Submitted by: Submitted by huynhn5
Views: 315
Words: 383
Pages: 2
Category: Science and Technology
Date Submitted: 02/19/2013 04:05 PM
Introduction:
When a capacitor (C) is connected to a dc voltage source like a battery, charge builds upon its plates and the voltage across the plate increases until it equals the voltage (V) of the battery. At any time (t), the charge (Q) on the capacitor plates is given by Q = CV. The rate of voltage rise depends on the value of the capacitance and the resistance in the circuit. Similarly, when a capacitor is discharged, the rate of voltage decay depends on the same parameters. Both charging and discharging times of a capacitor are characterized by a quantity called the time constant τ, which is the product of the capacitance (C) and the resistance (R), i.e.
τ = RC
Equipment and Material:
1. Power supply
2. Resistors
3. Wires and alligator clips
4. Circuit board
5. Conductors
6. Stopwatch
7. Capacitor
8. Switch
9. Voltmeter
Procedure:
R 560 kΩ
VO 5V VC Voltmeter
Measure VC vs. Time for every five seconds and list the data in a table.
Data
Time (second) | VC (V) |
0 | 0 |
5 | 1.12 |
10 | 1.97 |
15 | 2.61 |
20 | 3.15 |
25 | 3.49 |
30 | 3.81 |
35 | 4.00 |
40 | 4.18 |
45 | 4.32 |
50 | 4.38 |
55 | 4.41 |
60 | 4.52 |
65 | 4.56 |
70 | 4.59 |
75 | 4.62 |
80 | 4.64 |
85 | 4.65 |
90 | 4.66 |
Graph:
Calculation:
* Time for constant τ for RC circuit:
τ = RC
* τcal = 560 x 103 x 100x 10-6 = 56 sec
* τexp as measured from the VC vs. time graph.
* Q(t) = Q0 (1- e-t/RC)
* VC(t) = V0 (1- e-t/RC)
VC (t) = V0 (1- e-1) = 0.63
* When VC reaches 0.63 for V0 VC = 0.63 x 5 = 3.15 V
Error analysis:
% Error = | τcal-τexp|τcal x 100% = |56-20|56 x 100% = 64.29%
Conclusion:
We calculated the percentage errors and got 64.29% as the result. The possible sources of error could be machine malfunctions and human error. The machine could not...