8866 H1 Physics – J2/2011
7. D.C. Circuits
7. D.C. CIRCUITS
Content
• Practical circuits
• Series and parallel arrangements
Learning Outcomes
Candidates should be able to:
(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.
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8866 H1 Physics – J2/2011
7. D.C. Circuits
Types of electric current
•
Direct Current (D.C.):
•
Alternating Current: Flow of charges in the circuit reverses direction at
regular intervals (e.g. current from household mains).
Flow of charges in the circuit is in the same direction
all the time, from a higher potential to a lower potential
(e.g. current from battery).
Electric circuits consist of circuit components (e.g. batteries, resistors, and switches)
connected by conductors (e.g. copper cables).
For electric current to flow, the circuit components and conductors must form closed
loops. There must also be sources of electrical energy (e.g. batteries) and sinks of
electrical energy among the circuit components (e.g. resistors and lamps).
(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).
(a)
Recall and
use
Electrical Circuit Symbols
Symbol modifier
variable value
inherent non-linearity
preset value
Conductors and terminals
with connection between conductors
current-carrying conductor
single pole single
throw (SPST) switch
no connection between conductors
open terminals
single pole double
throw (SPDT) switch
ground/earth
double pole double
throw (DPDT) switch
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8866 H1 Physics – J2/2011
7. D.C. Circuits
D.C. sources
cell
battery (>1 cell)
d.c. source
photovoltaic cell
variable d.c. source
Measuring instruments
ammeter and milliammeter
voltmeter and millivoltmeter
galvanometer (alphabetical symbol, null deflection, current detected in
certain direction, current detected in another direction)
Resistors
fixed resistor
variable resistor (rheostat)
light-dependent resistor (LDR)
thermistor
potentiometer (voltage divider)
Other circuit components
diode
photodiode
light-emitting diode (LED)
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8866 H1 Physics – J2/2011
fuse
heating element
7. D.C. Circuits
indicator, lamp or light source
Other symbols found in TYS
generator
motor
loudspeaker
thermocouple
oscilloscope
aerial/antenna
electric bell
buzzer
(b)
draw and interpret circuit diagrams containing sources, switches, resistors,
ammeters, voltmeters, and/or any other type of component referred to in the
syllabus.
Note that, for a certain electric circuit, there are different ways of drawing its
circuit diagram.
Actual circuit
(b)
Draw and
interpret
Circuit diagram
Two other possible circuit diagrams for the above electric circuit are as
follows:
V
R2
R1
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8866 H1 Physics – J2/2011
Example 1:
7. D.C. Circuits
Draw the circuit diagram for the electric circuit shown below.
Actual circuit
Circuit diagram
Conservation of charge (must know in order to solve circuit problems)
Given that we are dealing with steady currents (i.e. no accumulation of charge at
circuit junctions), the sum of currents entering a circuit junction is equal to the sum of
currents leaving it.
I1 + I 2 = I 3 + I 4 + I 5
Taking currents entering circuit junction as positive and
currents leaving circuit junction as negative, we have:
I1 + I 2 − I 3 − I 4 − I 5 = 0
Taking currents leaving circuit junction as positive and
currents entering circuit junction as negative, we have:
I 3 + I 4 + I 5 − I1 − I 2 = 0
Example 2
The given diagrams show wires carrying currents I1, I2, I3, and I4, meeting at a junction.
Which of the following diagrams represents the equation I1 + I2 = I3 + I4?
A
B
C
D
I1 + I3 + I4 + = I2
I 1 + I 2 + I 3 = I4
I 1 + I 2 = I 3 + I4
(correct answer)
Impossible, all
currents are entering
and no current
leaving.
Conservation of energy (must know in order to solve circuit problems)
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8866 H1 Physics – J2/2011
7. D.C. Circuits
The algebraic sum of e.m.f. (i.e. sources of electrical energy) is equal to the algebraic
sum of p.d. (i.e. sinks of energy) for any closed loop within the circuit.
Example 3
Find I1 and I2 in terms of E1, E2, R1, R2 and R3.
Given that
E1 = 3.0 V
E2 = 1.5 V
R1 = R2 = R3 = 10 Ω
find the values of I1 and I2.
Solution:
E1 + E 2 = I1R1 + I1R2 = I1 (R1 + R2 )
Conservation of energy for the loop:
I1 =
E1 + E 2
R1 + R 2
E1 + E2 = I 2 R3
Conservation of energy for the loop:
I2 =
E1 + E 2
R3
Substituting E1 = 3.0 V, E2 = 1.5 V and R1 = R2 = R3 = 10 Ω,
I1 = 0.23 A and I2 = 0.45 A.
(c)
solve problems using the formula for the combined resistance of two or more
resistors in series.
Derivation of the effective resistance of resistors in series (for reference only):
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8866 H1 Physics – J2/2011
7. D.C. Circuits
For series connection, the effective resistance is the sum of individual resistance.
In general, for n similar resistors (each of resistance R) connected in series, the
effective resistance is Reff = nR .
Note: The effective resistance always increases when additional resistors are
connected in series.
1. Same water current flowing through each 2. Resistance to water flow increases
section of the tube, whether wide or
as number of narrow portions along
narrow (water cannot be compressed, no
the
tube
increases
(adding
accumulation, inflow = outflow).
obstruction to water flow).
Example 4
Calculate the effective resistance of a 4 Ω and two 3 Ω resistors connected in series.
Solution:
Reff = 4 + (2)(3) = 10 Ω
(d)
solve problems using the formula for the combined resistance of two or more
resistors in parallel.
Derivation of the effective resistance of resistors in parallel (for reference only):
For parallel connection, the reciprocal of effective resistance is the sum of
reciprocal of individual resistance.
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8866 H1 Physics – J2/2011
7. D.C. Circuits
In general, for n similar resistors (each of resistance R) connected in parallel, the
effective resistance is
R
1
n
= ⇒ Reff = .
n
Reff R
Note: The effective resistance always decreases when additional resistors are
connected in parallel.
The effective resistance of resistors in parallel is always less than the individual
resistance of each resistor.
1. The water current through the wide tube 2. Resistance to water flow decreases
as number of narrow tubes
is the same as the sum of the water
increases (adding channels for water
currents in each of the narrow tubes
flow).
(water cannot be compressed, no
accumulation, inflow = outflow).
Example 5
Calculate the effective resistance of a 2 Ω, a 3 Ω and a 4 Ω resistor connected in
parallel.
Solution:
1
1 1 1 6 + 4 + 3 13
=
= + + =
12
12
Reff 2 3 4
Reff = 0.903 Ω
(e)
solve problems involving series and parallel circuits for one source of e.m.f.
Example 6
A battery C of 1.5 V and negligible internal resistance is connected to the combination
of resistors as shown. Find the values of the currents I1, I2, and I3 in the diagram.
10 Ω
5Ω
C
I1
10 Ω
I1 =
5I 3 = 15I 2
5Ω
I3
1.5
= 0.109
10 + 3.75
= 0.11 A
⇒ I 3 = 3I 2
I2
I1 = I 2 + I 3 = I 2 + 3I 2 = 4I 2
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8866 H1 Physics – J2/2011
7. D.C. Circuits
10 Ω
C
⇒ I2 =
I1 0.109
=
= 0.0273
4
4
= 0.027 A
I 3 = I1 − I 2 = 0.109 − 0.0273 = 0.0817
= 0.082 A
3.75 Ω
I1
More worked examples
Find the equivalent resistances between points A and point B in Examples 7, 8, and 9.
Assume that all resistances in the circuits are 2 Ω each.
Example 7
E
B
A
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8866 H1 Physics – J2/2011
7. D.C. Circuits
Solution:
Step 1: Assign potential at each terminal of each resistors.
E
VB
VA
VC
VA
VF
VD
Step 2: Group resistors which have same potential at its terminals.
E
VA
VC
VB
VA
VF
VD
Step 3: Simplify the circuit
E
VA
VB
2 Ω //2 Ω
=
1
1 1
+
2 2
2 Ω //4 Ω
1Ω
=1Ω
VF
2 Ω
2 Ω
1.33 Ω
=
1
1 1
+
2 4
= 1.33 Ω
VD
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8866 H1 Physics – J2/2011
7. D.C. Circuits
E
VA
VB
3 Ω //2 Ω
=
1
1 1
+
3 2
= 1.2 Ω
1.33 Ω
1.2 Ω
VD
Hence, effective resistance between point A and point B = 1.2 + 1.33 = 2.53 Ω
Example 8
A
E
B
Solution:
Step 1: Assign potential at each terminal of each resistors
VF
VA
VD
E
VB
VC
VC
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8866 H1 Physics – J2/2011
7. D.C. Circuits
Step 2: Group resistors which have same potential at its terminals.
VF
VA
VD
E
2 Ω //2 Ω //4 Ω
=
VB
VC
VC
1
= 0.8 Ω
1 1 1
+ +
2 2 4
Step 3: Simplify the circuit
2 Ω //(2 Ω + 0.8 Ω )
=
1
1
1
+
2 2.8
= 1.2 Ω
Hence, effective resistance between point A and point B = 2 + 1.2 = 2.2 Ω
Example 9
A
B
E
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8866 H1 Physics – J2/2011
7. D.C. Circuits
Solution:
Step 1: Assign potential at each terminal of each resistors
VB
VC
VB
VA
VB
E
Step 2: Group resistors which have same potential at its terminals.
VB
VC
VB
VA
VB
E
Step 3: Simplify the circuit
VB
0 Ω (short circuit)
VA
VB
Hence, effective resistance between point A and point B = 2 Ω
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