TAP 129- 5: Discharging a capacitor
A capacitor is charged to a potential difference of 1.0 V. The potential difference is measured
at 10 s intervals, as shown in the table. When t = 15 s, a resistance of 1.0 Mis connected
across the capacitor terminals.
t/s
V/V
0
1.00
10
1.00
20
0.81
30
0.54
40
0.35
50
0.23
60
0.15
1.
What is the current in the resistor at t = 15 s?
2.
Plot a graph of V against t, and measure the rate of decrease of V immediately after
t = 15 s.
3.
Using the relationship I = d Q / d t = C d V / d t, calculate the value of the
capacitance.
4.
Plot also a graph of ln V against t to show the exponential decay of voltage, and use
the gradient to find the time constant (RC).
5.
From the time constant calculate the capacitance C.
6.
Explain which method gives a better value of C, and why.
Practical advice
This question could usefully follow an experiment where students see capacitors discharge. It
also shows that exponential behaviour can be demonstrated by the use of natural logarithms.
Answers and worked solutions
1.
1 A
2.
Approximately 3.8 × 10–2 V s–1
3.
Approximately 26 F
4.
24.5 s
5.
24 F
6.
The second method is better, because it avoids the difficulty of accurately drawing a
gradient for the V–t graph at 15 s.
Worked solutions
1.
I 
1 .0 V
V

 1.0 A
R 1.0  10 6 
2.
1.0
0.9
0.8
0.7
0.6
0.5
0.4
dV 0.8 V
=
= 38 mV s–1
dt
21 s
0.3
0.2
36 s – 15 s = 21 s
0.1
0
0
10
20
30
40
t/s
50
60
3.
I
1.0  10 6 A
dQ
dV
I
C
C

 26 F
dt
dt
dV / dt
0.038 V s 1
4.
t/s
ln V
0
0
10
10
20
–0.21
30
–0.62
40
–1.05
50
–1.47
60
–1.90
0.5
0 +
+
+
–0.5
+
–1
+
–1.5
+
–2
+
0
10
20
30
40
50
60
70
t/s
V  V0 e
t
RC
lnV  lnV0 
t
RC
gradient of line  
RC 
1
0.041 s –1
1
 2.0
 2.0


 0.041 s –1
RC 63 s - 14 s
49 s
 24 .5 s
5.
RC  24 .5 s
24.5 s
C
 24 μF
1  10 6 
6.
The second method is better, because it avoids the difficulty of accurately drawing a
gradient for the V–t graph at 15 s.
External references
This activity is taken from Advancing Physics Chapter 10, 80S