Topic 2.1 – Current, voltage, resistance and power.
Learning Objectives:
At the end of this topic you will be able to;
 recall and use R 
V
;
I
 apply Kirchoff’s laws;
 calculate the combined resistance of resistors connected in series
and or parallel;
 select appropriate values from the E24 series.
 perform calculations on voltage divider circuits;
 define power as VI and apply the formula to calculate power
dissipation.
1
Module ET2
Electronic Circuits and Components.
In module ET1 we discovered much about how logic circuits and some basic
voltage amplifiers were used. In this Module we are going to take a much
closer look at the way in which a number of components work and how we can
design circuits using a variety of components to meet a particular
specification.
Fundamental Circuit Concepts.
We will begin by revising some basic circuit ideas that you would have covered
during your GCSE Science course.
Types of Circuit.
There are two ways that components can be connected in circuits, these are
referred to as series and parallel.
Components in Series
Components connected in series are only connected to each other at one
point.
The circuit below shows 2 identical lamps connected in series
12 V
0V
• The same current flows through both components
• The sum of the voltages across each component in the circuit equals the
power supply voltage.
2
Topic 2.1 – Current, voltage, resistance and power.
Components in Parallel
The components sit side by side and connect to more than one other
component.
The circuit below shows 2 identical lamps connected in parallel
12 V
0V
• The voltage is the same across all parallel components.
• The sum of the currents in each branch of the circuit equals the
current leaving the power supply.
3
Module ET2
Electronic Circuits and Components.
Electric Current.
An electric current is a flow of charge carried by tiny particles known as
electrons.
Movement of charge is caused by the force from a voltage supply such as a
battery or mains supply acting on electrons which are free to move. When
electricity was first discovered scientists thought that current was a flow of
positive charge from the positive terminal of the battery to the negative
terminal as shown on the left hand diagram below.
A
A
Electron flow
Current flow
The discovery of the electron proved that this original idea was incorrect,
and that current is a flow of negatively charged particles from the negative
to the positive terminal. This is shown on the diagram on the right.
A decision was made not to change all the books that had been written, since
the effect of a negative charge moving in the opposite direction is the same
as a positive charge moving the other way. For most people the exact nature
of the flow of charge is not important, so the original decision remained and
we now call this conventional current.
Electric current is measured in units called Ampères (A), or Amps for short.
In our work in electronics we will usually be dealing with much smaller units of
current such as mA
 1

A or 103 A  and

 1000

μA
1


A or 106 A 

 1000000

The electric current is not used up by the components in a circuit but it
transfers energy from the voltage source to the various components making
up the circuit.
4
Topic 2.1 – Current, voltage, resistance and power.
Electric current is measured using an ammeter.
12 V
The ammeter is placed in series within the
circuit, as shown in the circuit diagram on the
right.
+
A
0V
Voltage.
Voltage is supplied by a battery, cell or power supply and is measured in volts
(V).
The voltage of a power supply is the driving force that pushes the current
through the components of a circuit. The larger the voltage the bigger the
push on the current.
The voltage across a circuit component, e.g. a bulb converting electrical
energy into heat and light is sometimes called a potential difference (p.d.).
Potential difference is also measured in volts (V).
Potential difference is measured using a voltmeter.
12 V
V
The voltmeter is placed in parallel to the component
as shown in the diagram opposite.
0V
From this point on we will simply refer to potential difference simply as
voltage.
5
Module ET2
Electronic Circuits and Components.
Resistance.
Resistance is the property of a material which restricts the flow of
electricity. Energy is converted into other forms (e.g. light, heat), as the
voltage across the component drives the current through it.
The unit of resistance is the Ohm (Ω). In electronics we usually deal with
much larger units of resistance as kΩ (1000Ω or 103Ω), and MΩ
(1000000Ω or 106Ω)
What causes resistance?
As an electric current flows, charged particles called electrons move
through a conductor. These moving electrons collide with the atoms of
the conductor, making it more difficult for the current to flow. This
causes the resistance.
What factors affect resistance?



Longer wires increase resistance. The electrons collide with atoms
more often in a long wire than in a short wire.
Thin wires increase resistance. A thin wire has fewer electrons to
carry the current than a thick wire.
The material the wire is made from e.g. nichrome is used for
heating elements, because it has a high resistance per metre, so
produces heat when current flows through it. Aluminium is used
for transmission lines because it has a very low resistance per
metre and so less energy is wasted during the transfer of
electricity from power stations to your home.
In 1826, based on his work on conduction in metal wires, Georg Ohm
formulated a law relating to the current passing through a wire, and the
voltage applied.
6
Topic 2.1 – Current, voltage, resistance and power.
He found that for a fixed temperature, the current flowing through a
resistor is directly proportional to the voltage across the resistor.
V=IR
This is known as Ohm’s Law. Ohm also introduced the SI unit of resistance,
as the Ohm (Ω).
Using Ohm’s Law, the resistance of any component can be calculated by
measuring the current in the circuit, and the voltage across the component.
The triangle opposite is often used to work
out the formula required to solve problems
of this type. Simply cover the term you want
to find and what is left is the formula to
use.
Example:
1.
A current of 0.1A flows through a 20Ω resistor. What is the voltage
across the resistor?
Formula required :
V  IR
Calculation :
V  IR
V  0.1 20
V  2V
7
Module ET2
Electronic Circuits and Components.
2.
A 10Ω resistor is connected across a 5 volt power supply. What is the
resulting current?
Formula required :
I
V
R
Calculation :
I
3.
V 5

 0.5 A  500mA
R 10
A resistor is placed in a circuit with a voltage of 12 volts and a current
flow of 100mA. What is the resistance?
Formula required :
R
V
I
Calculation :
R
V
12V
12V


 120
I 100mA 0.1A
Student Exercise 1 : Ohms Law Practice
1.
A 12V power supply is connected to a 3kΩ resistor. What is the current
flowing in the circuit?
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
8
Topic 2.1 – Current, voltage, resistance and power.
2.
A 9V battery provides a current of 1.5A to a resistor, What is the value
of the resistance?
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
3.
A lamp is marked with the following maximum ratings: 6V 0.3A. What is
the resistance of the lamp if it is operated at these maximum values ?
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
4.
A current of 4mA is driven through an 8 kΩ resistor. What is the
voltage of the power supply producing the current.
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
5.
A resistor of 15 Ω is connected to a battery of unknown voltage. The
current in the circuit is measured to be 0.1A. What is the voltage of the
battery ?
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
6.
A 4.5V battery is connected to a 36 Ω resistor. How much current
flows in the circuit ?
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
9
Module ET2
Electronic Circuits and Components.
7.
A current of 3A leaves a 24V power supply. What is the value of the
resistor connected to the power supply?
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
8.
A current of 0.2A flows through a 45 Ω resistor. What is the voltage
of the battery producing the current?
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
9.
A resistor of 24 Ω is connected to a power supply. The current in the
circuit is measured to be 0.05A. What voltage does the power supply
provide ?
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
10.
A 1.5V battery is connected to a 75Ω resistor. How much current flows
in the circuit ?
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
Note : More information about the different types of resistors available,
their properties and how to determine their value will be found in the
Supplementary Pack of Notes which are non-examinable called “Practical
Circuit Assistance.”
10
Topic 2.1 – Current, voltage, resistance and power.
The E24 Series of resistor values
If you look in a component suppliers’ catalogues for resistors e.g. Rapid
Electronics, you will find that resistors are available in the range from 1Ω to
10 MΩ. Clearly there are 10 million resistor values in between these two
values, and it would be completely impractical to expect any manufacturer to
produce every single value and make these available for sale.
The manufacturers actually produce a much smaller number of resistors but
cover the same range. This subset of resistors are perfectly adequate for
99.9% of all electronic circuits. When a situation requires an exact resistance
the most common solution is to use a variable resistor and adjust it to the
exact value required.
The resistor values available are as follows:
10, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 43, 47, 51, 56, 62, 68, 75, 82, 91
and any multiple of 10 these values, e.g. 130Ω, 3900Ω, 180kΩ, 1.2kΩ, 8.2MΩ
etc. If you count up the number of values you will find that there are 24 of
them, so this is where the name E24 Series comes from.
Kirchoff’s Laws.
Kirchhoff's circuit laws are two equalities that deal with the conservation of
charge and energy in electrical circuits, and were first described in 1845 by
Gustav Kirchhoff. We will consider some simple interpretations of these
here.
Kirchoffs First Law - This can be stated simply
as ‘the sum of currents entering a junction, is
equal to the sum of currents leaving a
junction’ as can be shown by the diagram
opposite where I1 = I2 + I3.
I2
I1
I3
11
Module ET2
Electronic Circuits and Components.
Now try these :
1.4A
1.8A
1.6A
2.3A
2.7A
5.2A
1A
3.8A
.....
.....
3A
3.3A
7.3A
.....
V1
Kirchoff’s Second Law – This can be
stated simply as ‘the sum of all
voltages in a series circuit is equal to
the voltage of the supply’ as shown in
the circuit opposite, where V1 = V2 + V3
V2
V3
For each of the following circuits complete the readings on the voltmeters.
(a)
9V
3V
12
Topic 2.1 – Current, voltage, resistance and power.
(b)
6V
5V
(c)
9V
2V
3V
13
Module ET2
Electronic Circuits and Components.
We will now look at the effect of connecting resistors in series and parallel.
Resistors in series
I
R1
R2
I
R
is equivalent to
V1
V2
V
The diagram shows two resistors of resistance R1 and R2 connected in series,
and a single resistor of resistance R equivalent to them. The current I in the
resistors, and in their equivalent single resistor, is the same. The total
voltage V across the two resistors must be the same as that across the single
resistor. If V1 and V2 are the voltages across each resistor,
V  V1  V2
But since voltage is given by multiplying the current by the resistance,
IR  IR1  IR2
IR  I ( R1  R2 )
Dividing by the current I (which is the same in both cases)
R  R1  R2
This equation can be extended so that the equivalent resistance R of several
resistors connected in series is given by the expression…
R  R1  R2  R3  R4  ...
Therefore, the combined resistance of resistors in series is just the sum
of all the individual resistances.
14
Topic 2.1 – Current, voltage, resistance and power.
Resistors in parallel
Now consider two resistors of resistance R1 and R2 connected in parallel, as
shown.
I
I1
R1
I2
R2
I
R
is equivalent to
V
V
The current through each will be different, but they will each have the same
voltage across them. The equivalent single resistor of resistance R will have
the same voltage across it, but the current will be the total current through
the separate resistors.
We know that
I  I1  I 2
and, using Ohm’s law, I 
V
, so
R
V V
V


R R1 R2
Dividing by the voltage V (which is same in both cases),
1
1
1


R R1 R2
This equation can be extended so that the equivalent resistance R of several
resistors connected in parallel is given by
1
1
1
1
1




 ...
R R1 R2 R3 R4
Therefore, the reciprocal of the combined resistance of resistors in
parallel is the sum of the reciprocals of all the individual resistances.
15
Module ET2
Electronic Circuits and Components.
Note that:
1. for two identical resistors in parallel, the combined resistance is equal to
half of the value of each one,
2. for n identical resistors in parallel, the combined resistance is equal to the
value of each one divided by n
3. for resistors in parallel, the combined resistance is always less than the
value of the smallest individual resistance.
4. for two resistors in parallel, there is an easier formula:
R
R1R2
R1  R2
Or in words
Rtotal 
Product of Resistors
Sum of Resistors
Electronics engineers often use this as a quick way to calculate the equivalent
of two resistors in parallel, and you don’t make errors with all those
reciprocals.
Be careful though, it only works for 2. If you have more, do the first 2
then put that with the next one etc.
16
Topic 2.1 – Current, voltage, resistance and power.
Student Exercise 2:
1. Calculate the equivalent resistance of the arrangement of resistors shown
below.
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
2.
Calculate the equivalent resistance of the arrangement of resistors.
(Hint: First find the resistance of the parallel combination.)
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
17
Module ET2
Electronic Circuits and Components.
3.
Calculate the effective resistance between the points A and B in the
network below.
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
4.
18
You have 3 resistors of 60Ω, 60Ω and 30Ω, Show how some or all of
these resistors can be connected to make the following resistor values.
(i)
150Ω
(ii) 20Ω
(iii) 90 Ω
(iv) 15 Ω
Topic 2.1 – Current, voltage, resistance and power.
Voltage dividers
Two resistors connected in series with a battery or power pack each have a
voltage across them. They may be used to divide the voltage of the supply.
This is illustrated below.
I
R1
V1
R2
V2
V
The current in each resistor is the same, because they are in series. Thus
V1  IR1 and V2  IR2
Dividing the first equation by the second gives
V1 R1

V2 R2
The ratio of the voltages across the two resistors is the same as the ratio of
their resistances.
If the voltage across the combination were 12V and R1 were equal to R2, then
each resistor would have 6V across it.
If R1 were twice the magnitude of R2, then V1 would be 8V and V2 would be 4V.
19
Module ET2
Electronic Circuits and Components.
Example:
If the voltage across the combination were 10 V and R1 were four as large as
R2, What would the voltage be across each resistor?
Voltage across R1 = ………………………………
Voltage across R2 = ………………………………
Also,
V  I ( R1  R2 )
and
V1  IR1
Dividing these gives,
V1
IR1

V I ( R1  R2 )
V1
R1

V ( R1  R2 )
So,
V1 
V  R1
( R1  R2 )
This is called the voltage divider formula (see next section)
A similar argument can be applied for the voltage across R2. You should be
able to show that V2 is given by the following formula:
V2 
V  R2
( R1  R2 )
Example: Calculate the voltage reading on the voltmeter in the following
circuit.
..............................................................................................
..............................................................................................
3
..............................................................................................
9V
24
?V
..............................................................................................
..............................................................................................
20
Topic 2.1 – Current, voltage, resistance and power.
Potentiometers
A potentiometer is a continuously-variable voltage divider.
A
9V
C
VOUT
B
In the diagram above a fixed voltage supply of 9 V is being used to provide a
current through a variable resistor, when used in this way the variable
resistor is called a potentiometer, and this can be used to produce a
continuously variable voltage.
All three connections to the variable resistor are used. The fixed ends are
connected across the battery so that there is the full battery voltage across
the ends AB of the resistor. As with the voltage divider, the ratio of the
voltages across AC and CB will be the same as the ratio of the resistances of
AC and CB.
When the sliding contact C is at the end A, the output voltage VOUT will be
9 V.
When the sliding contact is at end B, then the output voltage will be zero.
So, as the sliding contact is moved from A to B, the output voltage varies
continuously from the battery voltage down to zero.
21
Module ET2
Electronic Circuits and Components.
Student Exercise 3:
1.
Find the total resistance in each of the following combinations of
resistors.
(a)
10 
30 
80 
(b)
10k
20 
100k
1000 
20k
(d)
(c)
20k
20 
10k
2.0 
(e)
60k
40 
(f)
10k
15k
10 
2.
Given the information on this circuit diagram
find:
(a) the voltage of the battery.
(b) the current in the 20  resistor.
(c) the current supplied by the battery.
3.
Find the value of VOUT in the following circuits.
(a)
4
VIN = 20 V
12 
22
VOUT
Topic 2.1 – Current, voltage, resistance and power.
(b)
12 
VIN = 12 V
4
VOUT
(c)
200 
VIN = 6 V
800 
4.
5.
Plot a sketch graph showing how the
output voltage varies as the slider is
moved from the bottom to the top of the
variable resistor.
VOUT
V
Using only resistors from the E24 Series, show how you could make the
following resistance values using only two resistors in each case.
i)
60 Ω
ii)
500 Ω
iii)
165 Ω
iv)
670 Ω
v)
125 Ω
vi)
420 Ω
23
Module ET2
Electronic Circuits and Components.
6.
Determine the values of R1, V1, I1, I2 and R2 in the following circuit.
15 V
I2
0.4 mA
R2
R1
6V
I1
10 k
V1
0V
(a)
R1
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
(b)
V1
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
(c)
I1
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
(d)
I2
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
(e)
R2
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
(f)
What is the power dissipated in R1 ?
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
24
Topic 2.1 – Current, voltage, resistance and power.
7.
Determine the values of V1, I1, R1, I2 and R2 in the following circuit.
9V
I1
V1
1.5 k
2.5 mA
3V
R1
I2
R2
0V
(a)
V1
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
(b)
I1
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
(c)
R1
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
(d)
I2
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
(e)
R2
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
(f)
What is the effective resistance of R1 and R2 in parallel?
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
25
Module ET2
Electronic Circuits and Components.
Voltage at a point
It is sometimes more convenient to measure or calculate the voltage at a
point in a circuit rather the voltage across a component.
Voltages at a point are measured with reference to zero volts or ground.
A voltage at a point can be negative as well as positive
Example
Calculate the voltage at points X, Y, Z, P and Q, in the following circuits, with
respect to the zero volt line.
In the left hand circuit
In the right hand circuit
Voltage at X = +7V
Voltage at P = +5.4V
Voltage at Y = +6V
Voltage at Q = -6.6V
Voltage at Z = +4V
26
Topic 2.1 – Current, voltage, resistance and power.
Electrical Power
When electricity flows through a resistor or other components, some energy
is dissipated in the resistor or component as heat. The power dissipated in a
component can be calculated by using the following formula:
Power (W )  Voltage(V )  Current ( A)
P V  I
This equation can be combined with Ohm’s law to provide two alternative
forms of the equation, should R be known but V or I is not.
Substituting for V we get
P V  I
V  IR
P  ( I  R)  I
P  I2R
Alternatively, substituting for I we get
P V  I
P V 
P
I
V
R
V
R
V2
R
Example:
Calculate the power dissipated in a 10kΩ resistor, when the voltage across
the resistor is 10V.
P
V 2 10  10
100
1



 0.01W or 10mW
R 10,000 10000 100
27
Module ET2
Electronic Circuits and Components.
Student Exercise 4:
1.
Calculate the power dissipated in a light bulb, carrying a current of
60mA, when connected to a 6V battery.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
2.
A 220Ω resistor has a power rating of 0.4W. What is the maximum
current that can be allowed through the resistor if the power rating is
not to be exceeded?
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
3.
The current flowing through a resistor is 0.125A, and the voltage across
the resistor is 12V. Calculate
(i) the value of the resistance,
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
(ii) the power dissipated in the resistor.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
28
Topic 2.1 – Current, voltage, resistance and power.
Solutions to Student Exercises.
Exercise 1 : Ohms Law Practice
I
V 12

 4 mA
R 3k
2.
R
3.
R
V
6

 20
I 0.3
4.
V  IR
5.
V  IR
6.
I
8.
V  IR
1.
V  4  10 3  8  10 3
V  32V
V  0.1 15
V  1.5V
7.
9.
R
V 24

 8
I
3
V
9

 6
I 1.5
V 4.5

 0.125 A  125 mA
R 36
V  0.2  45
V  9V
10.
V  IR
I
V  0.05  24
V  1.2V
V 1.5

 0.02 A  20mA
R 75
Kirchoffs Law Exercises.
1.4A
1.8A
1.6A
2.7A
5.2A
0.1A
1A
3.8A
3A
3.3A
2.3A
4.8A
7.3A
3.3A
29
Module ET2
Electronic Circuits and Components.
(a)
9V
6V
3V
(b)
6V
5V
5V
1V
(c)
9V
2V
4V
3V
30
2V
Topic 2.1 – Current, voltage, resistance and power.
Student Exercise 3 :
1.
(i)
For each branch in the circuit there are two series resistors which
can be added together to give a total resistance in each branch of
6Ω.
(ii)
Effectively therefore we have two 6Ω resistors in parallel,
therefore,
R1R2
6  6 36


 3
R1  R2 6  6 12
the equivalent resistance of the arrangement of resistors is therefore
3Ω.
2.
For the parallel resistors.
R1R2
42 8

  1.33
R1  R2 4  2 6
Therefore for the whole network –
R  R1  R2
R  3  1.33
R  4.33
3.
The circuit is effectively a 20 Ω resistor and 10 Ω resistor in series,
connected in parallel with a 100 Ω resistor and a 50 Ω resistor also
connected in series.
So dealing with the series parts first we would have 20 + 10 =30Ω in one
branch, and 100 + 50 = 150Ω in the other branch, i.e. a parallel
combination of 30 Ω and 150 Ω. So the effective resistance between
the points A and B would be –
RAB 
R1R2
30  150 4500


 25
R1  R2 30  150 180
31
Module ET2
Electronic Circuits and Components.
4.
i)
60Ω + 60Ω + 30Ω in series = 150Ω
ii)
60Ω + 30Ω in parallel = 20Ω
iii)
60Ω + 30Ω in series = 90Ω
iv)
60Ω + 60Ω + 30Ω all in parallel with each other = 15Ω
Student Exercise 3:
1.
a)
R  R1  R2  R3
b)
R  R1  R2  R3
c)
R
R  10  30  80  120
R  10k  100k  1k  111k
R1R2
20  20 400


 10
R1  R2 20  20 40
d)
1 1
1
1



R R1 R2 R3
e)
1
1
1
1



R 20k 20k 10k
1 1 1 2

R
20k
1
4

R 20k
20k
R
 5 k
4
1 1
1
1



R R1 R2 R3
1 1 1
1
 

R 2 40 10
1 20  1  4

R
40
1 25

R 40
40
R
 1.6
25
32
Topic 2.1 – Current, voltage, resistance and power.
f)
For parallel circuit
R
R1R2
60k  15 k 900k


 12k
R1  R2 60k  15 k 75 k
For the whole circuit
R  R1  R2
R  12k  10k
R  22k
2.
(a)
the voltage of the battery will be the voltage across the series
combination of the 4Ω and 3Ω resistor.
V  IR
V  1.5  (3  4)
V  10.5V
(b)
the current in the 20  resistor, will be the same as the current in
the 8  resistor, and as these are in parallel with the 3  and 4 
resistor they will have the same voltage across them, therefore.
I 20  
(c)
V
10.5

 0.375 A  375mA
R 8  20
the current supplied by the battery will be the sum of the two
currents in each branch.
I battery  I 1  I 2
 1.5  0.375
 1.875 A
3.
(a)
Vout 
Vin  R2 20  12 240


 15V
R1  R2
4  12
16
(b)
Vout 
Vin  R2 12  4 48


 3V
R1  R2 4  12 16
33
Module ET2
Electronic Circuits and Components.
(c)
Vout 
Vin  R2
6  800
4800


 4.8V
R1  R2 200  800 1000
4.
Output Voltage
20V
0
Slider Position
Bottom
5.
Top
i)
27Ω + 33Ω in series = 60Ω
ii)
1kΩ + 1kΩ in parallel = 500Ω
iii)
330Ω + 300Ω in parallel = 165Ω
iv)
240Ω + 430Ω in series = 670Ω
v)
750Ω + 150Ω in parallel = 125Ω
vi)
30Ω + 390Ω in series = 420Ω
Note other solutions may be possible – if in doubt check with your tutor.
6.
34
6
 15 k
0.4 mA
a)
R1 
b)
V1  15  6  9V
c)
I1 
9
 0.9mA
10k
Topic 2.1 – Current, voltage, resistance and power.
7.
d)
I2  0.9mA  0.4mA  0.5mA
e)
R2 
f)
P  V  I  6  0.4mA  2.4mW
a)
V1  9  3  6V
b)
I1 
6
 4 mA
1.5 k
c)
R1 
3
 1 .2 k
2.5 mA
d)
I2  4mA  2.5mA  1.5mA
e)
R2 
6
 12k
0.5 mA
3
 2k
1.5 mA
f)
Reff 
1.2k  2k 2.4 k

 750
1.2k  2k 3.2k
or
Reff 
3
 750
4 mA
Student Exercise 4:
1.
P V  I
P  6  60
P  360mW
35
Module ET2
Electronic Circuits and Components.
2.
P  I2  R
P
R
P
0 .4
I

 1.818  10 3
R
220
I  0.042 A  42mA
I2 
3.
(i)
R
(ii)
Either:
Or
36
V
12

 96
I 0.125
P V  I
P  12  0.125  1.5W
P  I 2 R  0.125   96  1.5W
2
or P 
V 2 122 144


 1.5W
R
96
96
Topic 2.1 – Current, voltage, resistance and power.
The following questions have been taken from past examination papers.
1.
Use the information given in the circuit diagram, to determine the values of the quantities listed
below.
9V
I1
3V
1.5 k
0.6 mA
V1
R1
I2
R2
0V
(a)
I1
……………………………………………………………………………………….
………………………………………………………………………………………………
(b)
V1
……………………………………………………………………………………….
………………………………………………………………………………………………
(c)
R1
……………………………………………………………………………………….
……………………………………………………………………………………………….
(d)
I2
………………………………………………………………………………………
………………………………………………………………………………………………
(e)
R2
………………………………………………………………………………………
………………………………………………………………………………………………
[6]
37
Module ET2
Electronic Circuits and Components.
2.
Determine the values of V1, VS, I1, I2 and R2 in the following circuit.
VS
V1
3 mA
I2
1 k
R2
I1
9V
1.8 k
0V
(a)
V1
……………………………………………………………………………………….
………………………………………………………………………………………………
(b)
VS
……………………………………………………………………………………….
………………………………………………………………………………………………
(c)
I1
……………………………………………………………………………………….
……………………………………………………………………………………………….
(d)
I2
………………………………………………………………………………………
………………………………………………………………………………………………
(e)
R2
………………………………………………………………………………………
………………………………………………………………………………………………
[6]
38
Topic 2.1 – Current, voltage, resistance and power.
3.
Use the information given in the circuit diagram, to determine the values of the quantities listed
below.
9V
I1
3V
1.2 k
0.5 mA
V1
R1
I2
R2
0V
(a)
I1
……………………………………………………………………………………….
………………………………………………………………………………………………
(b)
V1
……………………………………………………………………………………….
………………………………………………………………………………………………
(c)
R1
……………………………………………………………………………………….
……………………………………………………………………………………………….
(d)
I2
………………………………………………………………………………………
………………………………………………………………………………………………
(e)
R2
………………………………………………………………………………………
………………………………………………………………………………………………
[6]
39
Module ET2
Electronic Circuits and Components.
4.
Use the information given in the circuit diagram, to determine the values of the quantities listed
below.
VS
I2
3 mA
4.5 V
R1
R2
I1
7.5 V
1 k
0V
(a)
VS
……………………………………………………………………………………….
………………………………………………………………………………………………
(b)
R1
……………………………………………………………………………………….
………………………………………………………………………………………………
(c)
I1
……………………………………………………………………………………….
……………………………………………………………………………………………….
(d)
I2
………………………………………………………………………………………
………………………………………………………………………………………………
(e)
R2
………………………………………………………………………………………
………………………………………………………………………………………………
(f)
What is the effective resistance of R1 and R2 in parallel.
……………………………………………………………………………………………….
……………………………………………………………………………………………….
[7]
40
Topic 2.1 – Current, voltage, resistance and power.
5.
Determine the values of V1, R1, I2, I3 and R3 in the following circuit. Hence determine the power
dissipated in the 15kΩ resistor.
12 V
0.5 mA
V1
R1
I2
3V
15 k
I3
R3
0V
(a)
V1
……………………………………………………………………………………….
………………………………………………………………………………………………
(b)
R1
……………………………………………………………………………………….
………………………………………………………………………………………………
(c)
I2
……………………………………………………………………………………….
……………………………………………………………………………………………….
(d)
I3
………………………………………………………………………………………
………………………………………………………………………………………………
(e)
R3
………………………………………………………………………………………
………………………………………………………………………………………………
(f)
What is the power dissipated in the 15kΩ resistor.
……………………………………………………………………………………………….
……………………………………………………………………………………………….
[7]
41
Module ET2
Electronic Circuits and Components.
6.
Determine the values of V1, I1, I2, R1, and R2 in the following circuit.
7.5 V
I2
0.2 mA
3V
R1
R2
I1
V1
10 k
0V
(a)
V1
……………………………………………………………………………………….
………………………………………………………………………………………………
(b)
I1
……………………………………………………………………………………….
………………………………………………………………………………………………
(c)
I2
……………………………………………………………………………………….
……………………………………………………………………………………………….
(d)
R1
………………………………………………………………………………………
………………………………………………………………………………………………
(e)
R2
………………………………………………………………………………………
………………………………………………………………………………………………
(f)
What is the effective resistance of R1 and R2 in parallel?
……………………………………………………………………………………………….
……………………………………………………………………………………………….
[7]
42
Topic 2.1 – Current, voltage, resistance and power.
7.
Determine the values of R1, V1, I1, I2 and R2 in the following circuit.
16 V
I2
I1
V1
8.2 k
R2
2 mA
7.8 V
R1
0V
(a)
R1
……………………………………………………………………………………….
………………………………………………………………………………………………
(b)
V1
……………………………………………………………………………………….
………………………………………………………………………………………………
(c)
I1
……………………………………………………………………………………….
……………………………………………………………………………………………….
(d)
I2
………………………………………………………………………………………
………………………………………………………………………………………………
(e)
R2
………………………………………………………………………………………
………………………………………………………………………………………………
(f)
What is the effective resistance of R1 and R2 in parallel?
……………………………………………………………………………………………….
……………………………………………………………………………………………….
[7]
43
Module ET2
Electronic Circuits and Components.
8.
Determine the values of R1, V1, I1, I2 and R2 in the following circuit.
15 V
I1
I2
V1
R2
3 k
5 mA
9V
R1
0V
(a)
R1
……………………………………………………………………………………….
………………………………………………………………………………………………
(b)
V1
……………………………………………………………………………………….
………………………………………………………………………………………………
(c)
I1
……………………………………………………………………………………….
……………………………………………………………………………………………….
(d)
I2
………………………………………………………………………………………
………………………………………………………………………………………………
(e)
R2
………………………………………………………………………………………
………………………………………………………………………………………………
(f)
What is the power dissipated in R1?
……………………………………………………………………………………………….
……………………………………………………………………………………………….
[7]
44
Topic 2.1 – Current, voltage, resistance and power.
Self Evaluation Review
Learning Objectives
recall and use R 
My personal review of these objectives:



V
;
I
apply Kirchhoff’s laws;
calculate the combined resistance of
resistors connected in series and or
parallel;
select appropriate values from the
E24 series.
perform calculations on voltage
divider circuits;
define power as VI and apply the
formula to calculate power
dissipation.
Targets:
1.
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
2.
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
45