I\'ve performed an experiment with RC circuits for my physics class. The circuit
ID: 1263779 • Letter: I
Question
I've performed an experiment with RC circuits for my physics class. The circuit consisted of two 1.5v batteries,a 100 ohm resistor, a capacitor, and a digital multimeter. I don't really understand capacitors, but I have to answer the following questions using the data I collected.
1. What was the voltage across the capacitor when the circuit was first closed?
----I'm guessing it was 3.07 V
2. Was the voltage across the capacitor the same as the voltage of the battery?
----As far as I can tell-- yes, at the very beginning while it was connected it had the same voltage.
3. What was the voltage across the capacitor when the battery was disconnected?
----2.49 and decreasing??
4. What is the time constant for this circuit?
---I have no idea
5. What is the internal resistance of your digital multimeter?
---I have no idea
Here is my data
Time (sec)
Voltage
Time (sec)
Voltage
0
3.07
110
0.39
10
2.49
120
0.33
20
2.09
130
0.27
30
1.70
140
0.23
40
1.43
150
0.19
50
1.18
160
0.16
60
0.97
170
0.13
70
0.81
180
0.11
80
0.67
190
0.09
90
0.56
200
0.08
100
0.47
210
0.06
Continue until V=0
Time (sec)
Voltage
Time (sec)
Voltage
0
3.07
110
0.39
10
2.49
120
0.33
20
2.09
130
0.27
30
1.70
140
0.23
40
1.43
150
0.19
50
1.18
160
0.16
60
0.97
170
0.13
70
0.81
180
0.11
80
0.67
190
0.09
90
0.56
200
0.08
100
0.47
210
0.06
Continue until V=0
Explanation / Answer
Initially when the capacitor and voltage battery are connnected , the voltage of the capacitor will be equal to the battery voltage ideally.
From the data, it can be seen that it is equal i.e (1.5 + 1.5 = 3.0 v ideally and the batteries in reality need not be exact 1.5v each so it is 3.07)
So the voltage across the capacitor when the circuit was first closed is 3.07 v (from the data given)
3)
the voltage across the capacitor when the battery was just disconnected will be 3.07 v and then it goes on deceasing as the capacitor discharges.
4)
the time constant is the time required to charge the capacitor, through the resistor, by ? 63 percent of the difference between the initial value and final value or discharge the capacitor to ?37 percent
37% of the initial value i.e 3.07v is 37/100 * 3.07 = 1.136 V = 1.14 V approx.
The voltage became 1.18V at 50 seconds and it might be 1.14 V just after this.
we cant say exact time when it comes to 1.14 V
it is around 52 seconds.
So the time constant will be 52 seconds approx.
time constant = R * C = 52
R = 100 ohm + R of digital multimeter
C is not mentioned in the question.
using this ui can find the resistance of multimeter
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