Math265 Course Project Part B redone
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DeVry University, Chicago *
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Course
265
Subject
Mathematics
Date
Jan 9, 2024
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Math265 Project Part B
Name: Taylor Johnson
In this part of the project, you will model the behavior of a resistor-inductor (RL) circuit in a transient
state.
Initially, the DC voltage source is off and there is no current in the circuit.
After the source ‘ε’ is turned
on, the current through the inductor quickly rises and produces an EMF that opposes the change in
current:
V
L
=
L
di
dt
where ‘L’ is the inductance
Using Kirchhoff’s Voltage Law and Ohm’s Law results in the equation
ε
−
IR
−
L
di
dt
=
0
This differential equation can be solved to determine an expression for the current through the circuit as
a function of time:
I
(
t
)
=
I
F
[
1
−
e
−
t
τ
L
]
where
I
F
=
ε
R
is the final current and
τ
L
=
L
R
is known as the
inductive time constant
.
The theoretical EMF produced by the inductor can be found as follows:
V
L
=
L
di
dt
=
L
d
dt
(
I
F
[
1
−
e
−
t
τ
L
]
)
=
ε e
−
t
τ
L
I.
Graphing Current vs Time (25 points)
1 |
P a g e
Consider an RL circuit that is being charged with a voltage source ε.
The voltage across the inductor is
shown as a function of time in the table below, with time in seconds (s) and voltage in volts (V).
The
magnitude of the current at time t
f
can be determined from the voltage (v) as
I
(
t
)
=
1
L
∫
t
i
t
f
v ∙dt
+
I
(
t
i
)
where I(t
i
) is the current at time t
i
.
We will use a technique to calculate this integral algebraically, to
determine the current from the voltage. By doing so, we are finding the approximate area under the
voltage versus time curve.
To determine the current at a time ‘2’, where time ‘1’ is the previous interval,
use
I
2
=
(
V
2
+
V
1
)/
2
∗
∆t
L
+
I
1
The numerator is the average voltage multiplied by the time interval and is an algebraic approximation
for the area underneath the voltage versus time curve between t
1
and t
2
.
In this example, for each interval,
∆t
=
3
×
10
−
7
s
∧
L
=
1
×
10
−
3
H
.
Referencing the voltage in the
table below, to determine the current for the first cell and knowing that at time t=0s,
I
1
=
I
(
t
=
0
s
)=
0
A
I
2
=
(
2.74
+
5
)/
2
∗(
3
×
10
−
7
)
1
×
10
−
3
+
0
=
0.0016
A
and for the next cell
I
3
=
(
1.51
+
2.74
)/
2
∗(
3
×
10
−
7
)
1
×
10
−
3
+
0.0016
A
=
0.0018
A
Repeat the process to fill in all of the cells in the table maintaining 3 significant figures for the current.
Time (s)
Voltage (V)
I(A)
0.0E+00
5.00
0
3.0E-07
2.74
0.00116
6.0E-07
1.51
0.00180
9.0E-07
0.83
0.00215
1.2E-06
0.45
0.00234
1.5E-06
0.25
0.00245
1.8E-06
0.14
0.00251
2.1E-06
0.07
0.00254
2.4E-06
0.04
0.00255
2.7E-06
0.02
0.00256
3.0E-06
0.01
0.00257
2 |
P a g e
Open Desmos by copying and pasting the following link into your web browser or use control+click
(
https://www.desmos.com/calculator/lhdbhowyyh
). Enter the data from the table for current in the
column ‘y
1
’.
You will fit the data using the theoretical equation for current as a function of time.
Adjust the fitting parameters ‘a’ and ‘b’ using the sliders so that the curve matches the points.
Since we
are using an approximating method for calculating the points, the curve will not be an exact match. To fit
the curve as well as possible, note the R
2
value.
R
2
is known as the
coefficient of determination
.
It is a
unit-less number between 0 and 1 and indicates how well the curve fits the points.
The closer R
2
is to 1,
the better the fit.
Adjust ‘a’ and ‘b’ so that R
2
is as close to 1 as possible
Record your fitting parameters below as well as your R
2
value:
a=0.0026
b=1950000
R
2
=0.9996
Comparing to the equation for current above, noting that
a
=
ε
R
and given that
ε
=
5
V
,
determine the resistor value.
Show your work
R=
ε
a
=
5
0.0026
=
1923.077
Ω
Noting that
b
=
1
τ
=
R
L
, determine the value of the inductor.
Show your work.
L=
τ
R
b
=
1923.077
1950000
=
9.862
x
10
−
4
H
Take a screenshot of Desmos showing current versus time and your data and paste below.
Screenshot of Desmos
3 |
P a g e
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II.
Graphing Voltage vs Time (25 points)
The magnitude of the voltage across the inductor was shown to be
V
L
=
ε e
−
t
τ
L
where
τ
L
=
L
R
Open Desmos by copying and pasting the following link into your web browser or use control+click
(
https://www.desmos.com/calculator/g3dkvlljti
)
Using Desmos and the quantities R, L and
ε
from Section I., plot the voltage as a function of time, using
the equation above.
In Desmos, type the values for ε, R and L to replace the # symbols in the equation
provided.
4 |
P a g e
Copy and paste your graph with the theoretical voltage versus time below.
Screenshot of Desmos:
III.
Summary of Section I (10 points)
Write a two paragraph summary of your findings from Section I.
Explain the setup and the results.
Section I Summary
As we can see in Section 1 is to determine the current from the voltage by to calculating the integral
algebraically. To demonstrate it algebraically I will explain how I got the current for I4.For the setup we
use an equation were we add the new voltage to the last and divide by 2. After we multiply it by 3 times
10 to the -7
th
power and divide by 1 times 10 to the -3
rd
power. Lastly we add that to the last current that
was found.
After doing the equation the answer I got for I4 was 0.00215A. After finding that answer I was able to
repeat the equation and get the answers for the rest of the table. Now we know the current from the
voltage source that is charging the RL circuit. After we use theoretical equation for current as a function
of time and determine the resistor value when
ε
=
5
V
. The answer that we get for that is
5 |
P a g e
1923.077
Ω
. The use this equation
b
=
1
τ
=
R
L
to find the value of the inductor and get
9.862
x
10
−
4
H .
IV.
Summary of Section II (10 points)
Write a two paragraph summary of your findings from Section II.
Explain the setup and the results.
Section II Summary
As we can see in Section II we have to graph the the magnitude of the voltage across the inductor. To do
that we have to use Desmos plot the voltage as a function of time by using this equation,
V
L
=
ε e
−
t
τ
L
where
τ
L
=
L
R
. To set up the equation we use the answers we got from determining the resistor value
and finding the value of the inductor with
ε
=
5
V
.
As we plug in the numbers is Desmos we come up with this equation
V
=
5
∗
e
−
1923.007
0.0009862
∗
t
. Plugging that
in Desmos and graphing shows that the graph curves downwards to the right. The graph shows the
theoretical voltage versus time .
6 |
P a g e
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