To investigate the discharging of a capacitor through a multimeter as voltmeter

Practical 4
Title: To investigate the discharging of a capacitor through a multimeter as voltmeter
Objective: To determine the internal resistance of a multimeter
Apparatus and Materials:
1. Capacitor
2. Multi-meter
3. Power supply
4. Switch
5. Stop watch
6. Connecting wires

Setup :

1 The circuit was constructed as Figure 5-1.
2 Ensure that the positive terminal of the power supply was connected to the positive terminal of the capacitor if an electrolytic capacitor.
3 Initially, switch S should be closed.

Figure 5-1

Theory:

When the switch is opened, the capacitor will be discharged through the resistor of the voltmeter.
The potential difference V across resistor will decrease exponentially with time t.
The formula which relates V and t is
℮

Therefore:

Procedure:

1 Initially the switch was closed.
2 The initial reading of the voltmeter was recorded.
3 Open the switch S, and simultaneously start a stop watch.
4 Values of potential difference V and time t was measure and recordedfor every 5 seconds until the reading on the multimeter becomes less than.
5 The reading of V, t and lnV was tabulated.
6 A graph of lnV against t was ploted.
7 The gradient of the graph was determined.
8 The answer from [7], and the value of capacitance Cwas used to determine a value for. The gradient,. Here C was the capacitance and not intercept value.

Data :

Time, t ± 0.1s
Voltage, Vc ± 0.01V
Average, Vc ± 0.01V
In Vc
0
11.14
11.95
11.55
2.45
5
8.34
7.82
8.08
2.09
10
6.03
4.69
5.36
1.68
15
3.43
7.08
3.26
1.18
20
2.15
1.96
2.06
0.72
25
1.39
1.28
1.34
0.29
30
0.95
0.84
0.90
- 0.11

Calculation:

Average of time, t =
=
= 15

Average of In Vc =
=
= 1.19

Centroid : ( 15 , 1.19 )

Gradient , m = ( = = - 0.09 vs-1

Experimental :
I =
Vc = Vmax ln Vc = ln Vmax
Y = C + mx

ln Vmax = C = 2.53 Vmax = 12.49 v

= gradient, m C = 1.0

Ri = = = 1.11 × 107 Ω