science-spark.co.uk
Module G485.2 Capacitors
student answer booklet
1
Lesson 22 - Capacitors in series and parallel questions
Answer the questions in the gap below and on the other side.
1.
Draw a diagram and then calculate the total capacitance of each of the following
arrangements of capacitors:
(a)
two capacitors of 100 F connected in series
(b)
two capacitors of 100 F connected in parallel
(c)
two capacitors, one of 100 F and the other of 200 F connected in series
(d)
two capacitors, one of 100 F and the other of 200 F connected in parallel
(e)
two capacitors, one of 100 F and the other of 4700 F connected in parallel with the
combination connected in series with another capacitor of capacitance 4700 F
2.
If you have several 2.0 F capacitors each capable of withstanding 240V without
breakdown how would you assemble a combination having an equivalent capacitance of:
(a)
0.40 F
(b)
1.2 F,
each capable of withstanding 1000V?
2
Answers and worked solutions
1
(a)
50 F
(b)
200 F
(c)
67 F
(d)
300 F
(e)
2400 F (2375 F)
2
(a)
Five capacitors in series
(b)
Three parallel sets of five capacitors in series
(both can tolerate 5  240V = 1200V)
External references
This activity is taken from Resourceful Physics http://resourcefulphysics.org/
3
Lesson 24 - Energy
stored in capacitors.
What to do
Work your way through the questions, preferably in the order in which they
appear. The first few questions will give you some practice in using the equations
relating to energy storage.
Questions
A 50 microfarad (F) capacitor is charged to a pd of 60 V.
1.
Calculate the charge on the capacitor.
2.
Calculate the energy stored.
3.
Calculate the energy stored when the pd is doubled to 120 V.
4.
Compare your answers to questions 2 and 3. What does this tell you
about the relationship between the energy stored by a capacitor and the
pd to which it has been charged?
5.
A 1000 F capacitor is charged so that its stores 2.0 J of energy.
Calculate the pd to which it has been charged.
6.
The incomplete table below contains values of capacitance, charge, pd
and energy for a series of charged capacitors. Carry out calculations and
fill in the blanks in the table.
Capacitanc Charge
e
Potential
difference
1000 F
16 V
Energy
4
10 mF
0.01 C
1.0 F
100 J
2.0 mC
5000 V
100 V
33 000 F
50 mJ
2.0 J
A 1.0 F capacitor is charged to a pd of 10 V.
7.
Calculate the charge on the capacitor.
8.
How much charge flowed through the battery during charging?
9.
How many electrons flowed through the battery during charging?
10.
Calculate the energy stored by the capacitor.
11.
How much energy was transferred from the battery during the charging
process?
12.
(Rather harder) You should have found different answers for questions 10
and 11. Explain this difference.
Hints
1.
You will have to do quite a bit of rearranging of equations.
6.
Think of the definition of potential difference when you are trying to work
out how much energy is involved in passing charge through a battery.
5
6
Practical advice
The range of questions is designed to give students practice in handling the
relevant equations. It is advisable that students work through the questions more
or less in the order they appear. Because there are three different equations
relating to energy storage in capacitors, there are several instances where
students have to choose which to use for their own convenience.
There are many pitfalls, not deliberate, into which students can fall because of
difficulties with unit prefixes and with standard form. From this point of view, the
questions provide some good practice in handling both of these issues.
If they have met only E = ½ Q V but not the other versions, it may be necessary
to spend some time with them developing ‘variations on the theme’. Question 12
is an issue familiar to teachers but difficult for most students to grasp; they might
need more help than is given in the hints. It is probably best left as extension
work for students aiming for higher grades.
Answers and worked solutions
1.
3.0 x 10–3 C
2.
0.090 J
3.
0.36 J
4.
If the pd is doubled, energy stored goes up by factor of four; energy is
proportional to t
he square of the pd
5.
63 V
6.
Capacitanc Charge
e
Potential
difference
Energy
1000 F
0.016 C
16 V
0.013 J
10 mF
0.01 C
1.0 V
5 mJ
1.0 F
14 C
14 V
100 J
4.0x 10-7 F
2.0 mC
5000 V
5J
10 F
1.0mc
100 V
50 mJ
33 000 F
0.36 C
11 V
2.0 J
7.
1.0 × 10–5 C
8.
1.0 × 10–5 C
7
9.
6.2 × 1013
10.
5.0 × 10–5 J
11.
10.0 × 10–5 J
12.
50% of the energy from the battery is used to heat the connecting wires
(warming the wires). If there were no resistance somewhere in the circuit
we would have a non-physical situation of the system discharging in an
infinitesimal time, with infinite currents. In practice inductive effects would
also come into play, even if we had a superconducting circuit. This raises
the issue of boundary conditions- you can’t have a non-physical system.
(See Physics Education July 2005 pages 370 and following.
For details how to obtain Physics Education see the IOP website.)
Worked solutions
1.
Q  CV  (50  106 F)  60 V  3.0  103 C
2.
E
1
QV
2

 (3.0  10 3 C)  60 V  0.090 J
1
2
3.
E  21 CV 2 
4.
1
2
 (50  106 F)  120 V   0.36 J
2
If the pd is doubled, the energy stored goes up by a factor of four; energy
stored is proportional to the square of the pd
5.
V 

2E
C
2  2.0 J
1000  10 6 F
 63 V
6.
Q  CV  (1000  10 6 F)  16 V  0.016 C
E  21 QV 
1
2
 0.016 C  16 V  0.13 J
V 
0.01 C
Q

 1 .0 V
C 10  10 3 F
E
1
QV
2

1
2
 0.01 C  1.0 V  5 mJ
8
2E
C
V 
2  100 J
 14 V
1 .0 F

Q  CV  1.0 F  14 V  14 C
Q
V
2.0  10 3 C

 4.0  10 7 F
5000 V
C
E  21 QV

1
2
 (2.0  10 3 C)  5000 V  5.0 J
2E
V
2  50  10 3 J

 1.0  10 3 C
100 V
Q
C
V
1.0  10 3 C

 10 F
100 V
Q
V 

2E
C
2  2 .0 J
33 000  10  6 F
 11 V
Q  CV  (33 000  10  6 F)  11 V  0.36 C
7.
Q  CV  (1.0  106 F)  10 V  1.0  105 C
8.
1.0  10 5 C
10.
E
1
QV
2

1
2
 (1.0  10 5 C)  10 V  5.0  10 5 J
11.


E  QV  1.0  105 C  10 V  1.0  104 J
12.
50% of the energy transferred from the battery is used to heat the
connecting wires.
External references
9
This activity is taken from Advancing Physics, Chapter 10, 110S
10
Lesson 25-28 questions - Capacitance
1.
In an experiment a capacitor is charged from a constant current supply by a 100
mA current pulse which lasts 25 s.
a)
Calculate the charge on the capacitor after this time.
Charge = ……2.5 C…. C (2)
b)
The pd across the capacitor is 6 V when it has been charged. Calculate the
capacitance of the capacitor. Use a suitable unit.
Capacitance = ……0.42 ……….. unit… F ………(3)
2.
A capacitor is charged at a constant current of 2.0 mA until the charge on the
capacitor is 0.010 C.
a)
How long did it take the capacitor to charge to this value?
Time = ………5……… s (2)
b)
The capacitance of the capacitor is 100 microfarad (μF). To what pd had it
been charged?
PD = …100…………. V (2)
3.
This question allows you to practise using the equation Q = C V. Fill in the gaps
in the table.
C
1000 μ
8μF
32 μ F
33 mF
50 μ F
Q
0.012C
2 mC
0.64 mC
0.2C
1.25 C
11
V
12 V
250 V
20V
6V
25 kV
10 mF
2.5 C
250V
(3)
4
In this circuit the capacitor is initially uncharged.
6V
470 k
2 F
a)
Calculate the current through the resistor when the switch is first closed.
Current = ……13 µA ……….. (2)
b)
What is the current after the switch has been closed for a long time?
Explain your answer.
Current = …… Zero ……….. A (2)
c)
is 2.0 V.
Calculate the current through the resistor when the pd across the capacitor
12
Current = ……8.5 µA ……….. (2)
d)
When this capacitor is charged through the fixed resistor, the graph shows
how the charge on the capacitor varies with time.
t
i)
What is the significance of the gradient of the graph?
……… The gradient is the current at that time…………………………………(1)
ii).
Explain why the gradient of the graph gradually falls?
As the capacitor charges up, the pd across it rises and the pd across the resistor falls. The
current through the resistor is equal to the pd across it divided by the resistance, so the
current, and therefore the gradient, gradually decreases.
……………………………………………………………………………………… (2)
ii)
Calculate the maximum charge on the capacitor.
Charge = ……12 µC ……… (2)
iii)
Use the same axes to sketch the curve you would expect if the
resistor were replaced by one with a smaller resistance.
(2)
5.
What is meant by the capacitance of a capacitor?
The ratio of the charge on the plates of the capacitor to the potential difference
across them.
C = Q/V.
……………………………………………………………………………………… (2)
6.
Define the farad.
13
The charge required to increase the potential difference across the plates by 1 V. (2)
7
(a)
A capacitor of capacitance 5 F is connected to a 6 V supply. What charge
is stored in the capacitor?
Q = CV = 5 x 10-6 x 6 = 30 x 10-6 = 30 C
Charge = ……30 ……….. C (2)
(b)
A 400 pF capacitor carries a charge of 2.5 x 10-8 C. What is the potential
difference across the plates of the capacitor?
V = Q/C = 2.5 x 10-8 / 400 x 10-12 = 62.5 V
Potential difference = ……62.5 …………. V (2)
8
A capacitor is charged such that there is a charge of +20 mC on the positive plate.
What is the charge on the negative plate?
20 mC equal and opposite
Charge = …20 mC equal and opposite ……….. C (2)
A 4700 F capacitor is connected as shown in the circuit diagram. When it is
fully charged:
9
4.5 V
C
(a)
what is the charge on the positive plate of the capacitor?
Q = CV = 4700 x 10-6 x 4.5 = 0.021 C
Charge = ……0.021 ……….. C (2)
14
(b)
what is the potential difference across the capacitor?
Potential difference = ……4.5 ……………. V (2)
(c)
how many additional electrons are on the negative plate?
0.021 / 1.6 x 10-19 = 1.32 x 1017
d)
diagram.
Number of electrons = ………1.32 x 1017…………. (2)
A resistor of 100 Ω is now added to the circuit as shown in the second
4.5 V
C
i)
Explain what effect this has on the time to charge up the capacitor.
…………………………………………………………………………………………..
………………… it will take longer ……………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (1)
ii)
What is the final charge on the plates?
it will be the same as before
iii)
Charge = …0.021 ……………… C (2)
What is the final potential difference across the capacitor?
4.5 V since no current is flowing finally there will be no potential difference across the
resistor whatever its value.
Potential difference = ………4.5 ………….. V(2)
A 250 F capacitor is charged through a 100 k
15
10.a) Calculate the time constant of the circuit.
time constant = ……25 …………. Unit …… s ……….. (3)
10b). The initial current is 100 A. What is the current after 30 s?
current = ………30  ……………… A (2)
Suggest values of R and C which would produce RC circuits with time constants
of (i) 1.0 s and (i) 20 s.
Time constant = … C = 10 F and R = 100 k;……………….
Time constant = …… C = 200 F and R = 100 k.……………. (4)
10c). The insulation between the plates of some capacitors is not perfect, and
allows a leakage current to flow, which discharges the capacitor. The capacitor is
thus said to have a leakage resistance. A 10 F capacitor is charged to a
potential difference of 20 V and then isolated. If its leakage resistance is 10 M
how long will it take for the charge to fall to 100 C?
t  RC ln 2  (10  106 )  (10  106 F)  0.693  69 s.
time = ………69 ………………. s (3)
A 100 F capacitor is charged and connected to a digital voltmeter (which has a
very high resistance). The pd measured across the capacitor falls to half its initial
value in 600 s.
10d). Calculate the time constant of the discharge process.
t1/2=ln2 x CR
RC 
600 s
 8.6  10 2 s.
ln 2
time constant = ………8.6 × 102 ……s……………. (2)
10e). Calculate the effective resistance of the capacitor insulation.
16
Resistance = ……8.6 M……………. (2)
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.
11.
I 
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
What is the current in the resistor at t = 15 s?
1 .0 V
V

 1.0 A
R 1.0  10 6 
current = ………1……………… A (2)
12.
Plot a graph of V against t, and measure the rate of decrease of V
immediately after
17
t = 15 s.
(4)
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
50
60
t/s
13.
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).
(4)
t/s
ln V
0
0
10
10
20
–0.21
30
–0.62
40
–1.05
50
–1.47
60
–1.90
18
0.5
0 +
+
+
–0.5
+
–1
+
–1.5
+
–2
+
0
10
20
30
40
50
60
70
t/s
14.
From the time constant calculate the capacitance C.
RC  24 .5 s
24.5 s
C
 24 μF
1  10 6 
Capacitance = ………………….. F (2)
15.
Explain which method gives a better value of C, and why.
The second method is better, because it avoids the difficulty of accurately
drawing a gradient for the V–t graph at 15 s
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
19
Worked solutions
1.
R C = 250  10–6 F  100  103 = 25 s
2.
I  I 0 e t / RC  100 A  e 30 / 25  30 A
3.
Any pair of values such that R C = 1.0 s and R C = 20 s is acceptable.
4.
Q0  CV  10 F  20 V  200 C. .
Since
Q
 e t / RC ,
Q0
 Q
t
 ln
RC
 Q0

 1
  ln 

 2

and
t  RC ln 2  (10  106 )  (10  106 F)  0.693  69 s.
5.
V
1
  e 600 s/RC
V0
2
so
 1   600 s
ln  
RC
2
and
RC 
600 s
 8.6  10 2 s.
ln 2
6.
R
8.6  10 2 s
100  10 6 F
 8.6 M
External references
This activity is taken from Advancing Physics Chapter 10, 70S
20