3.1 Electrons and energy
3.1a Monitoring and measuring alternating current
Alternating current
An alternating current (a.c.) periodically changes direction. Contrast this to
direct current (d.c.), which only flows in one direction. Although it is possib le
that current could vary in any way, we will consider a sinusoidal function, ie the
voltage constantly changes in the form of a sine function. This is the most
common form of a.c. since electricity generators work by spinning in circles,
which means that the current is pushed one way then the other, with the
instantaneous value of voltage constantly varying. Examples of each type of
current when displayed on an oscilloscope are shown below.
a.c. waveform
d.c. waveform
Measuring frequency and peak voltage
An oscilloscope plots a graph of changing voltage on the y-axis versus time on
the x-axis. Just like a graph we need to read the scale, except the numbers aren't
along the axis, they are on a dial underneath the screen. The voltage axis scale is
usually called the volts/div or volts/cm and the time axis scale is called the
timebase.
To calculate the frequency of an a.c. signal, we first of all have to find its period.
This is the time for one complete cycle of to and fro current, so we measure the
horizontal distance on the screen between crests.
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An alternating voltage varies between the same maximum negative and positive
value as the voltage pushes first one way (positive) then the other (negative).
This maximum is defined as the peak voltage. It can be measured from an
oscilloscope by measuring the vertical distance on the screen from the bottom of
the signal to the top. This then has to be halved to get the peak voltage.
Worked example
In the picture below each box on the CRO screen has a side of 1 cm.
Timebase = 5 ms cm–1
Volts/div = 2 V cm–1
The distance between crests is 4 cm.
The distance from bottom to top is 8 cm.
The time base was set at 5 ms cm –1 ,
then the period of the wave is:
T = 4 × 5 ms = 4 × 0.005 = 0.02 s
1
f 
T
1
f 
0  02
f = 50 Hz
The volts/div was set at 2 V cm –1 ,
then the peak voltage is:
V peak = ½ × 8 × 2
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V peak = 8 V
If the time base is switched off, the a.c. signal will not be spread along the xaxis. However, the voltage variation will continue, meaning a straight vertical
line will be displayed on the screen. The height of this line from bottom to top
can be processed in exactly the same way as above, ie halved then converted into
voltage using the volts/div.
Alternating current – peak and rms
Electricity is a method of transferring energy, so a.c. is capable of transferring
energy in the same way as d.c. The instantaneous amount of power being
supplied can be calculated using the value of voltage and/or current at a
particular point. However, in a.c. these values change constantly, completing 50
cycles every second in the UK. Our eye couldn’t possibly follow such quick
changes. We need an average, but the average value of the voltage during any
complete cycle is zero!
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The amount of energy transmitted by a.c. will depend on some ‘average’ value of
the voltage, but as we have seen above the average value is zero! A different
‘average’ must be used. The ‘average’ value of an a.c. voltage is called the root
mean square (rms) voltage (V rms ).
The definition of the rms voltage is that it is that value of direct voltage that
produces the same power (eg heating or lighting) as the alternating voltage. The
rms value is what is quoted on a power supply so that a fair comparison between
a.c. and d.c. can be made, eg a 6 V bat tery will produce the same brightness of
light bulb as a 6 V rms a.c. supply. The rms current has a similar definition.
Consider the following two circuits, which contain identical lamps.
The variable resistors are altered until the lamps ar e of equal brightness. As a
result the d.c. has the same value as the effective a.c. (ie the lamps have the same
power output). Both voltages are measured using an oscilloscope, giving the
voltage equation below. Also, since V = IR applies to the rms valves and to the
peak values, a similar equation for currents can be deduced.
V rms = 1 V peak
2
and
Irms = 1 Ipeak
2
Note: A multimeter switched to a.c. mode will display rms values.
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Graphical method to derive the relationship between peak and rms values of
alternating current
The power produced by a current I in a resistor of resistance R is given by I 2 R. A
graph of I 2 against t for an alternating current is shown below. A similar method
can be used for voltage.
2
The average value of I is
is
2
I peak
2
2
I peak
2
and therefore the average power supplied
R.
An identical heating effect (power output) for a d.c. supply is I rms 2 R (since I rms is
defined as the value of d.c. current that will supply equivalent power).
Setting both of these equal to each other gives:
2
I peak
2
2
I peak
2
I peak
2
2
R  I rms
R
2
 I rms
 I rms
Important notes:
1.
2.
Readings on meters that measure a.c. are rms values, not peak values.
For power calculations involving a.c. always use rms values:
2
P  I rmsVrms  I rms
R
2
Vrms
R
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3.
The mains supply is usually quoted as 230 V a.c. This is of course 230 V
rms. The peak voltage rises to approximately 325 V. Insulation must be
provided to withstand this peak voltage.
Example
A transformer is labelled with a primary of 230 V rms and secondary of 12 V rms.
What is the peak voltage which would occur in the secondary?
V peak = 2 × V rms
V peak = 1.41 × 12
V peak = 17.0 V
3.1b Current, voltage, power and resistance
Basic definitions
In the circuit below, when S is closed the free electrons in the conductor
experience a force that causes them to move. Electrons will tend to drift away
from the negatively charged end towards the positively charged end.
Current is a flow of electrons and is given by the flow of charge (coulombs) per
second:
I
Q
t
The energy required to drive the electron current aroun d this circuit is provided
by a chemical reaction in the battery. The electrical energy that is supplied by
the source is transformed into other forms of energy in the components that make
up the circuit.
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Voltage is defined as the energy transferred per unit charge.
V = W/Q
In this section W will be used for the work done, ie energy transferred, therefore
1 volt = 1 joule per coulomb (J C –1 ).
When energy is being transferred from an external source to the circuit, the
voltage is referred to as an electromotive force (emf). When energy is
transformed into another form of energy by a component in the circuit, the
voltage is referred to as a potential difference (pd).
Energy is supplied to the
circuit:
chemical  electrical
An emf (voltage) can be
measured
Energy is provided by
the circuit:
electrical  light + heat
A pd (voltage) can be
measured
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Sources of emf
Emfs can be generated in a great variety of ways. See the table below for
examples.
Chemical cell
Chemical energy drives the current
(eg battery)
Thermocouple
Heat energy drives the current
(eg temperature sensor in an oven)
Piezo-electric generator
Mechanical vibrations drive the current
(eg acoustic guitar pickup)
Solar cell
Light energy drives the current
(eg solar panels on a house)
Electromagnetic generator
Changes in magnetic field drive the current
(eg power stations)
Ohm’s law
In any circuit, providing the resistance of a compone nt remains constant, if the
potential difference V across the component increases, the current I through the
component will increase in direct proportion.
At a fixed temperature, for a given
conductor:
V  I, ie V/I is constant
Voltage
The constant of proportion is defined
to be the resistance,
ie R 
Current
V
or V  IR
I
This is Ohm’s law.
A component that has a constant resistance when the current through it is
increased is said to be ohmic. Some components do not have a constant
resistance; their resistance changes as the pd across the component is altered, for
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example a light bulb, a transistor or a diode. A graph of V versus I for such
components will not be a straight line.
Electric power
For a given component P = IV since:
power 
work done QV Q

  V  IV
time
t
t
Since V = IR
P = I × IR (substituting for V)
= I2R
or
P

so P  I 2 R 
V
 V (substitut ing for I )
R
V2
R
V2
R
The expression I 2 R gives the energy transferred in one second due to resistive
heating. Apart from obvious uses in electric fires, cookers, toasters etc,
consideration has to be given to heating effects in resistors, transistors and
integrated circuits, and care taken not to exceed the maximum ratings for such
components. The expression V 2 /R is particularly useful when the voltage of a
power supply is fixed and you are considering changes in resistance, eg different
power of heating elements.
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Example
An electric heater has two heating elements allowing three setting s: low, medium
and high. Show by calculation which element(s) would be switched on to provide
each power setting.
230 V
15 Ω
30 Ω
for 30 Ω: P 
V 2 2302
= 1800 W → low power

R
30
for 15 Ω: P 
V 2 2302
= 3500 W → medium power

R
15
for both:
1
1
1
1
 

Rp 15 30 10
R p = 10 Ω
P
10
V 2 2302
= 5300 W → high power

R
10
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Circuit rules
Series
Parallel
Current
IS  I1  I 2  I 3
I P  I1  I 2  I 3
Voltage
VS  V1  V2  V3
Resistance
RS  R1  R2  R3
1
1
1
1
 

RP R1 R2 R3
Addition of power
By conservation of energy, the total power used in both series and parallel
circuits is the sum of the power used in each component.
Useful to remember
Identical resistors in parallel
Two resistors of the same value (R) wired in parallel:
R P = half the value of one of them (½R).
eg
2 × 1000 Ω in parallel, R P = 500 Ω
2 × 250 Ω in parallel, R P = 125 Ω
R
≡ ½R
R
Voltage and current by proportion
The voltage across an individual resistor in a series circuit is proportional to the
resistance. The current through one of two resistors in parallel is inve rsely
proportional to the resistance, ie proportional to the other resistance.
Examples
Current in parallel
Voltage in series
12 V
2Ω
3A
4Ω
8Ω
4Ω
12 Ω
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3A will split in the ratio:
R S = 8 + 4 + 12 = 24 Ω
2
4
1 2
:
, ie :
42 42
3 3
V1 
8
 12  4 V
24
1
 3A=1A
3
V2 
4
 12  2 V
24
2
3 A = 2 A
3
V3 
12
 12  6 V
24
The smallest resistor takes the largest current, so:
I 1 = 2 A, I 2 = 1 A
Worked example 1
40 V
In the circuit shown opposite calculate:
(a)
(b)
(c)
the current in each resistor
the pd across each resistor
the power dissipated in each resistor.
6Ω
4Ω
5Ω
10 Ω
(a)
4 Ω and 6 Ω in series are equivalent to 10 Ω so the resistor network can be
represented as:
1
1 1 1
  
Rp 10 10 5
RP = 5 Ω
R T = 5 + 5 = 10 Ω
IT 
12
VT 40

 4 A (current through 5 Ω)
RT 10
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I 1 meets a resistance of 10 Ω and so does I 2 .
so I 1 = I 2 = 2 A
(b)
For the 5 Ω resistor: V = IR = 4 × 5 = 20 V
for the 4 Ω resistor: V = IR = 2 × 4 = 8 V
for the 6 Ω resistor: V = IR = 2 × 6 = 12 V
for the 10 Ω resistor: V = IR = 2 × 10 = 20 V
(c)
For 5 Ω: P = IV = 20 × 4 = 80 W
for 4 Ω: P = I 2 R = 2 2 × 4 = 16 W
for 6 Ω:
P
V 2 12 2

 24 W
R
6
for 10 Ω: P = VI = 20 × 2 = 40 W
Worked example 2
A 10 W model car motor operates on a
12 V supply. In order to slow the car
it is connected in series with a controller
(rheostat).
12 V
The controller is adjusted to reduce the
motor’s power to 4 W.
M
Assuming that the resistance of the motor
does not change, calculate:
(a)
(b)
the resistance of the controller at this setting
the power wasted in the controller.
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(a)
We cannot find the resistance of the controller directly. We need to find the
resistance of the motor and the total resistance of the circuit when the
motor is operating at 10 W.
For the motor at 10 W, P = 10 W, V = 12 V
P
V2
V 2 12 2

, so R 
R
P
10
R = 14.4 Ω
For the motor at 4 W, P = I 2 R, so I =
P
4

R
14  4
= 0.527 A
For the whole circuit: RT 
VT
12

 22.8 Ω
I 0.527
Resistance of controller = 22.8 – 14.4 = 8.4 Ω
(b)
Power in controller = I 2 R = 0.527 2 × 8.4
= 2.3 W
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The potential divider
A potential divider provides a convenient way of obtaining a variable voltage
from a fixed voltage supply.
Consider first two fixed resistors, R 1 = 10 Ω and R 2 = 20 Ω, connected in series
across a 6 V supply:
6V
10 Ω
20 Ω
V1
V2
The current in this circuit can be calculated using Ohm’s law (total R = 10 + 20 =
30 Ω).
so,
I
VS
6

R 30
= 0.2 A
The voltage across R 1
V 1 = IR 1 = 0.2 × 10
=2V
The voltage across R 2
V 2 = IR 2 = 0.2 × 20
=4V
If you look closely you can see that the voltages across the resistors are in the
ratio of their resistances (½ in this case),
ie,
V1 R1

V2 R2
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The voltage across R 1 is given by:
V1 
R1
 VS
R1  R2
similarly, the voltage across R 2 is
V2 
R2
V
R1  R2
This is known as the potential divider rule.
A variable resistor connected as shown below provides another way of changing
the ratio R 1 /R 2 . The resistance between A and W represents R 1 and that between
W and B represents R 2 .
A variable voltage (eg from 0 to 6 V in the example shown below) is available
between A and W as the wiper (or slider) is rotated. For example, it will be 3 V
when the wiper is halfway round the track.
B
6V
W
W
A
A
Note that the above only holds true as long as the load connected across AW has
a very high resistance. If it has a resistance comparable with that of AW the
situation becomes more complicated. The effective resistance at the output must
then be calculated before the potential divider rule is applied.
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Worked example
20 kΩ
12 V
Y
10 kΩ
A
X
(a)
Calculate the voltage across XY.
R1
R2
VS
V1
=
=
=
=
10 kΩ
20 kΩ
12 V
?
R1
VS
R1  R2
10
V1 
12
10  20
V1 
V1 = 4 V
(b)
The circuit attached to XY has a resistance of 10 kΩ. Calculate the
effective resistance between XY and thus the voltage across XY.
The resistance between XY comes from the two identical 10 kΩ resistors in
parallel. The combined resistance is therefore 5 kΩ. Use this resistance as
R 1 in the potential divider formula.
R1
R2
VS
V1
=
=
=
=
5 kΩ
20 kΩ
12 V
?
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R1
VS
R1  R2
5
V1 
12
5  20
V1 
V 1 = 2.4 V
3.1c Electrical sources and internal resistance
Up to this stage in your study of physics we have assumed that power supplies
are ideal. This means that their voltage remains constant and they can supply any
current we wish if we connect the correct resistance. In most cases these are
good assumptions, but if you take a small battery and connect bulbs one after
another in parallel you will notice that eventually their brightness dims. A
similar effect can be noticed if you start a car with the headlights on. Both these
situations have one thing in common: they involve significant current being
supplied by the battery. If you are able to touch the battery you may notice it
getting warm when in use. This is a wasted transfer of energy and can be
modelled by considering that the battery, like all conductors, has some resistance
of its own. This model explains why some of the chemical energy converted is
dissipated as heat and is not available to the circuit – resistors convert electrical
energy to heat energy. We say that the power suppl y has an internal resistance, r.
In most cases this is so small (a few ohms) that it can be considered negligible.
However, when the current in the circuit is large, meaning the external resistance
is small, its effects can be significant. In the past we h ave assumed that any cell
could deliver an unlimited current, but clearly this is not the case. The greater
the current the more energy will be dissipated in the power supply until
eventually all the available energy (the emf) is wasted and none is availab le
outside the power supply.
Energy will be wasted in getting the charge through the supply (this energy
appears as heat) and so the energy per unit charge available at the output (the
terminal potential difference or tpd) will fall. There will be ‘lost v olts’. The lost
volts = Ir.
Note: A common confusion is that Ir stands for internal resistance. r is the
symbol for internal resistance and Ir is the value of the lost volts.
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The fact that a power supply has an internal resistance is not a very difficul t
problem to deal with because we can represent a real supply as:
real cell = ideal cell + a resistance r
Internal
resistance
Ideal cell
r
E
Real cell
Special case – open circuit (I = 0)
We now need to consider how we can measure the key quantities. Firstly, it
would be very useful to know the voltage of the ideal cell. This is called the emf
of the cell, which was defined previously as the energy (eg chemical energy)
supplied per unit of charge. It is found by finding the voltage across the cell on
‘open circuit’, ie when it is delivering no current. This can be done in practice by
using a voltmeter or oscilloscope.
Why should this be so? Look at this diagram:
Since I = 0
r
I=0
V across r = I × r = 0 × r
E
=0
So no voltage is dropped across the internal resistanc e and a voltmeter across the
real cell would register the voltage of the ideal cell, the emf.
Under load
When a real cell is delivering current in a circuit we say that it is under load and
the external resistance is referred to as the load.
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Consider the case of a cell with an internal resistance of 0.6 Ω delivering current
to an external resistance of 11.4 Ω:
Cell emf. = 6 V
r
0.6 Ω
E
current = 0.5 A
I = 0.5 A
V
11.4 Ω
The voltage measured across the terminals of the cell will be the voltage across
the 11.4 Ω resistor, ie,
V = IR = 0.5 × 11.4
= 5.7 V
and the voltage across the internal resistance:
V = Ir = 0.5 × 0.6
= 0.3 V
so the voltage across the terminals is only 5.7 V (this is the terminal potential
difference, or tpd) and this happens because of the vol tage dropped across the
internal resistance (0.3 V in this case). This is called the lost volts.
If Ohm’s law is applied to a circuit containing a battery of emf E and internal
resistance r with an external load resistance R:
r
I
E
R
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and if I is the current in the circuit and V is the pd across R (the tpd.), then using
conservation of energy:
emf = tpd + lost volts
E = V across R + V across r
and: E = V + Ir
This is the internal resistance equation, which can clearly b e rearranged into
different forms, for example:
V = E – Ir
or, since V = IR,
E = IR + Ir
ie E = I(R + r)
Special case – short circuit (R = 0)
The maximum current is the short-circuit current. This is the current that will
flow when the terminals of the supply are joined with a short piece of thick wire
(ie there is no external resistance).
By substituting R = 0 in the above equation we get:
E = I(0 + r)
ie E = Ir
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Measuring E and r by a graphical method
When we increase the current in a circuit the tpd will decrease. We can use a
graph to measure the emf and internal resistance. If we plot V on the y-axis and I
on the x-axis, we get a straight line of negative gradient. From above:
V = E – Ir
V = (–r) × I + E
Comparing with the equation of a straight line
y = mx + c
we can see that the gradient of the line ( m) is equivalent to –r and the y-intercept
(c) is E.
Voltage
Intercept (c)
= E (emf)
Gradient (m)
= –r (internal
resistance)
Current
This graph can also be modified to show that the external resistance ( R) and
internal resistance (r) act as a potential divider. As R decreases then lost volts
must increase.
Voltage
r (internal
resistance)
R (external
resistance)
Current
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We can then use the potential divider relationships:
emf = tpd + lost volts
lost volts r

tpd
R
lost volts 
tpd 
r
 emf
Rr
R
 emf
Rr
Second graphical method
When we change the external resistance (R) in a circuit, there will be a
corresponding change in the current flowing. If we plot R on the y-axis and 1/I
on the x-axis, we get a straight line of positive gradient and negative intercept.
From above:
E = IR + Ir
Divide both sides by I:
E/I = R + r
R = (E × 1/I) – r
Comparing with the equation of a straight line:
y = mx + c
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we can see that the gradient of the line ( m) is equivalent to E and the y-intercept
(c) is –r.
Voltage
Gradient (m)
= E (emf)
Intercept (c)
= –r (internal
resistance)
1 .
Current
Worked example
A cell of emf 1.5 V is connected in series with a 28 Ω resistor. A voltmeter
measures the voltage across the cell as 1.4 V.
r
E
V
28 Ω
Calculate:
(a)
(b)
(c)
the internal resistance of the cell
the current if the cell terminals are short circuited
the lost volts if the external resistance R is increased to 58 Ω.
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(a)
E = V + Ir
In this case we do not know the current, I, but we do know that the voltage
across the 28 Ω resistor is 1.4 V, and I = V/R , so I = 0.05 A.
1.5 = 1.4 + 0.05r
r
(b)
0.1
 2Ω
0.05
Short circuit:
r
Total resistance= 2 Ω
1.5 V
I
(c)
2Ω
I
E 1.5

 0.75 A
r
2
E = I(R + r)
1.5
= I(58 + 2)
I
= 1.5/60 = 0.025 A
lost volts = Ir
= 0.025 × 2 = 0.05 V
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3.1d Capacitors
Capacitance is the ability (or capacity) to store charge. A device that stores
charge is called a capacitor.
Practical capacitors are conductors separated by an insulator. The simplest type
consists of two metal plates with an air gap between them. The symbol for a
capacitor is based on this:
Relationship between charge and pd
The capacitor is charged to a chosen
voltage by setting the switch to A. The
charge stored can be measured directly by
discharging through the coulometer with
the switch set to B. In this way pairs of
readings of voltage and charge are
obtained.
It is found that the charge stored on a capacitor and the pd (voltage) across it are
directly proportional:
Q
0
V
so Q = a constant
V
This constant is defined as the capacitance, C:
C
26
Q
V
or
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Q = CV
The formal definition of capacitance is therefore the charge stored per unit
voltage. Notice that this means that capacitance is the gradient of the above
graph. The unit of capacitance is the farad (F).
From the above formula: 1 F = 1 coulomb per volt.
The farad is too large a unit for practical purposes and the following
submultiples are used:
1 mF (microfarad) = 1 × 10 –6 F
1 nF (nanofarad) = 1 × 10 –9 F
Note: When a capacitor is charging, the current is not constant (more on this
later). This means the formula Q = It cannot be used to work out the charge
stored.
Worked example
A capacitor stores 4 × 10 –4 C of charge when the potential difference across it is
100 V. Calculate the capacitance.
C
Q 4  10 4

 4  10 6 F (4 mF)
V
100
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Energy stored in a capacitor
Why work must be done to charge a capacitor
Consider this:
-
Vc
+
+
Electron
current
Electron
current
Vs
Vs
When current is switched
on electrons flow onto
one plate of the capacitor
and away from the other
plate.
This results in one plate
becoming negatively
charged and the other
plate positively charged.
Vc = V s
+
+
+
+
Vs
Eventually the current
ceases to flow. This
happens when the pd
across the plates of the
capacitor is equal to the
supply voltage.
The negatively charged plate will tend to repel the electrons approaching it. In
order to overcome this repulsion work has to be done and energy supplied. This
energy is supplied by the battery. Note that current does not flow through the
capacitor, electrons flow onto one plate and away from the other plate.
For a given capacitor the pd across the plates is directly proportional to the
charge stored. Consider a capacitor being charged to a pd of V and holding a
charge Q.
Charge
The energy stored in the capacitor is given by the area under
graph
the
Q
V p.d.
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Area under graph= 1 Q × V
2
Energy stored = 1 Q × V
2
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If the voltage across the
capacitor was constant work
done = Q × V, but since V is
varying, the work done =
area under graph.
This work is stored as electrical energy, so:
E = ½QV
(Contrast this with the work done moving a charge in an electric field where W =
QV. In a capacitor the amount of charge and voltage are constantly changing
rather than fixed and therefore the ½ is needed as an averaging factor.)
Since Q = CV there are alternative forms of this relationship:
energy = ½CV 2
and energy = ½
Q2
C
Worked example
A 40 mF capacitor is fully charged using a 50 V supply. Calcu late the energy
stored in the capacitor.
energy = ½CV 2
= ½ × 40 × 10 –6 × 50 2
= 0.05 J
Charging and discharging a capacitor
Charging
Consider the following circuit:
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When the switch is closed the current flowing in the circuit and the voltage
across the capacitor behave as shown in the graphs below.
pd across
capacitor
current
Supply voltage
0
0
time
time
Consider the circuit at three different times.
0
0
A
A
As soon as the switch is
closed there is no charge on
the capacitor. The current is
limited only by the
resistance in the circuit and
can be found using Ohm’s
law.
30
A
+ +
+ +
+ +
-
--
-
As the capacitor charges a
pd develops across the
plates which opposes the
pd of the cell. As a result
the supply current
decreases.
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0
--
The capacitor becomes
fully charged and the pd
across the plates is equal
and opposite to that across
the cell.and the charging
current becomes zero.
Consider this circuit when the capacitor is
fully charged, switch to position B
A
Discharging
-- -Consider the circuit opposite in which
the capacitor is fully charged:
Consider this circuit when the capacitor is
fully charged, switch to position B
A
If the cell is shorted out of the circuit
-- -the capacitor will discharge.
A
++ ++
A
B
A
++ ++
If the cell is taken out of the circuit and the
switch is set to A, the capacitor will
discharge
A
B
A
B
While the capacitor is discharging, the current in the circuit and the voltage
across the capacitor behave as shown in the graphs below:
pd across
capacitor
Current
0
Supply voltage
0
time
time
Although the current/time graph has the same shape as that during charging, the
currents in each case are flowing in opposite directions. The discharging current
decreases because the pd across the plates decreases as charge leaves them.
A capacitor stores charge, but unlike a cell it has no capability to supply more
energy. When it discharges, the energy stored will be used in the circuit, eg in
the above circuit it would be dissipated as heat in the resistor.
Factors affecting the rate of charge and dischar ge
The time taken for a capacitor to charge is controlled by the resistance of the
resistor R (because it controls the size of the current, ie the charge flow rate) and
the capacitance of the capacitor (since a larger capacitor will take longer to fill
and empty). As an analogy, consider charging a capacitor as being like filling a
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If the cell is taken
switch is set to A
discharge
31
jug with water. The size of the jug is like the capacitance and the resistor is like
the tap you use to control the rate of flow.
The values of R and C can be multiplied together to form what is known as the
time constant. Can you prove that R × C has units of time, seconds? The time
taken for the capacitor to charge or discharge
is related to
the time
constant.
Current
large
capacitor
small capacitor
Large capacitance and large resistance both increase the charge or discharge
time.
The I/t graphs for capacitors of different value during charging are shown
Time below:
Current
large capacitor
small capacitor
Time
The effect of capacitance on charge
Current
small resistor
current
Current
small resistor
large resistor
Time
The effect of resistance on charge
current
large resistor
Note that since the area under the I/t graph is equal to charge, for a given
capacitor the area under the graphs must be equal.
Time
32
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Worked example
The switch in the following circuit is closed at time t = 0.
VS = 10 V
1 MΩ
2 μF
(a)
Immediately after closing the switch what is
(i)
the charge on C?
(ii) the pd across C?
(iii) the pd across R?
(iv) the current through R?
(b)
When the capacitor is fully charged what is
(i)
the pd across the capacitor?
(ii) the charge stored?
(a)
(i)
The initial charge on the capacitor is zero.
(ii)
The initial pd across the capacitor is zero since there is no charge.
(iii) pd across the resistor is 10 V
(V R = V S – V C = 10 – 0 = 10 V)
(b)
V 10

 1 × 10 –5 A
R 10 6
(iv)
I
(i)
Final pd across the capacitor equals the supply voltage, 10 V.
(ii)
Q = VC = 2 × 10 –6 × 10 = 2 × 10 –5 C
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