportional to the voltage so if
voltage increases, the current increases.
b.
How does the resistance of the resistor change? (answer, explain)
When the current is constant, the resistance is directly porportional to the voltage so if
voltage increases, the resistance increases.
3. If you change the resistance of the resistor:
a.
How does the current through the circuit change? (answer, explain)
When the voltage is constant, the resistance is inversely porportional to the current so if
resistance increases, the current decreases.
b.
How does the voltage of the battery change? (answer, explain)
When the current is constant, the resistance is directly porportional to the voltage so if
resistance increases, the voltage increases.
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Physics Lab (Online Simulation)
7
4. Consider the two circuits below.
Use your understanding of voltage, resistance, and current to answer these questions:
a.
What do you think will happen when the switches are closed?
(answer, explain)
When the switches are closed and the resistance is constant, the left model will light the
bulbs and run. However, the right model doubles the voltage which doubles the current
and brightness in the circuit
.
b.
How do you think the lights’ brightness will compare?
The left model will have half the brightness of the right model and the right will have
double
the brightness of the left model.
Physics Lab (Online Simulation)
8
c.
Open the
Intro
screen of Circuit Construction Kit DC. Build 2 circuits.
Turn on “values”.
An ammeter is used to measure current in a circuit. Use the ammeter to compare the
current in the two circuits. Compare and contrast the two circuits. Explain the difference
in brightness from the two circuits by relating it to Ohm
’s Law.
Insert a capture of the circuits with the switch closed for supporting evidence.
Model 1 (left circuit):
Current: .45A
Voltage: 9V
Brightness: The brightness here is at a normal level
Model 2 (right circuit):
Current: .90A
Voltage: 18V
Brightness: The brightness is two times brighter than model 1
Physics Lab (Online Simulation)
9
5. Consider the two circuits below.
Use your understanding of voltage, resistance, and current to answer these questions:
•
What do you think will happen when the switches are turned closed?
(answer, explain, evidence)
When both switches are closed, both circuits will light up but the right model will have
half the current and brightness of the left model.
•
How do you think the lights’ brightness will compare?
The right model will have half the brightness of the left model
•
Open the
Intro
screen of Circuit Construction Kit DC. Build the 2 circuits and check your
answers.
Use the values obtatined from the ammeter and Ohm’s Law to explain the
difference in brightness from the two circuits.
Insert a capture of the circuits with the switch closed for supporting evidence.
Model 1 (left circuit):
Current: .90A
Voltage: 9V
Brightness: The brightness here is at a normal level
Model 2 (right circuit):
Current: .45A
Voltage: 9V
Brightness: The brightness is half of model 1
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Physics Lab (Online Simulation)
10
Part B: Kirchoff’s Law
Objectives:
To investigate Kirchhoff’s Laws: Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law
(KVL).
Simulation Tools:
DC ‐ Power supply, voltmeter, ammeter, resistors, and connecting wires.
Theory and Background:
Kirchhoff’s laws follow from the laws of conservation of energy and conservation of charge. These laws
are used to analyze electrical circuits, which contain combinations of batteries, resistors and capacitors. In
this experiment, we are interested in inve
stigating Kirchhoff’s laws for a direct current (DC) circuit, for
which the electrical currents are constants in magnitude and direction.
The two Kirchhoff’s
laws are referred to as Kirchhoff’s Current Law (KCL)
, also called Junction Rule
and Kirchhoff’s
Voltage Law (KVL), also called Loop Rule.
Kirchhoff’s Current Law (KCL):
The sum of the currents entering any junction, in a closed circuit, must equal the sum of the currents
leaving it; or the algebraic sum of all currents at that point is zero
,
I = 0
……..………….. (1)
This law is a restatement of charge conservation.
Kirchhoff’s Voltage Law (KVL):
The algebraic sum of the changes in potential around any closed path of a closed circuit is equal to zero
.
In mathematical terms, this statement can be expressed as:
+
IR = 0
……………… (2)
Procedure
Circuit 1:
1)
Click on the following link from PHET Colorado Simulation to open the lab
https://phet.colorado.edu/sims/html/circuit-construction-kit-dc/latest/circuit-construction-kit-
dc_en.html
Physics Lab (Online Simulation)
11
2)
Choose Conventional Current
3)
Use the components in the left side to build the circuit shown below: Note* there is a larger
resistor in the components menu.
4)
Click on the resistor (R
1
) and fix
it at 1000Ω. That is R
1
= 1000Ω.
5)
R
2
= 1000Ω
and
R
3
= 100Ω
6)
Click on the Battery to the left (
1
) and fix it at 9V,
1
= 9V.
7)
Click on the Battery to the right (
2
) and fix it at 6V,
2
= 6V
8)
Click on the Voltmeter from the right side and drag it to measure (V
1
) the voltage across R
1
, (V
2
)
the voltage across R
2
and (V
3
) the voltage across R
3
.
R
1
R
2
R
3
1
2
Physics Lab (Online Simulation)
12
9)
Click on the Ammeter from the right side and drag it and put it in series with R
1
to measure (I
1
),
with R
2
to measure (I
2
) and with R
3
to measure (I
3
).
10)
Record the values (I
1
, I
2
, I
3
, V
1
, V
2
and V
3
) into table 1.
Table 1
Show your mathematical calculations here:
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Physics Lab (Online Simulation)
13
I1 + I2 = I3
V=IR
0 = (R3)(I3) - (E1) + (R1)(I1)
0 = (R3)(I3) - (E2) + (R2)(I2)
I1 = (-(R3)(I3) + (E1))/R1
I2 = (-(R3)(I3) + (E2))/R2
I3 = ((-(R3)(I3) + (E1))/R1) + ((-(R3)(I3) + (E2))/R2)
I3 = ((E1)(R2) +(E2)(R1))/((R1)(R2) + (R1)(R3) + (R2)(R3))
I3 = .0125A
I2 = .00475A
I1 = .00775A
V3 =
(.0125A)(100) = 1.25V
V2 = (.00475A)(1000) = 4.75V
V1 = (.00775A)(1000) = 7.75V
Physics Lab (Online Simulation)
14
Questions:
1)
Using your experimental results of I
1
, I
2
and I
3
from Table 1 and Circuit 1, verify the
Kirchhoff’s current law (KCL)
.
I1 + I2 = I3
I3 =
.0125A
I2 = .00475A
I1 = .00775A
.00775A + .00475A = .0125A
2)
Using your experimental results of V
1
, V
2
and V
3
from Table 1 and Circuit 1, verify
Kirchhoff’s voltage law (KVL)
.
E1
–
V1
–
V3 = 0
9V
–
7.75v -1.25v = 0
E2
–
V2
–
V3 = 0
6v
–
4.75v - 1.25v = 0
Both paths fit Kirchhoff’s voltage law and when you add all the emf’s and voltages, you get the
sum of 0
.
Conclusions:
•
(for Ohm’s Law)
•
(for Kirchoff’s laws)
Physics Lab (Online Simulation)
15
Circuit 2: (Extra credit: 15 points)
Build the follwing cicuit. Note* you can reset the screen if it’s easeier by clicking the reset
button on
the bottom right corner.
R
1
R
3
R
2
1
2
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Physics Lab (Online Simulation)
16
1)
Click on the resistor (R
1
) and fix it at 5600Ω. That is R
1
= 5600Ω.
2)
R
2
= 3300Ω
and
R
3
= 1800Ω
3)
Click on the Battery to the left (
1
) and fix it at 120V,
1
= 120V.
4)
Click on the Battery to the right (
2
) and fix it at 100V,
2
= 100V
5)
Click on the Voltmeter from the right side and drag it to measure (V
1
) the voltage across R
1
, (V
2
)
the voltage across R
2
and (V
3
) the voltage across R
3
.
6)
Click on the Ammeter from the right side and drag it and put it in series with R
1
to measure (I
1
),
with R
2
to measure (I
2
) and with R
3
to measure (I
3
).
Record the values (I
1
, I
2
, I
3
, V
1
, V
2
and V
3
) into table 2.
Table 2
Show your calculations here:
1
= 120V
2
= 100V
Experimental results
Calculated results (use the loops)
R (
)
I (A)
V (volt)
R (
)
I (A)
V (volt)
5600
.02
88.61
5600
.0158
88.63
3300
.01
31.39
3300
.0095
31.37
1800
.01
11.39
1800
.0063
11.37
Physics Lab (Online Simulation)
17
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