3/7/2024
PHYS 262
Slow Decay Lab Report
Measured Capacitance = 0.930 µF
Actual Resistance of 22 MΩ resistor = 22.613 MΩ
Measured Voltage=13.897v
Average observed half life = 24.8 seconds
Calculated 1/e lifetime= 37.77s
Product of Resistance and Capacitance= 21.03 V/A
Time(s)
10
17.5
25
32.5
40
47.5
55
62.5
70
77.5
85
92.5
100
107.5
115
122.5
130
137.5
145
152.5
Voltage
(V)
10.309
8.218
6.571
5.27
4.247
3.416
2.723
2.211
1.8
1.468
1.203
0.974
0.804
0.665
0.55
0.457
0.38
0.314
0.263
0.215
Voltage vs Time
12
10
Voltage(v)
Measurement
#
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
y = 12.759e-0.027x
8
In this graph 12.759 represents
our initial voltage, and 0.027
represents our 1/RC value.
6
4
2
0
0
50
100
Time(s)
The point (33.997, 4.995) in
the graph represents the
value for the 1/e life, 33.40 s.
The point (25.67, 6.54) in the
graph represents the value
for the 1/2 life, 25.67s.
Since 0.027 is 1/RC, then
RC= 37.03.
150
200
3/7/2024
PHYS 262
Slow Decay Lab Report
A negative linear slope was
obtained from the graph, which
represents value of -1/RC.
Hence, RC=36.76
The values obtained with the
previous graph match the
desired data.
ln(V) vs Time
3
2.5
y = -0.0272x + 2.5463
Ln of Voltage(v)
2
1.5
1
0.5
0
-0.5 0
20
40
60
80
100
120
140
160
180
-1
-1.5
-2
Time(s)
CONCLUSION/QUESTIONS
• Why was the DMM placed in series with the 22 megaohm resistor instead of in parallel across it?
If the DMM is in parallel, then the R total value will be lowered. This would lead to lower RC,
which is not desired because it decays quickly.
•
Start with the defining equations for the Ohm and the Farad and prove that the quantity RC has dimensions
of time.
o
𝑅=
𝑉
𝐼
→ 𝐼=
𝑞
𝑡𝐶
=
𝑞
𝑉
→𝑅=
(𝑉)(𝑡)
𝑞
→ 𝑅𝐶 =
(𝑉)(𝑡)(𝑞)
(𝑞)(𝑉)
= 𝑡
•
Use your Cartesian graph of V vs. t and write down the mathematical relationship between V (across the
DMM) and t and Vo. Assume that the switch is opened at time t=0 so Vo is the voltage across the DMM at
t=0.
𝑉 = 12.759𝑒 −0.027𝑡
𝑉 = 12.759
𝑡
1
−
= −0.027𝑡 → 𝑅𝐶 =
𝑅𝐶
0.027
•
Use your mathematical result from question 3 and find a mathematical expression for I(t), the current
through the DMM.
12.759𝑒 −0.027𝑡
𝐼(𝑡) =
𝑅
•
Assume that the switch opens at t=0. Sketch I(t) vs. t and find a value for the maximum current. Express in
microamperes.
12.759
𝐼(𝑡) =
= 0.34 ∗ 10−3 𝑚𝐴
36.76 ∗ 106
•
Determine the value of the maximum voltage on the capacitor? Hint: The value is not anywhere close to 14
45V
What is the value of the maximum charge stored in the capacitor?
𝑄𝑚𝑎𝑥 = 12.759(0.930 µF) = 11.86 ∗ 10−6 𝐶
•
3/7/2024
PHYS 262
Slow Decay Lab Report
•
U (stored energy) = (Q2 )/2C. Find the maximum value of the energy stored in the capacitor. Units!!
(11.86 ∗ 10−6 )2 𝑉
→ 75.6 ∗ 10−6 𝐽
2(0.930 µ)𝐹