# The physics of motion by Ka99e0d

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```									                                 North Berwick High School

Department of Physics

Higher Physics

Unit 2 Electricity
Section 1       Voltage, Current and
Resistance
Section 1                   Voltage, Current and Resistance

Note Making
Make a dictionary with the meanings of any new words.

Alternating current
1.    State the difference between a.c. and d.c. and the main way in which
a.c. is produced.
2.    Copy both graphs.

Measuring frequency and peak voltage
1.    Copy the worked example.

Alternating current – peak and rms
1.    Define the term rms and copy both formulae.

Graphical method
1.    Copy the 3 important notes.

Current, voltage, power and resistance
1.    Define the terms current and voltage.
2.    State the meanings of emf and pd (not just what the letters
stand for).

Sources of emf
1.    State some sources of emf.
Ohm’s law
1.   You do not need to define Ohm's law but you must be able to
use the equation.

Electric power
1.   State the 3 equations for power.
2.   State the significance of I 2 R.
3.   Copy the example if you think it will help.
4.   You should already know the circuit rules but copy the table if
you don't.
5.   Copy the addition of power sentence.

Voltage and current by proportion
1.   Copy the examples. Make sure that you understand them.

The potential divider
1.   State the function of a potential divider circuit.
2.   Copy the example and calculations of the potential divider
circuit. Note that you can also use the formulae on the next
page.
3.   Copy the diagram with the variable resistor and explain how it
works.
4.   You should also copy the worked example (a former exam
question).

Electrical sources and internal resistance
1.   Describe what is meant by internal resistance, lost volts and
tpd.
2.   Explain how the emf of a cell may be measured.
terms.)
4.   Show how the internal resistance equation is produced.
5.   Show how the short circuit equation is produced.
Measuring E and r by a graphical method
1.   Copy both graphs and questions 14 and 16 to explain both
methods. (pages 53 and 55)

Capacitors
1.   What is meant by capacitance.
2.   Copy the graph, equation and the formal definition of
capacitance.

Energy stored in a capacitor
1.   Copy the diagrams to show a capacitor charging.
2.   Explain why work has to be done when charging a capacitor.
3.   Copy the graph of Q v V and the energy equations. (Read the

Charging and discharging a capacitor
1.   Copy the charging and discharging graphs. You must be able to
clearly explain why they are this shape.

Factors affecting the rate of charge and
discharge
1.   Copy both graphs and again, explain their shapes.
Section 1                   Voltage, Current and Resistance
Contents
Content Statements ......................................................................... 1
Alternating current .......................................................................... 2
Measuring frequency and peak voltage ............................................ 2
Alternating current – peak and rms .................................................. 4
Graphical method to derive the relationship between peak and rms
values of alternating current ............................................................ 6
Current, voltage, power and resistance ............................................ 9
Sources of emf ............................................................................... 10
Ohm’s law ...................................................................................... 11
Electric power ................................................................................ 12
Circuit rules ................................................................................... 14
Voltage and current by proportion ................................................. 15
The potential divider ...................................................................... 19
Electrical sources and internal resistance ....................................... 23
Measuring E and r by a graphical method ....................................... 27
Second graphical method ............................................................... 28
Capacitors ...................................................................................... 31
Relationship between charge and pd .............................................. 31
Energy stored in a capacitor ........................................................... 33
Charging and discharging a capacitor ............................................. 35
Factors affecting the rate of charge and discharge ......................... 37
Problems ....................................................................................... 39
Solutions ........................................................................................ 68
Content Statements

Content                 Notes                               Context

a) Monitoring and       a.c. as a current which changes     Using a multimeter as an
measuring a.c.       direction and instantaneous         ammeter, voltmeter and
value with time. Monitoring a.c.    ohmmeter.
signals with an oscilloscope,       Oscilloscope as a voltmeter
including measuring frequency,      and waveform monitor.
and peak and r.m.s. values.

b) Current, potential   Current, potential difference and   Investigating simple a.c. or
difference (p.d.),   power in series and parallel        d.c. circuits with switches
power and            circuits. Calculations involving    and resistive components.
resistance           p.d., current and resistance may    Potential dividers in
involve several steps.              measurement circuits and
Potential dividers as voltage       used with variable resistors
controllers.                        to set and control voltages in
electronic circuits.
c) Electrical sources   Electromotive force, internal
and internal         resistance and terminal potential   Investigating the reduction
resistance.          difference. Ideal supplies, short   in voltage as additional
circuits and open circuits.         devices are connected.
Determining internal resistance     Investigating internal
and electromotive force using       resistance of low voltage
graphical analysis.                 power supplies.

d) Capacitors           Capacitors and the relationship
between capacitance, charge and
potential difference.               Energy storage. Flash
The total energy stored in a        photography.
charged capacitor is the area       Smoothing and suppressing.
under the charge against            Capacitance based touch
potential difference graph. Use     screens.
the relationships between
energy, charge, capacitance and
potential difference.
Variation of current and
potential difference against time
for both charging and
discharging. The effect of
resistance and capacitance on
charging and discharging curves.

1
Section 1                   Voltage, Current and
Resistance

Alternating current
Electric current is the rate of electron flow past a given point in an
electric circuit, measured in Coulombs/second which is named Amperes.

An alternating current (a.c.) periodically changes direction. Contrast this
to direct current (d.c.), which only flows in one direction. Although it is
possible that current could vary in any way, we will consider a sinusoidal
function, ie the voltage constantly changes in the form of a sine wave.
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

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screen. The voltage axis scale (y) is usually called the volts/div or
volts/cm and the time axis scale (x) 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.

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 ,        The volts/div was set at 2 V cm –1 ,

3
then the period of the wave is:             then the peak voltage is:

T = 4 × 5 ms = 4 × 0.005 = 0.02 s           V peak = ½ × 8 × 2

1
f 
T                                     V peak = 8 V
1
f 
0  02

f = 50 Hz

If the time base is switched off, the a.c. signal will not be spread along
the x-axis. However, the voltage variation will continue, meaning a
straight vertical line will be displayed on the screen. The peak voltage
can still be calculated 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

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couldn’t possibly follow such quick changes. We need an average, but
the average value of the voltage during any complete cycle is zero!

We know that homes, schools and industry use a.c. supplies, so there
must be an average that is not zero. 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 battery 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.

5
The variable resistors are altered until the lamps are 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        and    Irms = 1 Ipeak
2                             2

Note: A multimeter switched to a.c. mode will display rms values.

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
I peak
The average value of I 2 is              and therefore the average power
2
2
I peak
supplied is            R.
2

6
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
R  I rms R
2

2
2
I peak
 I rms
2

2

I peak
 I rms
2

Important notes:

1.        Readings on meters that measure a.c. are rms values, not peak
values.

2.        For power calculations involving a.c. always use rms values:

2
Vrms
P  I rmsVrms  I rms R 
2

R

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.

7
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 = √2 × 12

V peak = 17.0 V

8
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:

Q
I
t

The energy required to drive the electron current around 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.

Voltage is defined as the energy transferred (work done) 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).

9
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

Sources of emf
Emfs can be generated in a great variety of ways. See the table below
for examples.

Chemical energy drives the current
Chemical cell
(eg battery)

Heat energy drives the current
Thermocouple
(eg temperature sensor in an oven)

Mechanical vibrations drive the current
Piezo-electric generator
(eg acoustic guitar pickup)

Light energy drives the current
Solar cell
(eg solar panels on a house)
Changes in magnetic field drive the current
Electromagnetic generator
(eg power stations)

10
Ohm’s law
In any circuit, providing the resistance of a component 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
Voltage                                    given conductor:

V  I, ie V/I is constant

The constant of proportion is
defined to be the resistance,

V
Current       ie R      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 example a light bulb, a tran sistor or a diode. A
graph of V versus I for such components will not be a straight line.

11
Electric power
For a given component P = IV since:

power           =      work done   =    QV     =      Q xV         =IV
time             t            t

Since V = IR

P = I × IR (substituting for V)

= I2R

or

P =        V xV
R

P =        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.

12
Example
An electric heater has two heating elements allowing three settings:
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 =      V2           = 230 2    1800 W → low power
R              30

for 15 Ω: =           P =     V2           = 230 2    3500 W → medium power
R              15

for both:

1      = 1 + 1 = 1
Rp       15  30  10

R p = 10 Ω

power

P =     V2            = 230 2   5300 W → high power
R               10

13
Circuit rules

Series                 Parallel

Current        IS  I1  I 2  I 3   I P  I1  I 2  I 3

Voltage       VS  V1  V2  V3      V s = V B1 = V B2 = V B3
1  1  1   1
Resistance     RS  R1  R2  R3           
RP R1 R2 R3

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
R
Two resistors of the same value (R) wired in parallel:
≡ ½R
R P = half the value of one of them (½R).
R
eg     2 × 1000 Ω in parallel, R P = 500 Ω

2 × 250 Ω in parallel, R P = 125 Ω

14
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 inversely proportional to the resistance, ie proportional to
the other resistance.

Examples
Current in parallel                   Voltage in series

2Ω                                          12 V
3A

4Ω                                  8Ω       4Ω     12 Ω

3A will split in the ratio:           R S = 8 + 4 + 12 = 24 Ω

2    4      1 2                              8
:     , ie :                       V1        12  4 V
4 2 4 2     3 3                             24

1                                              4
 3A=1A                              V2        12  2 V
3                                             24

2                                             12
3 A = 2 A                           V3        12  6 V
3                                             24

The smallest resistor takes
the largest current, so:

I 1 = 2 A, I 2 = 1 A

15
Worked example 1

40 V
In the circuit shown opposite calculate:

6Ω          4Ω
(a)    the current in each resistor
5Ω
(b)    the pd across each resistor                            10 Ω

(c)    the power dissipated in each resistor.

Solution
(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
=     +  =
p       10 10 5

RP = 5 Ω

R T = 5 + 5 = 10 Ω

(current through 5 Ω)

I 1 meets a resistance of 10 Ω and so does I 2 .

so I 1 = I 2 = 2 A                (current through 4 Ω, 6 Ω and 10 Ω)

(b)    For the 5 Ω resistor:         V = IR        =4×5               = 20 V

for the 4 Ω resistor:         V = IR        =2×4               =8V

for the 6 Ω resistor:         V = IR        =2×6               = 12 V

for the 10 Ω resistor:        V = IR        = 2 × 10           = 20 V

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(c)    For 5 Ω:       P = IV        = 20 × 4       = 80 W

for 4 Ω:       P = I2R       = 22 × 4       = 16 W
2            2
for 6 Ω:       P= V          = 12           = 24 W
R            6
for 10 Ω:      P = IV        = 20 × 2       = 40 W

(note: you can use any of the electrical power formulae that are appropriate.)

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                   12 V
(rheostat).

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)    the resistance of the controller at this setting

(b)    the power wasted in the controller.

Solution
(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, R = ?

V2          V 2 122
P      , so R     
R           P    10

R = 14 Ω

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P     4
For the motor at 4 W,      P = I 2 R,          so      I=      
R   14  4

= 0.53 A

For the whole circuit:     RT       = VT            = 12       = 23 Ω
I              0.527

Resistance of controller            = 22.8 – 14.4              = 8.4 Ω

(b)   Power in controller        = I2R        = 0.527 2 × 8.4        = 2.3 W

18
The potential divider
A potential divider provides a convenient way of obtaining a variable
voltage from a fixed voltage supply.

Example
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).

VS   6
so,            I      
R 30

= 0.2 A

The voltage across R 1        V1     =    IR 1    = 0.2 × 10

=2V

The voltage across R 2        V2     =    IR 2    = 0.2 × 20

=4V

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If you look closely you can see that the voltages across the resistors are
in the ratio of their resistances (½ in this case),

V1 R1
ie,     
V2 R2

The voltage across R 1 is given by:

R1
V1             VS
R1  R 2

similarly, the voltage across R 2 is

R2
V1            VS
R1  R2

This is known as the potential divider rule.

A variable resistor connected as shown on the next page 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 .

20
A variable voltage (eg from 0 to 6 V) 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.

21
Worked example

20 kΩ

12 V                                   Y
10 kΩ
A
X

(a)    Calculate the voltage across XY.

R 1 = 10 kΩ           R 2 = 20 kΩ   V S = 12 V   V1 = ?

R1
V1             VS
R1  R 2

10
V1            12
10  20

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.

R 1 = 5 kΩ             R 2 = 20 kΩ   V S = 12 V   V1 = ?

R1
V1             VS
R1  R 2

5
V1           12
5  20

V 1 = 2.4 V

22
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 supply 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 have 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 available 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 volts’. 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.

23
The fact that a power supply has an internal resistance is not a very
difficult problem to deal with because we can represent a real supply as:

real cell = ideal cell + a resistance r

Ideal cell                                      Internal
resistance
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:

I=0                 Since I = 0
r
E                                         V across r = I × r = 0 × r

= 0

So no voltage is dropped across the internal resistance and a voltmeter
across the real cell would register the voltage of the ideal cell, the emf.

24
When a real cell is delivering current in a circuit we say that it is under

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
E            0.6 Ω
V
I = 0.5 A
current = 0.5 A
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 voltage
dropped across the internal resistance (0.3 V in this case). This is called
the lost volts.

25
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
E                        I

R

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 be 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

26
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
(internal resistance)

Intercept (c) =
E (emf)

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.

27
Voltage
Ideal cell
r (internal
resistance)

R (external
R (external
resistance)
resistance)

Current

We can then use the potential divider relationships:

emf = tpd + lost volts

lost volts r

tpd      R

r
lost volts          emf
R r

R
tpd          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

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Comparing with the equation of a straight line:

y = mx + c

we can see that the gradient of the line (m) is equivalent to E and the y-
intercept (c) is –r.

Resistance
= E (emf)

Intercept (c)
= –r (internal
1
resistance)
I

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)   the internal resistance of the cell

(b)   the current if the cell terminals are short circuited

(c)   the lost volts if the external resistance R is increased to 58 Ω.

29
Solution
(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/ , so I = 0.05 A.

1.5 = 1.4 + 0.05r                 or use

0.1                         V1 = R1
r         2Ω
0.05                         V2 R2

(b)   Short circuit:
r
1.5 V         2Ω         I
Total resistance = 2 Ω

E 1.5
I         0.75 A
r   2

(c)                       E = I(R + r)

1.5= I(58 + 2)

= 1.5/60 = 0.025 A

lost volts = Ir

= 0.025 × 2 = 0.05 V

30
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

so Q = a constant
V
0                    V

31
This constant is defined as the capacitance, C:

Q
C            or          Q = CV
V

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.

Q 4  10 4
C            4  10 6 F (4 μF)
V   100

32
Energy stored in a capacitor
Why work must be done to charge a capacitor

Consider this:

Vc                      Vc = Vs
-        +                   -    +
-    +
-        +                   -    +
-    +
Electro                      Electro
curren
n                            curren
n
t                            t

Vs                           Vs                            Vs

When current is             This results in one plate    Eventually the current
switched on electrons       becoming negatively          ceases to flow. This
flow onto one plate of      charged and the other        happens when the pd
the capacitor and away      plate positively charged.    across the plates of the
from the other plate.                                    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.

33
charge

Q

V      p.d.

The energy stored in the capacitor is given by the area under the 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

Q2
and energy = ½
C

Worked example
A 40 mF capacitor is fully charged using a 50 V supply. Calculate the
energy stored in the capacitor.

energy = ½CV 2

= ½ × 40 × 10 –6 × 50 2

= 0.05 J

34
Charging and discharging a capacitor
Charging
Consider the following circuit:

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
current                               capacitor

Supply voltage

0                                         0
time                                       time

Consider the circuit at three different times.
0                            0                              0

A                            A                              A            + +
+     +                      + +
+
+                            +
-     -                      -
-   -
-

As soon as the switch is
As soon as the switch           As the capacitor ges
As the capacitor               The capacitor
closed there is no             pd develops across a
char
The capacitor the
fully charged andbecomes
becomes
is capacitor. is no charge
closed, there The
the                              charges, a opposes
plates which pd                  fully charged is
across the platesand the
chargecharge on The            the                            pd
limited only by
on the capacitor.             pd of the cell. As a the
develops across
the                            equalacross to plates is
and oppositethethat
pd
current is in the circuit
resistanceis limited only by   the supply                     the cell.and opposite
the
current                       result which
plates                       across and the hargin to
equal
can
and be found using the
the resistance in
decreases
current
opposes the pd of
current becomesg
c that across the cell and
Ohm’s                          .                              zero.
circuit and can be found       the cell. As a result          the charging current
using Ohm's law.               the supply current             becomes zero.
decreases .

35
Consider this circuit when the capacitor is        If the cell is t
fully charged, switch to position B                switch is set
discharge
Discharging
A                                                  A
-- --
A         B
Consider the circuit opposite in which                  ++ ++
the capacitor is fully charged:

Consider this circuit when the capacitor is        If the cell is taken out of the circuit and the
fully charged, switch to position B                switch is set to A, the capacitor will
discharge
A                                                 A
- - is
If the cell - - shorted out of the circuitB                                            A            B
A
the capacitor will discharge.
++ ++

While the capacitor is discharging, the current in the circuit and the
voltage across the capacitor behave as shown in the graphs below:

Current                                         pd across
capacitor           Supply voltage

0                                                  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.

36
Factors affecting the rate of charge and
discharge
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 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.

large charge or
Large capacitance and large resistance both increase the capacitor
Current
discharge time.
small capacitor
The I/t graphs for capacitors of different value during charging are
shown below:
Time

Current        large capacitor         Current        small resistor
small capacitor                        large resistor

Time                                  Time

Current         small resistor
The effect of capacitance on                  The effect of resistance on
charge current large resistor                 current

Note that since the area under the I/t graph is equal to charge, for a
Time
given capacitor the area under the graphs must be equal.

37
Worked example
The switch in the following circuit is closed at time t = 0.

V S = 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?

Solution
(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)

V   10
(iv)   I         6
 1 × 10 –5 A
R 1x10

(b)   (i)    Final pd across the capacitor equals the supply voltage, 10 V.

(ii)   Q = VC = 2 × 10 –6 × 10 = 2 × 10 –5 C

38
Problems

Monitoring and measuring a.c.
1.   (a)   What is the peak voltage of the 230 V mains supply?

(b)   The frequency of the mains supply is 50 Hz. How many times
does the voltage fall to zero in 1 second?

2.   The circuit below is used to compare a.c. and d.c. supplies.

A B

The variable resistor is used to adjust the brightness of the lamp
until the lamp has the same brightness when connected to either
supply.

(a)   Explain why the brightness of the lamp changes when the
setting on the variable resistor is altered.

(b)   What additional apparatus would you use to ensure the
brightness of the lamp is the same when connected to either
supply?

(c)   The time-base of the oscilloscope is switched off. Diagram 1
shows the oscilloscope trace obtained when the switch is in
position B. Diagram 2 shows the oscilloscope trace obtained
when the switch is in position A.

39
Y gain set to 1 V cm –1

Using information from the oscilloscope traces, find the
relationship between the root mean square (r.m.s.) voltage
and the peak voltage of a voltage supply.

(d)   The time-base of the oscilloscope is now switched on. Redraw
diagrams 1 and 2 to show what happens to the traces.

3.   The root mean square voltage produced by a low voltage power
supply is 10 V.

(a)   Calculate the peak voltage of the supply.

(b)   An oscilloscope, with its time-base switched off, is connected
across the supply. The Y-gain of the oscilloscope is set to
5 V cm –1 . Describe the trace seen on the oscilloscope screen.

4.   (a)   A transformer has a peak output voltage of 12 V.

Calculate the r.m.s. value of this voltage.

(b)   An oscilloscope, with the time base switched off, is connected
across another a.c. supply. The Y gain of the oscilloscope is
set to 20 V cm –1 . A vertical line 6 cm high appears on the
oscilloscope screen.

Calculate:

(i)    the peak voltage of the input
(ii)   the r.m.s. voltage of the input.

40
5.   An oscilloscope is connected across a signal generator. Th e time-
base switch is set at 2·5 ms cm –1 .

The diagram shows the trace on the oscilloscope screen.

(a)   (i)    What is the frequency of the output from the signal
generator?

(ii)   What is the uncertainty in the frequency to the nearest
Hz?

(b)   The time base switch is now changed to:

(i)    5 ms cm –1

(ii)   1·25 ms cm –1 .

Sketch the new traces seen on the screen.

6.   An a.c. signal of frequency 20 Hz is connected to an oscilloscope.
The time-base switch on the oscilloscope is set at 0.01 s cm –1 .

Calculate the distance between the neighbouring peaks of this
waveform when viewed on the screen.

41
Current, voltage, power and resistance

1.    There is a current of 40 mA in a lamp for 16 s.
Calculate the quantity of charge that passes any point in the circuit
in this time.

2.    A flash of lightning lasts for 1 ms. The charge transferred between
the cloud and the ground in this time is 5 C.
Calculate the value of the average current in this flash of lightning.

3.    The current in a circuit is 2.5 × 10 –2 A. How long does it take for
500 C of charge to pass any given point in the circuit?

4.    There is a current of 3 mA in a 2 kΩ resistor. Calculate the p.d.
across the resistor.

5.    Calculate the values of the readings on the meters in the following
circuits.
+           –                               +           –
A                          12 V          A
10 V

70          (a)           20              15 
(b)
10                                         9

V
+               –
A

6V             3                  (c)
5

V

42
6.        Calculate the unknown values R of the resistors in the following
circuits.
2A
+           –                           +              –
A                                       A
40 V                                    40 V

5         (a)                          4          (b)         R=?
R=?

10                                         10 

V
20 V

7.        Calculate the total resistance between X and Y for the following
combinations of resistors.

20                                    10 
X                                         X

(a)            10           10          (b)         10           10 

Y                                         Y
10 

10 
X                                                      5
X
4         20 
(d)            20            25 
(c)
1         5
Y
10 
10 

43
10                                            5
X                                               X

(e)          5           10        25            (f)    6           12    3

10                                            8
Y                                                   Y

8.     In the following circuit the reading on the ammeter is 2 mA.
Calculate the reading on the voltmeter.

+       –
A

3 k

5 k

V

9.     Calculate the power in each of the following situations.

(a)   A 12 V battery is connected to a motor. There is a current of
5 A in the motor.

(b)   A heater of resistance 60 Ω connected across a 140 V supply.

(c)   A current of 5 A in a heater coil of resistance 20 Ω.

44
10.   The heating element in an electric kettle has a resistance of 30 Ω.

(a)   What is the current in the heating element when it is
connected to a 230 V supply?

(b)   Calculate the power rating of the element in the kettle.

11.   A 15 V supply produces a current of 2 A in a lamp for 5 minutes.
Calculate the energy supplied in this time.

12.   Calculate the readings on the ammeter and the voltmeter in the
circuit shown below.

+        –
6V           A

V         6                  6           6

13.   Each of the four cells in the circuit shown is identical.

A

2                5             20 

6V

Calculate

(a)   the reading on the ammeter

(b)   the current in the 20 Ω resistor

(c)   the voltage across the 2 Ω resistor.

45
14.   A voltage of 12 V is applied across a resistor. The current in the
resistor is 50 mA. Calculate the resistance of the resistor.

15.   The LED in the circuit below is to emit light.

A

R

B

(a)   What is the required polarity of A and B when connected to a
5 V supply so that the LED emits light?

(b)   What is the purpose of the resistor R in the circuit?

(c)   The LED rating is 20 mA at 1·5 V. Calculate the valu e of
resistor R.

16.   Write down the rules which connect the (a) potential differences
and (b) the currents in series and parallel circuits.

17.   What is the name given to the circuit shown?
Write down the relationship between V 1 , V 2 , R 1 and R 2 .

R1                 V1
10V
R2                 V2

46
18.   Calculate the values of V 1 and V 2 of the circuit in question 17
when:

(a)     R 1 = 1 kΩ               R 2 = 49 kΩ

(b)     R 1 = 5 kΩ               R 2 = 15 kΩ

19.   The light dependent resistor in the
circuit is in darkness.                                                R1                  V1
Light is now shone on the LDR.      10V
Explain what happens to the
readings on V 1 and V 2 .                                                                  V2

20.   Calculate the p.d. across resistor R 2 in each of the following
circuits.

(a)                       +5 V        (b)                     +5 V          (c)                     +5 V

R1                                                                          t
2 k                                 4 k                                500 

R2                                   R2                                    R2
8 k                                       1 k                              750 
0V                                     0V                                 0V

21.   Calculate the p.d. across AB (voltmeter reading) in each of the
following circuits.
+12 V                                +5 V                              +10 V

3 k           9 k                  5 k          10 k                                     6 k
3 k
A         V          B             A          V        B            A               V   B

3 k           3 k                  2 k          8 k                2 k                  4 k

0V                                   0V                                0V
(a)                                (b)                               (c)

47
22.   A circuit consisting of two potential dividers is set up as shown.

0V
+9 V
6 k   A 9 k

V

3 k B     6 k

(a)   Calculate the reading on the voltmeter.

(b)   (i)     Suggest a value of a resistor to replace the 9 kΩ resistor
that would give a reading of 0 V on the voltmeter.

(ii)    Suggest a value of resistor to replace the 3 kΩ resistor
that would give a reading of 0 V on the voltmeter.

23.   In the circuits shown the reading on the voltmeters is zero.
Calculate the value of the unknown resistors X and Y in each of the
circuits.

12 V                          12 V

120                         4 k          15 k
X

V                             V

120               9       12 k               Y

48
Electrical sources and internal resistance

1.   State what is meant by:

(a)   the e.m.f. of a cell
(b)   the p.d. between two points in a circuit.

2.   A circuit is set up as shown.

3                 X
A1          A2

12 V         12                 4

Y

(a)   Calculate the total resistance of the circuit.
(b)   Calculate the readings on the ammeters.
(c)   What is the value of the p.d. between X and Y?
(d)   Calculate the power supplied by the battery.

3.    The circuit shown uses a 230 V alternating mains supply.

12                S

230 V          8              24 

Calculate the current in each resistor when:

(a)      switch S is open
(b)      switch S is closed.

49
4.   An electric cooker has two settings, high and low. This involves two
heating elements, R 1 and R 2 . On the low setting the current from
the supply is 1 A. On the high setting the current from the supply is
3 A.

230 V                     R1                     R2

(a)   Calculate the resistance of R 1 and R 2 .

(b)   What is the power consumption at each setting?

5.   A lamp is rated at 12 V, 36 W. It is connected in a circuit as shown.

(a)   Calculate the value of the resistor R that allows the lamp to
operate at its normal rating.

(b)   Calculate the power dissipated in the resistor.

50
6.   In the circuit shown, r represents the internal resistance of the cell
and R represents the external resistance (or load resistance) of the
circuit.

r

V

A
S      R

When S is open, the reading on the voltmeter is 2 0 V.

When S is closed, the reading on the voltmeter is 1 6 V and the
reading on the ammeter is 0 8 A.

(a)   What is the value of the emf of the cell?
(b)   When S is closed what is the terminal potential difference
across the cell?
(c)   Calculate the values of r and R.
(d)   The resistance R is now halved in value. Calculate the new
readings on the ammeter and voltmeter when S is closed.

7.   The battery in the circuit shown has an emf of 5 0 V.

r

V

A

The current in the lamp is 0.20 A and the reading on the voltmeter
is 3.0 V. Calculate the internal resistance of the battery.

51
8.    A battery of emf of 4.0 V is connected to a load resistor with a
resistance of 15 Ω. There is a current of 0.2 A in the load resistor.
Calculate the internal resistance of the battery.

9.    A signal generator has an emf of 8.0 V and an internal resistance of
4. 0 . A load resistor is connected across the terminals of the
generator. The current in the load resistor is 0.50 A. Calculate the

10.   A cell is connected in a circuit as shown below.

(a)   Calculate the terminal p.d. across the cell.

(b)   The resistance of the variable resistor R is now increased.

(i)    Describe and explain what happens to the current in the
circuit.

(ii)   Describe and explain what happens to the p.d. across
the terminals of the cell.

11.   A cell has an emf of 1·5 V and an internal resistance of 2·0 Ω.
A 3·0 Ω resistor is connected across the terminals of the cell.
Calculate the current in the circuit.

52
12.   A student is given a voltmeter and a torch battery. When the
voltmeter is connected across the terminals of the battery the
reading on the voltmeter is 4·5 V.
When the battery is connected across a 6·0 Ω resistor the reading
on the voltmeter decreases to 3·0 V.

(a)   Calculate the internal resistance of the battery.

(b)   What value of resistor when connected across the battery
reduces the reading on the voltmeter to 2·5 V?

13.   In the circuit shown, the battery has an emf of 6·0 V and an
internal resistance of 1·0 Ω.

6V
r

1

V

A
S        R

When the switch is closed, the reading on the ammeter is 2·0 A.

What is the corresponding reading on the voltmeter?

14.    To find the internal resistance of a cell a load resistor is connected
across the terminals of the cell. A voltmeter is used to measure
V tpd , the voltage measured across the terminals of the cell. An
ammeter is used to measure I, the current in the variable resistor.
The table below shows the results obtained as the resistance of
the variable resistor is changed.

V tpd (V)        1·02   0·94     0·85    0·78    0·69   0·60
I (A)            0·02   0·04     0·06    0·08    0·10   0·12

53
(a)   Draw a diagram of the circuit used to produce these results.

(b)   Plot a graph of the results and from it determine:

(i)     the emf of the cell

(ii)    the internal resistance of the cell

(iii) the short circuit current of the cell.

15.   A variable resistor is connected across a power supply. A voltmeter
is used to measure V tpd , the voltage measured across the terminals
of the supply. An ammeter is used to measure I, the current in the
variable resistor. The table below shows the results o btained as
the resistance of the variable resistor is changed.

V t.p.d. (V)      5·5     5·6    5·7    5·8    5·9
I (A)             5·0     4·0    3·0    2·0    1·0

Plot a graph of V tpd against I.

(a)   What is the value of the open circuit p.d.?

(b)   Calculate the internal resistance of the power supply.

(c)   Calculate the short circuit current of the power supply.

(d)   The variable resistor is now removed from the circuit and a
lamp of resistance 1·5  is connected across the terminals of
the supply.

Calculate:

(i) the terminal p.d.

(ii) the power delivered to the lamp.

54
16.     A circuit is set up as shown to investigate the properties of a
battery.

E
r

V

A
S         R

The variable resistor provides known values of resistance R.
For each value of resistance R, the switch is closed and the current
I noted.
The table shows the results obtained.

(Ω)         0        2        4              6              8      10     12

I (A)         6·80     3·78     2·62           2·00           1·62   1·36   1·17

1/I (A –1 )

(a)   Show that the relationship E = I(R + r) can be put in the form:

E
R        r
I

(b)   Complete the third row in the table.
(c)   Use the values of R and 1/I to plot a graph.
(d)   Use the information in the graph to find:
(i)    the internal resistance of the battery
(ii)   the e.m.f. of the battery.
(e)   The battery is now short circuited. Calculate the current in
the battery when this happens.

55
17.     A student uses the following circuit to investigate the condit ions
for transferring the maximum power into a load resistor.

12 V
r

4

V

A
S       R

For each setting of the variable resistor the current in the circuit is
recorded.

The table below shows the results obtained.

R (Ω)           1         2          3             4       5          6

I (A)           2·40      2·00       1·71          1·50    1·33       1·20

Power in R
(W)

(a)   Complete the table by calculating the power in the load for
each value of R.
(b)   Sketch a graph to show how the power in the load resistor R
varies with R.
(c)   In order to achieve maximum transfer of power, what is the
relationship between the internal resistance of the power
source and the resistance of the load resistor?

18.     An automotive electrician needed to accurately measure the
resistance of a resistor.

She set up a circuit using an analogue milliammeter and a digital
voltmeter.

56
The two meter readings were as shown in the diagram below.

1.7
1.8
1
0                     2

mA
OO1.3 V

(a)   What are the readings on the ammeter and the voltmeter?
(b)   What is the nominal resistance calculated from these
(c)   What is the smallest division on the milliammeter?
(d)   What is the absolute uncertainty on the milliammeter?
(e)   What is the absolute uncertainty on the voltmeter?
(f)   What is the percentage uncertainty on the milliammeter
(g)   What is the percentage uncertainty on the voltmeter
(h)   Which is the greatest percentage uncertainty?
(i)   What is the percentage uncertainty in the resistance?
(j)   What is the absolute uncertainty in the resistance?
(k)   Express the final result as (resistance ± uncertainty) Ω
(l)   Round both the result and the uncertainty to the relevant
number of significant figures or decimal places.

57
Capacitors

1.   A 50 µF capacitor is charged until the p.d. across it is 100 V.

(a)   Calculate the charge on the capacitor when the p.d. across it
is 100 V.

(b)   (i)    The capacitor is now ‘fully’ discharged in a time of 4·0
milliseconds.

Calculate the average current during this time.

(ii)   Why is this average current?

2.   A capacitor stores a charge of 3·0 × 10 –4 C when the p.d. across its
terminals is 600 V.

What is the capacitance of the capacitor?

3.   A 30 µF capacitor stores a charge of 12 × 10 –4 C.

(a)   What is the p.d. across its terminals?

(b)   The tolerance of the capacitor is ± 0·5 µF. Express this
uncertainty as a percentage.

4.   A 15 µF capacitor is charged using a 1·5 V battery.

Calculate the charge stored on the capacitor when it is fully
charged.

5.   (a)   A capacitor stores a charge of 1·2 × 10 –5 C when there is a
p.d. of 12 V across it. Calculate the capacitance of the
capacitor.

58
(b)   A 0·10 µF capacitor is connected to an 8·0 V d.c. supply.
Calculate the charge stored on the capacitor when it is fully
charged.

6.   A circuit is set up as shown.

The capacitor is initially uncharged. The switch is now closed.

The capacitor is charged with a constant charging current of

2·0 × 10 –5 A for 30 s.

At the end of this time the p.d. across the capacitor is 12 V.

(a)   What has to be done to the value of the variable resistor in
order to keep the current constant for 20 s?

(b)   Calculate the capacitance of the capacitor.

7.   A 100 µF capacitor is charged using a 20 V supply.

(a)   How much charge is stored on the capacitor when it is fully
charged?

59
(b)   Calculate the energy is stored in the capacitor when it is fully
charged.

8.   A 30 µF capacitor stores 6·0 × 10 –3 C of charge. How much energy
is stored in the capacitor?

9.   The circuit below is used to investigate the charging of a capacitor.

A
10 k
12 V

2000 µF

The battery has negligible internal resistance.

The capacitor is initially uncharged. The switch is now closed.

(a)   Describe what happens to the reading on the ammeter from
the instant the switch is closed.

(b)   How can you tell when the capacitor is fully charged?

(c)   What would be a suitable range for the ammeter?

(d)   The 10 k Ω resistor is now replaced by a larger resistor and
the investigation repeated.

What is the maximum voltage across the capacitor now?

60
10.   In the circuit below the neon lamp flashes at regular intervals.

R

120 V dc                     C

The neon lamp requires a potential difference of 100 V across it
before it conducts and flashes. It continues to glow until the
potential difference across it drops to 80 V. While lit, its resistance
is very small compared with the resistance of R.

(a)   Explain why the neon bulb flashes.

(b)   Suggest two methods of decreasing the flash rate.

11.   In the circuit below the capacitor C is initially uncharged.

A
S
+
9V                       C                V
–

Switch S is now closed. By carefully adjusting the variable resistor R
a constant charging current of 1·0 mA is maintained.

The reading on the voltmeter is recorded every 10 seconds. The
results are shown in the table below.

61
Time (s) 0       10    20    30       40

V (V)   0        1·9   4·0   6·2      8·1

(a)     Plot a graph of the charge on the capacitor against the p.d.
across the capacitor.
(b)     Use the graph to calculate the capacitance of the capacitor.

12.   The circuit below is used to charge and discharge a capacitor.

A                    B                VR
1      2
100 V
VC

The battery has negligible internal resistance.

The capacitor is initially uncharged.

V R is the p.d. across the resistor and V C is the p.d. across the
capacitor.

(a)     What is the position of the switch:
(i)      to charge the capacitor
(ii)     to discharge the capacitor?

(b)     Sketch graphs of V R against time for the capacitor charging
and discharging. Show numerical values for the maximum and
minimum values of V R .

(c)     Sketch graphs of V C against time for the capacitor charging
and discharging. Show numerical values for the maximum and
minimum values of V C .

(d)     (i)   When the capacitor is charging what is the direction of
the electrons between points A and B in the wire?

62
(ii)   When the capacitor is discharging what is the direction
of the electrons between points A and B in the wire?

(e)   The capacitor has a capacitance of 4·0 µF. The resistor has
resistance of 2·5 MΩ.

Calculate:

(i)    the maximum value of the charging current

(ii)   the charge stored by the capacitor when the capacitor is
fully charged.

13.   A capacitor is connected in a circuit as shown.

S        3 M

+                                   3 F
3V
–

The power supply has negligible internal resistance. The capacitor
is initially uncharged.

V R is the p.d. across the resistor and V C is the p.d. across the
capacitor.

The switch S is now closed.

(a)   Sketch graphs of:

(i)    V C against time during charging. Show numerical values
for the maximum and minimum values of V C .

(ii)   V R against time during charging. Show numerical values
for the maximum and minimum values of V R .

63
(b)   (i)    What is the p.d. across the capacitor when it is fully
charged?
(ii)   Calculate the charge stored by the capacitor when it is
fully charged.

(c) Calculate the maximum energy stored by the capacitor.

14.   A capacitor is connected in a circuit as shown.

S      6 k

+                                   20 F
12 V
–

The power supply has negligible internal resistance.

The capacitor is initially uncharged. The switch S is now closed.

(a)   Calculate the value of the initial current in the circuit.

(b)   At a certain instant in time during charging the p.d. across the
capacitor is 3 V. Calculate the current in the resistor at this
time.

(c)   Calculate the current in the circuit when the charge on the
capacitor is 80 µC.

64
15.               The circuit shown is used to investigate the charge and discharge
of a capacitor.

A
VR
2         1                                                1 k

12 V
VC
10 mF

The switch is in position 1 and the capacitor is uncharged.

The switch is now moved to position 2 and the capacitor charges.

The graphs show how V C , the p.d. across the capacitor, V R , the p.d.
across the resistor, and I, the current in the circuit, vary with time.

VC versus time                                                     VR versus time

14                                                                14
12                                                                12
10                                                                10
VC / V

8
VR / V

8
6                                                                 6
4
2                                                                 4
0                                                                 2
0              20                40            60             0
0            20              40    60
time / s
time / s

Current versus time

14
12
10
I / mA

8
6
4
2
0
0       20              40        60                                                        65
time / s
(a)          The experiment is repeated with the resistance changed to
2 kΩ.

Sketch the graphs above and on each graph sketch the new
lines which show how V C , V R and I vary with time.

(b)   The experiment is repeated with the resistance again at 1 kΩ
but the capacitor replaced with one of capacitance 20 mF.
Sketch the original graphs again and on each graph sketch the
new lines which show how V C , V R and I vary with time.

(c)   (i)      What does the area under the current against time
graph represent?

(ii)     Compare the areas under the current versus time graphs
in the original graphs and in your answers to (a) and (b).
Give reasons for any differences in these areas.

(d)   At any instant in time during the charging what should be the
value of (V C + V R ) ?

(e)   The original values of resistance and capacitance are now
used again and the capacitor fully charged. The switch is now
moved to position 1 and the capacitor discharges.

Sketch graphs of V C , V R and I from the instant the switch is
moved until the capacitor is fully discharged.

16.   A student uses the circuit shown to investigate the charging of a
capacitor.

S      1 k

+                                  10 F
12 V
–

The capacitor is initially uncharged.

66
The student makes the following statements:

(a)   When switch S is closed the initial current in the circuit does
not depend on the internal resistance of the power supply.

(b)   When the capacitor has been fully charged the p.d. across the
capacitor does not depend on the internal resistance of the
power supply.

Use your knowledge of capacitors to comment on the truth or
otherwise of these two statements.

67
Solutions
Monitoring and measuring a.c.

1.   (a)      325 V
(b)      100 times

2    (c)      V r.m.s . = 0.71 V peak

3.   (a)      14 V

4.   (a)      8.5 V
(b)      (i)      60 V
(ii)     42 V

5.   (a)      (i)      100 Hz
(ii)     ±2 Hz

6.   5 cm

Current, voltage, power and resistance

1.   0.64 C

2.   5 × 10 3 A

3.   2·0 × 10 4 s

4.   6V

68
5.    (a)      I = 0·1 A
(b)      I = 0·5 A, V = 4·5 V
(c)      I = 2 A, V = 10 V

6.    (a)      5
(b)      6

7.    (a)      25 
(b)      25 
(c)      24·2 
(d)      13·3 
(e)      22·9 
(f)      14·7 .

8.    3·75 V

9.    (a)      60 W
(b)      327 W
(c)      500 W

10.   (a)      7·7 A
(b)      1763 W

11.   9000 J

12.   I = 0·67 A, V = 4 V

13.   (a)      0·67 A
(b)      0·13 A
(c)      1·34 V

14.   240 

69
15.   (a)    A = +, B = -

(b)    To reduce the current through/in the resistor to the correct
value (and also to reduce the voltage across the resistor to the
correct operating value.)

(c)    175 

16.   See notes.

17.   R1/R2 = V1/V2

18.   (a)    V 1 = 0·2 V, V 2 = 9·8 V
(b)    V 1 = 2·5 V, V 2 = 7·5 V

19.   V 2 decreases (as R_LDR decreases as light increases). V 1 increases
(as V 1 +V 2 =V supply )

20.   (a)    4V
(b)    1V
(c)    3V

21.   (a)    3V

(b)    - 0·8 V
(c)    0V

22.   (a)    0·6 V
(b)    (i)       12 k
(ii)      4 k

23.   X = 9  Y = 45 

70
Electrical sources and internal resistance
1.    See notes

2.    (a)    6
(b)    A 1 = 2 A, A 2 = 1·5 A
(c)    6V
(d)    24 W

3.    (a)    In 12 Ω current is 11·5 A; in 8 Ω current is 11.5A; in 24 Ω
current is 0 A
(b)    In 12 Ω current is 12·8 A; in 8 Ω current is 3·2 A; in 24 Ω current
is 9·6 A

4.    (a)    R 1 = 230       R 2 = 115 
(b)    Low 230 W        high 690 W

5.    (a)    4
(b)    36 W

6.    (a)    2·0 V
(b)    1·6 V
(c)    r = 0·5  R = 2 
(d)    1·3 A, 1·3 V

7.    10 

8.    5

9.    12 

10.   (a)    1·3 V
(b)    (i)     As R is increased, current decreases (since emf = I x total
resistance and total resistance of circuit has increased)

71
(b)      (ii)     As R is increased, the p.d. across the terminals
INCREASES.

Since lost volts = Ir and I decreases, lost volts also
decreases so

t.p.d = emf-lost volts increases (i.e. less energy is lost as
heat through cell).

11.   0·30 A

12.   (a)      3·0 
(b)      3·75 

13.   4·0 V

14.   (a)      Very similar to diagram for Q3.1.42 except battery is replaced
by a cell and R is replaced by variable resistor. Switch not
necessary .
(b)            (i)      1·1 V, the intercept on the y-axis
(ii)     4·2  the gradient of the line
(iii)    0·26 A

15    (a)      6V
(b)      0·1 
(c)      60 A
(d)      (i)      5·6 V
(ii)     21 W

16.   (a)      Divide through by I: E/I = R+r ==> rearrange R = E/I -r

(b)      0·147, 0·264,0·382, 0·500, 0·617, 0·735, 0·855
(d)      (i)      2·5 
(ii)     17 V

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(e)      6·8 A

18.   (a)      Ammeter 1·76 mA, voltmeter 1·3 V
(b)      740 
(c)      0·02 mA
(d)      ±0·01 mA
(e)      ± 0.1 V
(f)      0·6 %
(g)      8%
(h)      8%
(i)      8%
(j)      59 
(k)      (740± 59) 
(l)      (740± 60) 

Capacitors

1.    (a)      5·0 × 10 –3 C
(b)      (i)       1·25 A

2.    0·5 F

3     (a)      40 V
(b)      1·7%

4.    2·25 × 10 –5 C

5.    (a)      1·0 F
(b)      0·8 C

6.    (b)      50 F

73
7.    (a)      2·0 × 10 –3 C
(b)      0·020 J

8.    0·60 J

9.    (a)       Initially the ammeter goes from 0 to a large (maximum)
reading. This then rapidly decreases and eventually falls back to
zero as the capacitor becomes fully charged

(b)      Reading on ammeter is 0 A

(c)      0 to 2 mA (max. current 1·2 mA)

(d)      12 V

10.   (a)      The capacitor takes a short time to charge up. When the p.d.
across the capacitor is greater than 100V, the neon lamp
becomes conducting. Charge rapidly flows off the capacitor
plates which causes the bulb to flash. When the p.d. drops
below 80V the resistance of the neon bulb is again very high.
No charge flows off the plates and the capacitor charges up
again. This cycle repeats causing the bulb to flash

(b)      Increase the capacitance of the capacitor or increase the
resistance of the resistor.

11.

74
(b)     4·9 mF

12.   (a)     (i)      Charging Position 1

(ii)     Discharging position 2

(b,c)   See notes

(d)     (i)      While charging, electrons flow from B to A (attracted
towards + terminal)

(ii)     While discharging, electrons flow in the opposite
direction, from A to B

(e)     (i)      40 A
(ii)     4·0 × 10 2 C

13.   (a)     (i) 0V when switch is closed rises to maximum Vc = 3 V

(ii) Maximum V R = 3V (initial reading) falls to 0V

(b)     (i)      3V
(ii)     9 C
(c)     1·35 × 10 –5 J

14.   (a)     2 mA
(b)     1·5 mA
(c)     First find voltage across capacitor = 4V  V R = 8V 
I = 1.3 mA (1.3x10 -3 A)
15.   (a)     V c will take longer to rise to the same value. V R will fall more
slowly from the same initial value. The current will start from
6mA (half-height) but will take longer to fall to zero.

(b)     Very similar to graphs in (a) V c will take longer to rise to the
same value. V R will fall more slowly from the same initial value.
The current will start from the same value 12mA but will take
twice as long to fall to zero.

(c)     (i)      The charge that is stored on the capacitor.

(ii)     Area under graphs for (a) is the same as the original
experiment. (total capacitance has not changed). Area

75
under graph (b) is twice that of the original graph (as
the capacitance has doubled it has twice the capacity to
store charge for the same applied voltage.)

(d)   12V

(e)   V c will start at 12 V and gradually fall to zero. The capacitor is
now effectively acting as the power supply in a circuit with a
resistor and so the voltage across the resistor will equal the
voltage across the capacitor. (the graph of V R will be the same
as the graph of V c ). The current is now flowing in the opposite
direction and so it will start from -12mA and gradually rise to
zero!

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