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This corresponds to the "D.C. Circuits" component of Electricity and Magnetism in the GCE 'O' Level - Physics (Pure) syllabus.
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This presentation was produced by the Science department of Temasek Secondary School. Redistribution or reproduction of this resource is prohibited by copyright regulations. This resource should be used for educational purposes.
D.C. CircuitsIn this section, we will learn to
draw circuit diagrams,
calculate the current, potential difference and resistance in a series and parallel circuit.
18.1Electric Circuits
Lamp
d.c. supply
Switch
1
23
4Conductors
18.1 Electric Circuits1. An electric circuit is a closed path in
which electric charges can flow from one terminal to another.
2. In every electric circuit, the following must be present:
• A e.m.f. source that drives charge and produces a current, e.g a battery.
18.1 Electric Circuits• Components that “feed” on current and do a job, e.g. a bell that makes a sound.
• Conductors that join the source and various components together, e.g. copper wires.
• Switches that break or complete a circuit.
18.2 Circuit DiagramsA circuit diagram is used to represent an electric circuit. Symbols are used to represent the devices or components used in electric circuits.
(Ref. textbook pg 321)
switch
18.2 Circuit Diagrams
resistor (fixed)
18.2 Circuit Diagrams
variable resistorrheostat
18.2 Circuit Diagrams
fuse
18.2 Circuit Diagrams
ammeter
A18.2 Circuit Diagrams
voltmeter
V18.2 Circuit Diagrams
galvanometerG
1. A galvanometer is used to measure small currents while an ammeter is used to measure greater currents.
2. A galvanometer must be connected in series to a circuit.
18.2 Circuit Diagrams
lamp
18.2 Circuit Diagrams
cell
18.2 Circuit Diagrams
bell/buzzer
18.2 Circuit Diagrams
ground/earth
18.2 Circuit Diagrams
power supply
240V240V
18.2 Circuit Diagrams
wires joinedat a point
18.2 Circuit Diagrams
wires crossed over each other but not joined
18.2 Circuit Diagrams
18.3 Series Circuit1. In a series circuit, there is ONLY
ONE path in which charge can flow.
AResistor
Lamp
Ammeter
CellI I
II
18.3 Series Circuit2. If there is a break anywhere in
the circuit, current will stop flowing.
AResistor Ammeter
Cell
No Current Flow
18.3 Series Circuit Let’s practice:
Draw a circuit diagram containing a d.c supply, a lamp, an ammeter, a switch and a variable resistor all connected in series. (J92 P2)
18.3 Series CircuitAnswer:
A
Lamp
Ammeter
d.c. supply
Switch
Variable resistor
18.4 Parallel Circuit1. In a parallel circuit, there is MORE
THAN ONE path in which charge can flow.
Lamps
Cell
AB
Current splits up at junction A and join together again at B
18.4 Parallel Circuit2. Even if one of the lamps is
removed, there will still be a flow of current in the other lamp.
Lamp
Cell
18.4 Parallel Circuit Let’s practice:
Draw a circuit diagram containing a d.c supply, two lamps connected in parallel and a switch that can turn off only one of the lamps.
18.4 Parallel CircuitAnswer:
Lamps
d.c. supply
Series + Parallel Circuit
AV
18.5 Current In Series Circuit
1. In a series circuit, the current through every component is the SAME.
A
I1 I4
I3I2
I1 = I2 = I3 = I4
Example 1
12 V
3 A
Q1: What is the name given to the device X?
X
A1: Lamp
Q2: The current passing through X is 3A, what is the current passing through the resistor and the cell?
A2: 3A
Recall: All the resistors are connected in series, the current passing through each resistor is the same.
R1 R2 R3 R4
V1 V2 V3 V4
I I
18.5 P.D In Series Circuit
18.5 P.D In Series Circuit
2. Let V be the potential difference across the combination of the 4 resistors in series.
V = V1 + V2 + V3 + V4
R1 R2 R3 R4
V1 V2 V3 V4
I I
V
18.5 P.D In Series Circuit
Points to note:
3. V1 = V2 = V3 = V4 if only the resistors are identical i.e. having the same resistance.
4. In a series circuit, the component that has the LARGEST resistance will have the LARGEST potential difference across it.
Example 2
12 V
Q1: What is the total p.d across the resistor and the lamp? A1: P.d across resistor and lamp
= e.m.f of cell, E = 12 V
Q2: If the p.d across the resistor is 9 V, what is the p.d across the lamp?A2: P.d across lamp, V = 12 V - 9 V = 3 V
18.5 Resistance In Series Circuit
Let R be the combined/total/effective resistance of the 4 resistors.
R1 R2 R3 R4
V1 V2 V3 V4
I I
V
18.5 Resistance In Series Circuit
How is R related to R1, R2, R3 and R4?
RI I
V
The arrangement can then be simplified from this …
R1 R2 R3 R4
V1 V2 V3 V4
I I
V
to…
18.5 Resistance In Series Circuit
Using equation V = IR, we have…
V1 = IR1 V2 = IR2 V3 = IR3 V4 = IR4
and… V = IR
R1 R2 R3 R4
V1 V2 V3 V4
I I
V
RI I
V
18.5 Resistance In Series Circuit
5. V = V1 + V2 + V3 + V4
IR = IR1 + IR2 + IR3 + IR4
R = R1 + R2 + R3 + R4
R1 R2 R3 R4
V1 V2 V3 V4
I I
V
RI I
V
18.5 Resistance In Series Circuit
R = R1 + R2 + R3 + R46. For resistors in series, the
combined /total/effective resistance is the SUM of the individual resistances.
7. The effective resistance is GREATER than any of the individual resistance.
Example 3
12 V
1) Calculate the effective resistance of the circuit.
2) What is the current passing through the resistor?
0.5 3 2.5
R = 0.5 + 3 + 2.5 = 6
I = V/R = 12/6 = 2 A
Examples on Series CircuitExample 4:
Find the combined resistance of the three resistors in series given that R1 = 1 , R2 unknown and R3 = 2 The current I recorded in the ammeter is 1 A and the voltmeter reading V across R2 = 3 V.
I
R2
A
R3 R1
V
R2 = V / I
= 3 / 1 = 3
Combined resistance = R1 + R2 + R3
= 1 + 3 + 2 = 6
I = 1 A
R2
A
R3 (2) R1 (1)
V3 V
Examples on Series CircuitExample 5:
A cell of e.m.f. 1.5 V was connected in series with two resistors, as shown below
1.5 V
6 4
Calculate
i) the effective resistance of the circuit,
ii) the current flowing in the circuit,
iii) the potential difference across the 4 resistor.
1.5 V
6 4
i) Effective resistance, R = R1 + R2 = 4 + 6 = 10
ii) Current flowing in circuit, I = V / R = 1.5 / 10
= 0.15 A
1.5 V
6 4
iii) P.d across 4 resistor,
V4 = IR
= 0.15 4
= 0.6 V
1.5 V
6 4 0.15 A
WAIT! That is not all….
What is the p.d across the 6 resistor?
18.6 Current In Parallel Circuit
In a parallel circuit, the current from the source is shared by 2 or more branches.
Lamp 1 Current from source is shared by lamp 1 and 2
Lamp 2 A B
Current In Parallel Circuit
I1 = I2 + I3 = I4
Lamp 1
Lamp 2
I1
I2
I3
I4
Current In Parallel Circuit
1. The SUM of the currents in the separate branches of a parallel circuit is EQUAL to the current from the source.
I1 = I2 + I3 = I4
Current In Parallel CircuitPoints to note:
2. In a parallel circuit, the component with the SMALLEST resistance will allow the LARGEST current to pass.3. I1 = I2 if only the lamps are identical i.e. having the same resistance.
Example 8
Which ammeter shows a faulty reading?
4A
2A
2A
4A
2A
PS
R
Q
T
P.D In Parallel Circuit
V3
V2
V1
4. Each component joined in parallel have the same potential difference across it.
V1 = V2 = V3
Example 9
Fill in the blanks using = , or <.
V1 _____V2 I1 ______I2
I1
I2
10
5
V1
V2
Resistance In Parallel Circuit
I = I1 + I2
V = V1 = V2
Let R be the combined/total/ effective resistance of the 2 resistors.
R1
R2
V2
V1
V
I
I1
I2
Resistance In Parallel Circuit
The arrangement can then be simplified from this … to…
R
V
V
I
I R1
R2
V2
V1
V
I
I1
I2How is R related to R1 and R2?
Resistance In Parallel Circuit
4. I = I1 + I2
R1
R2
V2
V1
V
II1
I2
R
V
V
I
I
2
2
1
1
R
V
R
V
R
V
Resistance In Parallel Circuit
21
111
RRR
5. For resistors in parallel, the reciprocal of the combined /total/effective resistance is the SUM of the reciprocal of individual resistances.
6. The effective resistance is SMALLER than any of the individual resistance.
Example 10 What is the effective resistance of the three resistors?
Let effective resistance be R
1/R = 1/3 +1/6 + 1/6
= 2/3
R = 3/2 = 1.5
3
6
6
Examples on Parallel Circuit
3.0
6.0
4.0 V
The circuit diagram shows a 6.0 resistor and a 3.0 resistor in parallel and connected to a 4.0 V battery.
Example 11:
3.0
6.0
4.0 V
1) Calculate the effective resistance of the parallel resistors.
1/R = 1/3.0 + 1/6.0
= 1/2.0
R = 2.0
3.0
6.0
4.0 V
Effective resistance = 2.0
2) Calculate the current flowing through the battery.
Ibattery = V/R = 4.0/2.0
= 2.0 A
3.0
6.0
4.0 V
3) Calculate the current flowing through in each resistor.
P.d across the resistors = 4.0 V
I3.0 = V/R = 4.0/3.0 = 1.3 A
I6.0 = Ibattery - 1.3 = 2.0 – 1.3 = 0.7 A
3.0
6.0
4.0 V
4) A third resistor is connected in parallel with the original pair. Is the current through the battery larger, smaller or the same as before. Explain. Current will be larger because the effective resistance of the circuit is smaller than before.
Examples on Parallel Circuit
Example 12:
Textbook pg 328 Example 18.3
Series + Parallel Circuit
AR1 R2
R3
R2 is connected in parallel with R3.
R1 is connected in series with the ammeter and the resistor combination of R2 and R3.
Resistance In Series + Parallel Circuit
Example 13:
What is the effective resistance of the circuit below?
5
3
3
Resistance In Series + Parallel Circuit
First, add the resistors in parallel.
Let R1 be the effective resistance of the resistors in parallel.
1/R1 = 1/3 + 1/3
R1 = 1.5 5
3
3
Resistance In Series + Parallel Circuit
This is then reduced to two resistors in series.
5
3
3
5 R1 = 1.5
Resistance In Series + Parallel Circuit
Effective resistance of circuit,
R = 5 +1.5
= 6.5
5 R1 = 1.5
Examples on Series + Parallel Circuit
Example 14:
The diagram shows three resistors connected to a 6.0 V battery supply
A6.0
6.0
8.0 6.0 V
1) Calculate the combined resistance of the 8.0 and the 6.0 resistors in series. Let combined resistance be Rc
Rc = 8.0 + 6.0 = 14.0
A6.0
6.0
8.0 6.0 V
2) Calculate the effective resistance of the circuit.
Let effective resistance of circuit be R
1/R = 1/14.0 + 1/6.0 = 5/21
R = 21/5 = 4.2
A6.0
6.0
8.0 6.0 V
3) Calculate the current through the ammeter. Effective resistance of circuit = 4.2
Current through ammeter = Current in circuit = V/R = 6.0/4.2
= 1.4 A
A6.0
6.0
8.0 6.0 V
Examples on Series + Parallel Circuit
Example 15:
Textbook pg 329 Example 18.4
Examples on Series + Parallel Circuit
Example 16:
Textbook pg 334 Q3
Example 17:
Textbook pg 335 Q4
Short Circuit
Short Circuit occurs in a closed circuit when there is an alternative path for current to flow. This alternative path is of a much lower resistance than the original path.
Short Circuit
The lamp will not light up because, compared to the copper wire, it offers more resistance and thus current by-passed it.
ALamp
Ammeter
CellI I
II Copper wire
Short Circuit
During a short circuit, the current flowing is very large as the resistance is very small. This can be dangerous as a large current may cause heating resulting in a fire.
Example 18:
Textbook pg 334 Q2
Examples on Series + Parallel Circuit