= Cexperimental R T -= Now, measure the actual value of the capacitance with the digital multimeter and compare it to the value obtained from the graph: 12.4 Actual capacitance is: FI
= Cexperimental R T -= Now, measure the actual value of the capacitance with the digital multimeter and compare it to the value obtained from the graph: 12.4 Actual capacitance is: FI
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Question
the actual resistance is 0.80 mega ohms
![Experim
Time
[s]
3
na 6
ait
12
15
18
21
24
29
35
Ischarging):
Voltage across the
resistor, AVR [v]
5.9
4.3
byly 3,1
2,3
1.6
1.2
0.8
0.6
0.4
0,2
In (AVR)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F50351a65-5e52-4545-9e5a-caf58e03b0d0%2F55479d66-0283-48a1-b8c3-41d2a3699383%2F75bz6zq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Experim
Time
[s]
3
na 6
ait
12
15
18
21
24
29
35
Ischarging):
Voltage across the
resistor, AVR [v]
5.9
4.3
byly 3,1
2,3
1.6
1.2
0.8
0.6
0.4
0,2
In (AVR)
![Interpret
In (AVR) = In (AV) -t as y(t)=b+mt
T
and get the result: graph of In (AVR) versus / (called a semi-log graph) should be linear
with y-intercept b = ln (AV) and slope m =
1
=-²/
T
We can determine t and hence C from an experimental measurement of the slope.
Draw a graph of In (AVR) (log of measured voltage across the resistor, vertical axis)
vs time (horizontal axis).
From the y-intercept of the best-fit line, find AVo = 9[V], as expected.
Using the slope, find T =
Using this calculate the unknown capacitance:
T
Cexperimental R
-
-
1
slope
=
Now, measure the actual value of the capacitance with the digital multimeter and
compare it to the value obtained from the graph:
Actual capacitance is: 12,4
FI](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F50351a65-5e52-4545-9e5a-caf58e03b0d0%2F55479d66-0283-48a1-b8c3-41d2a3699383%2F7822lx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Interpret
In (AVR) = In (AV) -t as y(t)=b+mt
T
and get the result: graph of In (AVR) versus / (called a semi-log graph) should be linear
with y-intercept b = ln (AV) and slope m =
1
=-²/
T
We can determine t and hence C from an experimental measurement of the slope.
Draw a graph of In (AVR) (log of measured voltage across the resistor, vertical axis)
vs time (horizontal axis).
From the y-intercept of the best-fit line, find AVo = 9[V], as expected.
Using the slope, find T =
Using this calculate the unknown capacitance:
T
Cexperimental R
-
-
1
slope
=
Now, measure the actual value of the capacitance with the digital multimeter and
compare it to the value obtained from the graph:
Actual capacitance is: 12,4
FI
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