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Learning Outcomes
a. Recall and use appropriate circuit symbols as set out in SI Units, Signs, Symbols and
Abbreviations (ASE, 1981) and Signs, Symbols and Systematics (ASE, 1995).
b. Draw and interpret circuit diagrams containing sources, switches, resistors,
ammeters, voltmeters, and/or any other type of component referred to in the
syllabus.
c. Solve problems using the formula for the combined resistance of two or more
resistors in series.
d. Solve problems using the formula for the combined resistance of two or more
resistors in parallel.
e. Solve problems involving series and parallel circuits for one source of e.m.f.
f.* Show an understanding of the use of a potential divider circuit as a source of variable
p.d.
g.* Explain the use of thermistors and light-dependent resistors in potential dividers to
provide a potential difference which is dependent on temperature and illumination
respectively.
h.* Recall and solve problems using the principle of the potentiometer as a means of
comparing potential differences.
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The Electrical Circuit Symbols
Symbol Meaning Symbol Meaning
Cell / battery thermistor
power supply diode
switch potential divider
A
ammeter earth
V
voltmeter aerial/antenna
galvanometer wires crossing with
no connection
filament lamp
wires crossing with
resistor
connection
variable resistor
Loudspeaker
light dependent
resistor
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Principles Involved in Solving Circuits
Problems
Conservation of charges states that charges cannot be created nor destroyed.
Hence, the total charge that enters a junction per unit time (i.e. current entering)
must be equal to the total charge that leaves the same junction per unit time (i.e.
current leaving).
Principle 1: The sum of currents entering any junction in an electric circuit is always
equal to the sum of currents leaving that junction (aka. Kirchhoff’s First Law)
V
I = I 1 + I2 I1 I
R1
I
I2
R2
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Principles Involved in Solving Circuits
Problems
Resistors In Parallel Resistors In Series
V E
I1 I
R1 I R1 R2
I
I2
R2 V1 V2
From Principle 1: I = I1 + I2 From Principle 2: E = V1 + V2
V V V
Which implies that: = + Which implies that: I Rc = I R1 + I R2
Rc R1 R2
1 1 1
Thus: = + Thus: Rc = R1 + R2
Rc R1 R2
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Principles Involved in Solving Circuits
Problems
E.g. 1 The diagrams show connected wires which carry currents I1, I2, I3 and I4.
The currents are related by the equation I1 + I2 = I3 + I4.
To which diagram does this equation apply? [N98/P1/Q16]
A B C D
I1 I3 I4
I4 I3
I2
I1 I2
I4 I1 I1 I2
I2 I3 I4 I3
C
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Principles Involved in Solving Circuits
Problems
Extra: Calculate the currents IA, IB, IC, ID, IE, and IF as shown in the diagram below.
IF
IA = 8.4 – 3.7 = 4.7 mA
IB = 8.4 mA 1.3 mA
IC = 1.3 mA 2.4 mA IE
ID IC
ID = 8.4 – 1.3 = 7.1 mA IB
IE = 7.1 – 2.4 = 4.7 mA
IA
IF = 8.4 mA 3.7 mA
8.4 mA
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Principles Involved in Solving Circuits
Problems
E.g. 2 Referring to the circuit drawn, determine the value of I1, I and R, the
combined resistance in the circuit.
2V
Applying Ohm’s Law,
I1 160 Ω
I1 = 2 ÷ 160 = 0.0125 A
I2 = 2 ÷ 4000 = 5 × 10-4 A I
I3 = 2 ÷ 32000 = 6.25 × 10-5 A I2 4000 Ω
Since I = I1 + I2 + I3, I3 32000 Ω
I = 13.1 mA
Applying Ohm’s Law, Note: Rc of resistors in // is always
R = 2 ÷ (13.1 × 10-3) smaller than smallest R in the // network
= 153 Ω
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Principles Involved in Solving Circuits
Problems
Conservation of energy states that energy cannot be created nor destroyed.
Therefore, the electrical energy produced by the source is equal to the sum of
electrical energy consumed by all the components.
From the conservation of energy,
Power supplied by = Power dissipated by E
the source the resistors
I R1 R2
IE = I2 R1 + I2 R2
E = I R1 + I R2
i.e. E = V1 + V 2 V1 V2
Principle 2: In any closed loop in an electric circuit, the total potential rise equals
to the total potential drop along that loop.
i.e., in the circuit shown above the EMF of the battery, E = V1 + V2
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Principles Involved in Solving Circuits
Problems
Resistors In Parallel Resistors In Series
V E
I1 I
R1 I R1 R2
I
I2
R2 V1 V2
From Principle 1: I = I1 + I2 From Principle 2: E = V1 + V2
V V V
Which implies that: = + Which implies that: I Rc = I R1 + I R2
Rc R1 R2
1 1 1
Thus: = + Thus: Rc = R1 + R2
Rc R1 R2
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Principles Involved in Solving Circuits
Problems
E.g. 3 Referring to the circuit drawn, determine the value of the current I and the
potential difference across each resistor. 2V
Total resistance = 100 + 600 + 300
= 1000 Ω I
I = 2 ÷ 1000 100 Ω 600 Ω 300 Ω
= 0.002 A
Applying Ohm’s law,
Potential difference across the 100 Ω resistor = 0.002
x 100
= 0.2 V
Potential difference across the 600 Ω resistor = 0.002
x 600
= 1.2 V
Potential difference across the 300 Ω resistor = 0.002
x 300
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Principles Involved in Solving Circuits
Problems
Extra: A battery with an EMF of 20 V and an internal resistance of 2.0 Ω is
connected to resistors R1 and R2 as shown in the diagram. A total current of 4.0
A is supplied by the battery and R2 has a resistance of 12 Ω. Calculate the
resistance of R1 and the power supplied to each circuit component.
E - I r = I 2 R2
2Ω
20 – 4 (2) = I2 (12)
I2 = 1 A
20 V
Therefore, I1 = 4 – 1 = 3 A
4A R1
E – I r = I 1 R1 I1
12 = 3 R1 R2
Therefore, R1 = 4 Ω
I2
Power supplied to R1 = (I1)2 R1
= 36 W
Power supplied to R2 = (I2)2 R2
= 12 W
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Principles Involved in Solving Circuits
Problems
E.g. 4 The diagram shows a network of three resistors. Two of which have a
resistance of 2.5 Ω and one of which has a resistance of 5.0 Ω, as shown.
Determine the resistance between X and Y. [J91/P1/Q14]
Y
Ω
5.0
2.5 Ω
Resistance between X and Y
Z
1 1
= ( + ) −1
2.5 2.5 + 5
Ω
= 1.875 Ω
2.5
X
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Principles Involved in Solving Circuits
Problems
E.g. 5 Three resistors of resistance 2 Ω, 3 Ω and 4 Ω respectively are used to
make the combinations X, Y and Z shown in the diagrams. List the combinations
in order of increasing resistance.
X Y Z
2Ω 4Ω 3Ω
2Ω 2Ω
4Ω 3Ω 3Ω
4Ω
1 1 −1
Resistance for X = ( + ) = 1.56 Ω
2 4+3
1 1 −1 Therefore, the combination of
Resistance for Y = 2 + ( + ) = 3.71 Ω resistors in order of increasing
4 3 resistance is Z X Y.
1 1 1 −1
Resistance for Z = ( + + ) = 0.923 Ω
3 2 4
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Principles Involved in Solving Circuits
Problems
E.g. 6 Calculate the resistance across XY for the network shown.
2Ω 3Ω
5Ω
4Ω
6Ω
X Y
−1
1 −1
1
−1
1
Resistance = + + 5 + = 3.277 Ω
2 + 3 4
6
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The Potential Divider
A potential divider is a circuit in which two (or more) resistors are connected in
series with a supply voltage (EMF).
The resistors “divide” the EMF proportionally according to their resistance.
R1 = 6 kΩ V1
E = 12 V
(negligible
internal R2 = 4 kΩ V2
resistance)
R1 R2
V1 = ×E Similarly, V2 = ×E
R1 + R 2 R1 + R 2
6 4
= ×12 = ×12
4+6 4+6
= 7.2 V = 4 .8 V
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The Potential Divider
E.g. 7 Two resistors, of resistance 300 kΩ and 500 kΩ respectively, form a
potential divider with outer junctions maintained at potentials of +3 V and –15 V.
+3 V 300 kΩ X 500 kΩ -15 V
Determine the potential at the junction X between the resistors.
300
The potential difference across the 300 kΩ resistor = [ 3 − ( − 15) ]
300 + 500
= 6.75 V
The potential at X = 3 - 6.75
= - 3.75 V
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The Potential Divider
E.g. 8 The diagram below shows a potential divider circuit which by adjustment
of the position of the contact X, can be used to provide a variable potential
difference between the terminals P and Q.
Determine the limits of this potential difference. X
4 kΩ P
25 V Q
6 kΩ
When X is at the upper end of the resistor, the potential difference across PQ
4
= × 25 = 10 V
4+6
When X is at the lower end of the resistor, the potential difference across PQ
0
= × 25 = 0 V
0 + 10
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The Potential Divider
E.g. 9 A 20 V battery with an internal resistance of 2.5 kΩ is connected to a 50 kΩ
device. A voltmeter with an internal resistance of 120 kΩ and an ammeter with
negligible internal resistance are used to measure the potential difference across the
device and the current passing through it respectively.
(a) Sketch the circuit diagram for this setup.
(b) Determine the reading on the voltmeter and the ammeter.
−1
Rc (parallel to the voltmeter) =
1 1
+
50 k 120 k
20 V 2.5 kΩ = 35 kΩ
35.3 k
I Voltmeter reading, V = × 20 = 19 V
50 kΩ 35.3 k + 2.5 k
A 20
I= = 0.53 mA
35.3 k + 2.5k
V
120
120 kΩ Ammeter reading = × 0.529 = 0.37 mA
120 + 50
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The Use of a Thermistor in a Potential
Divider
A thermistor is a resistor whose resistance varies greatly with temperature. Its
resistance decreases with increasing temperature. It can be used in
temperature
potential divider circuits to monitor and control temperatures.
E.g. 10 In the figure below, the thermistor has a resistance of 800 Ω when hot,
and a resistance of 5000 Ω when cold. Determine the potential at W when the
temperature is hot.
+7 V
When thermistor is hot, potential difference across it
800
= × ( 7 − 2) 1.7 kΩ
800 + 1700
W
= 1.6 V
The potential at W = 2 + 1.6 V = 3.6 V
2V
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The Use of a Light Dependent Resistor
(LDR) in a Potential Divider
An LDR is a resistor whose resistance varies with the intensity of light falling on it.
Its resistance decreases with increasing light intensity. It can be used in a
intensity
potential divider circuit to monitor light intensity.
E.g. 13 In the figure below, the resistance of the LDR is 6.0 MΩ in the dark but
then drops to 2.0 kΩ in the light. Determine the potential at point P when the
LDR is in the light.
+18
In the light the potential difference across the LDR V
2k
= × (18 − 3)
3k +2k
P
=6V
3.0 kΩ
The potential at P =
18 – 6
3V
= 12 V
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The Potentiometer
The potentiometer is a device used to compare PD or EMF. It can measure an
unknown EMF by comparing this EMF with a known standard EMF.
Driver cell, E
Note:
1) E > V
L2 2) Same polarity
L1
B A
Potentiometer wire (Usually, 1 m)
Sliding Contact
p.d. source, V
Potential difference along wire ∝ length of the wire.
Sliding contact will move along AB until galvanometer shows 0 (i.e. current = 0)
potentiometer is said to be “balanced”.
If the cell has negligible internal resistance, and if the potentiometer is balanced,
L1
EMF / PD of the unknown source, V = ×E
L1 + L 2
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The Potentiometer
E.g. 12 In the circuit shown, the potentiometer wire has a resistance of 60 Ω.
Determine the EMF of the unknown cell if the balanced point is at B.
20 Ω 12 V
Resistance of wire AB
0.65 0.35 m B 0.65 m
= × 60
0.65 + 0.35 C A
= 39 Ω
500 Ω 30 Ω
EMF of the test cell
39
= × 12
60 + 20
= 5.85 V
