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series and parallel - PEARSON

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Heinemann Physics 1 4
​e
Chapter 4 Practical electrical circuits
Section 4.1 Series and parallel circuits
Worked example: Try yourself 4.1.1
Calculating an equivalent series resistance
A string of Christmas lights consists of 20 light bulbs connected in series. Each bulb has a resistance of 8 Ω. Calculate
the equivalent series resistance of the Christmas lights.
Thinking
Working
Recall the formula for equivalent series resistance.
RT = R1+ R2+ … + Rn
Substitute in the given values for resistance.
RT = 8 + 8 + 8 + …
As there are 20 equal globes, the problem is easier to
answer by multiplying 8 Ω by 20 globes.
= 20 × 8
= 160 Ω
Worked example: Try yourself 4.1.2
USING EQUIVALENT SERIES RESISTANCE FOR CIRCUIT ANALYSIS
Use an equivalent series resistance to calculate the current flowing in the series circuit below and the potential
difference across each resistor.
12 V
100 Ω
690 Ω
330 Ω
Thinking
Working
Recall the formula for equivalent series resistance. Find
the total resistance In the circuit.
RT = R1 + R2 + R3 + …
Use Ohm’s law to calculate the current in the circuit.
Whenever calculating current in a series circuit, use RT
and the voltage of the power supply.
Use Ohm’s law to calculate the potential difference across
each separate resistor.
= 100 + 690 + 330 = 1120 Ω
I=
V
R
=
12
1120
= 0.011 A
V = IR
Therefore:
V1 = 0.011 A × 100 Ω = 1.1 V
V2 = 0.011 A × 690 Ω = 7.6 V
V3 = 0.011 A × 330 Ω = 3.6 V
Use the loop rule to check the answer.
VT = V1 + V2 + V3
= 1.1 + 7.6 + 3.6
= 12.3 V
Since this is approximately the same as the voltage
provided by the cell, the answer is reasonable.
Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd)
ISBN 978 14 8861 12​6 ​1
Heinemann Physics 1 4
​e
Worked example: Try yourself 4.1.3
CALCULATING AN EQUIVALENT PARALLEL RESISTANCE
A 20 Ω resistor is connected in parallel with a 50 Ω resistor. Calculate the equivalent parallel resistance.
Thinking
Working
Recall the formula for equivalent effective resistance.
1
RT
Substitute in the given values for resistance.
1
RT
Solve for RT.
1
RT
=
=
=
=
=
RT =
1
R1
1
+
1
1
5
1
50
2
+
100
+…
50
+
20
1
R3
1
+
20
+
R2
100
7
100
100
7
= 14.3 Ω
Worked example: Try yourself 4.1.4
USING EQUIVALENT PARALLEL RESISTANCE FOR CIRCUIT ANALYSIS
10 V
se an equivalent parallel resistance to calculate the
current flowing in the parallel circuit below and through
each resistor of the circuit.
30 Ω
50 Ω
Thinking
Working
Recall the formula for equivalent parallel resistance.
1
RT
Substitute in the given values for resistance.
1
RT
Solve for RT .
1
RT
=
=
=
=
=
RT =
1
R1
1
+
1
30
1
30
+…
50
+
5
1
R3
1
+
150
+
R2
1
50
+
3
150
8
150
150
8
= 19 Ω
Use Ohm’s law to calculate the current in the circuit. To
calculate I, use the voltage of the power supply and the
total resistance.
Use Ohm’s law to calculate the current through each
resistor. Remember that the voltage through each resistor
is the same as the voltage of the power supply, 10 V in
this case.
Use the junction rule to check the answers.
Icircuit =
V
R
=
10
19
= 0.53 A
30 Ω resistor: I30 =
50 Ω resistor: I50 =
V
R
V
R
=
=
10
30
10
50
= 0.33 A
= 0.2 A
Icircuit = I30 + I50
0.53 A = 0.33 A + 0.20 A
Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd)
ISBN 978 14 8861 12​6 ​1
Heinemann Physics 1 4
​e
Worked example: Try yourself 4.1.5
COMPLEX CIRCUIT ANALYSIS
Calculate the potential difference across, and current through, each resistor in the circuit below.
R1 = 20.0 Ω
R5 = 5.0 Ω
R6 = 10.0 Ω
100.0 V
R7 = 30.0 Ω
R2 = 50.0 Ω
R3 = 25.0 Ω
R4 = 100.0 Ω
Thinking
Working
Find an equivalent parallel resistance for each
of the parallel groups of resistors.
For the 10 Ω and 5 Ω resistor
group:
1
R5−6
=
=
=
1
+
10
1
+
10
Similarly, for the group of three
parallel resistors:
1
1
5
R2−4
2
=
10
3
=
10
R5−6 =
10
R2−4 =
3
= 3.33 Ω
Find an equivalent series resistance for the
circuit. Note that the circuit can now be
thought of as containing four resistors in series.
Use Ohm’s law to calculate the current in
the circuit. Use the supply voltage and total
resistance to do this calculation.
Use Ohm’s law to calculate the potential
difference across each resistor (or parallel
group of resistors) in series.
=
1
+
25
4
100
1
50
+
+
2
100
1
100
+
1
100
7
100
100
7
= 14.3 Ω
RT = 20 + 3.3 + 30 + 14.3 = 67.6 Ω
V = IR
∴R=
V
I
=
100
67.6
= 1.48 A
V = IR
∆V1 = 1.48 × 20.0 = 29.6 V
∆V5−6 = 1.48 × 3.3 = 4.9 V
∆V7 = 1.48 × 30.0 = 44.4 V
∆V2−4 = 1.48 × 14.3 = 21.2 V
Check:
29.6 + 4.9 + 44.4 + 21.2 ≈100 V
(with some slight rounding error).
Use Ohm’s law where necessary to calculate
the current through each resistor.
I1 = I7 = 1.48 A
I=
V
R
4.9
I5 =
I6 =
5
4.9
10
= 0.98 A
= 0.49 A
Check: 0.98 + 0.49 ≈1.48 A
(with some slight rounding
error).
This confirms that the junction
rule holds for this section.
Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd)
I2 =
I3 =
I4 =
21.2
50
21.2
25
21.2
100
= 0.42 A
= 0.85 A
= 0.21 A
Check:
0.42 + 0.85 + 0.21 = 1.48 A
This confirms that the junction
rule holds for this section.
ISBN 978 14 8861 12​6 ​1
Heinemann Physics 1 4
​e
Worked example: Try yourself 4.1.6
COMPARING POWER IN SERIES AND PARALLEL CIRCUITS
Consider a 200 Ω and a 800 Ω resistor wired in parallel with a 12 V cell. Calculate the power drawn by these resistors.
Compare this to the power drawn by the same two resistors when wired in series.
Thinking
Working
Calculate the equivalent resistance for the parallel circuit.
1
RT
1
=
R1
1
=
1
RT
+
200
4
=
800
1
+…+
R2
+
+
1
Rn
1
800
1
800
5
=
800
RT = 160 Ω
Calculate the total current drawn by the parallel circuit.
V=IR
I=
=
V
R
12
160
= 0.075 Ω
Use the power equation to calculate the power drawn by
the parallel circuit.
P = VI
= 12 × 0.075
= 0.9 W
Calculate the equivalent resistance for the series circuit.
RT = R1 + R2
= 200 + 800
= 1000 Ω
Calculate the total current drawn by the series circuit.
V = IR
I=
=
V
R1
12
1000
Use the power equation to calculate the power drawn by
the series circuit.
P = VI
Compare the power drawn by the two circuits.
Pparallel
= 0.012 A
= 12 × 0.012 = 0.144 W
Pseries
=
0.9
0.144
= 6.25
The parallel circuit draws 6.25 times as much power as
the series circuit.
Section 4.1 Review
KEY QUESTIONS SOLUTIONS
1
B
RT = R1 + R2
= 20 + 20 = 40 Ω
1=
=
V
R1
6
40
= 0.15 A
V across each resistor:
V = IR = 0.15 × 20 = 3 V
(Or, as the resistors are equal, the same voltage will be lost across each and will add to 6 V, so 3 V must be lost across
each resistor.)
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ISBN 978 14 8861 12​6 ​1
Heinemann Physics 1 4
​e
2
a RT = R1 + R2 + R3
= 100 + 250 + 50 = 400 Ω
V
IT =
3
=
RT
= 0.0075 A or 7.5 mA
400
b R = 100 Ω and I = 0.0075 A
V100 = IR
= 0.0075 × 100 = 0.75 V
1
3
RT
1
=
1
+
R1
R2
1
=
R1
+
1
R1
(the resistors are identical, so R1 = R2)
2
=
RT
R1
∴ RT =
2
R1 = 2 × RT = 2 × 68 = 136 Ω
4
a
1
RT
1
=
=
=
=
1
+
R1
1
20
1
20
R2
1
+
10
2
+
20
3
20
20
RT =
3
= 6.67 Ω
V
I =
R
IT =
5
6.67
= 0.75 A
b I20 =
c I20 =
5
V20
5
=
R
V10
20
5
=
R
10
= 0.25 A
= 0.5 A
a V = IR
V40 = 0.3 × 40 = 12 V (300 mA = 0.3 A)
Since the components are in parallel the voltage across the 40 Ω resistor (or the 60 Ω resistor) is also the voltage of
the battery.
b I60 =
6
V60
12
=
R
60
= 0.2 A (or 200 mA)
First determine the total resistance of the circuit:
1
R3−4
1
=
=
1
+
R3
1
10
R4
1
+
10
R3−4 = 5 Ω
RT = 20 + 15 + 5 = 40 Ω
IT =
VT
RT
=
12
40
= 0.3 A (or 300 mA)
I1 = I2 = IT = 0.3 A (since these are in series)
V1= I1R1 = 0.3 × 20 = 6 V
V2 = I2R2 = 0.3 × 15 = 4.5 V
V3 = V4 = I3−4R3−4 = 0.3 × 5 = 1.5 V
I3 = I4 =
7
V3
R3
=
V4
R4
=
1.5
10
= 0.15 A
Rtop row = 3 + 4 = 7 Ω
Rbottom row = 5 + 6 = 11 Ω
1
Rparallel
=
1
7
+
1
11
=
11
77
+
7
77
=
18
77
Rparallel = 4.278 Ω
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ISBN 978 14 8861 12​6 ​1
Heinemann Physics 1 4
​e
8
a RT = R1 + R2 + … + Rn
= 20 + 20 + 20 + 20
= 80 Ω
V = IR
∴I=
=
V
R
10
80
= 0.125 A
P = VI
= 10 × 0.125
= 1.25 W
b
1
RT
=
=
=
RT
1
=
1
R1
1
20
4
+
+
1
R2
1
20
…+
+
1
20
1
Rn
+
1
20
20
20
4
RT = 5 Ω
V = IR
∴I=
V
R
=
10
5
=2A
P = VI
= 10 × 2
= 20 W
9
C. Parallel wiring allows each appliance to be switched on and off independently (and also receive mains
voltage supply).
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ISBN 978 14 8861 12​6 ​1
Heinemann Physics 1 4
​e
Section 4.2 Using electricity
Worked example: Try yourself 4.2.1
VOLTAGE DIVIDER
A voltage divider is constructed from a 12 V battery, a 40 Ω resistor and a 20 Ω resistor as shown. Calculate the
voltage output, Vout, of the circuit.
12 V
40 Ω
+
R1
Vin
20 Ω
R2
Vout
Thinking
Working
Calculate the total resistance of the circuit.
RT= 40 + 20
= 60 Ω
Calculate the current flowing through the circuit.
V
I=
=
R
12
60
= 0.2 A
Calculate the potential difference across the second
resistor.
Vout = IR
= 0.2 × 20
=4V
Worked example: Try yourself 4.2.2
THERMISTOR IN A VOLTAGE DIVIDER CIRCUIT
A voltage divider circuit includes a thermistor (R1) and a fixed resistor (R2). The characteristic curve of the thermistor and
the circuit are shown in Figure 4.2.7. Using the graph and the information included on the circuit diagram, determine
the following:
a
the resistance of the thermistor at 20°C
Thinking
Working
The resistance of the thermistor can be read straight from
the graph at the point where the temperature is 20°C.
R = 20 kΩ
b the current in the circuit
Thinking
Working
Find the effective resistance of the circuit, note that the
fixed resistor is 5 kΩ.
RT = 20 kΩ + 5 kΩ
Find the current.
= 25 kΩ
I=
=
V
R
5
25 000
= 0.20 mA
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ISBN 978 14 8861 12​6 ​1
Heinemann Physics 1 4
​e
c
the output potential difference, Vout.
Thinking
Working
Use Ohm’s Law to calculate the voltage across the fixed
resistor.
V = IR
= 0.0002 × 5 000
=1V
Worked example: Try yourself 4.2.3
LDR IN A VOLTAGE DIVIDER CIRCUIT
A voltage divider circuit includes an LDR (R1) and a fixed resistor (R2).
Using the information included on the circuit diagram in Worked example 4.2.3 and the fact that the LDR has a
resistance of 3.0 kΩ at a light intensity of 2000 mW m−2, determine the following:
a
the total resistance of the circuit
Thinking
Working
The total resistance equals the resistance of the LDR plus
the 6.8 kΩ fixed resistor.
R = 3.0 + 6.8 = 9.8 kΩ
b the current in the circuit
Thinking
Working
Find the current using Ohm’s law.
I=
=
V
R
9
9800
= 0.92 mA
c
the output potential difference, Vout.
Thinking
Working
Use Ohm’s Law to calculate the voltage across the fixed
resistor.
V = IR
= 0.00092 × 6800
= 6.3 V
Worked example: Try yourself 4.2.4
DIODES
A student is investigating the current−voltage characteristics of a diode using the same circuit diagram and I−V graph
shown in Worked example 4.2.4.
a
What is the potential difference across the diode?
Thinking
Working
The potential difference across the diode can be read
directly from the I−V graph.
At 5.0 mA the voltage across the diode will be 0.6 V.
b What is the value of R1?
Thinking
Working
The battery voltage is 9.0 V, the voltage drop across
the diode is 0.6 V and the current is 5.0 mA. Use this
information to find R1.
voltage across R1 = 9.0 − 0.6
resistance R1 =
=
V
I
= 8.4 V
8.4
5.0 × 10−3
= 1680 Ω
Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd)
ISBN 978 14 8861 12​6 ​1
Heinemann Physics 1 4
​e
Worked example: Try yourself 4.2.5
LEDs
An LED has the following optimum operating characteristics:
Id = 30 mA when Vd = 1.9 V
Use the diagram from Worked example 4.2.5 to determine the value of the current-limiting resistor (R) if the current
through the diode is to be limited to 30 mA when powered by a 6 V battery
Thinking
Working
Voltage divider circuit so the 6 V is divided between the
resistor, R, and the LED.
VR = 6 − 1.9 = 4.1 V
Use Ohm’s Law to calculate the value of R.
R=
=
V
I
4.1
30 × 10−3
= 137 Ω
Section 4.2 Review
KEY QUESTIONS SOLUTIONS
1
b A
a C
RT = 100 + 300 = 400 Ω
RT = 100 + 300 = 400 Ω
V = IR
V = IR
I=
=
V
I=
R
10
=
400
= 0.025 × 100
= 0.025 × 300
= 2.5 V
= 7.5 V
Alternatively:
R300
2
R300 + R400
400
= 0.025 A
V100 = IR100
= 0.025 A
V300 = IR300
Vout =
V
R
10
× Vin =
300
400
Alternatively, V100 + V300 = 10 V
× 10 = 7.5 V
∴ V100 = 10 − V300 = 10 − 7.5 = 2.5 V
RT = 1000 + 2000 = 3000 Ω
V = IR
I=
=
V
R
24
3000
= 0.008 A
V2000 = IR2000
= 0.008 × 2000
= 16 V
3
Input transducer
Signal-processing component
Output transducer
LDR / microphone / thermistor
diode / potentiometer
LED / light globe / speaker
4
A and B. Diodes only allow current flow in one direction. An LED is a type of diode.
5
B. As the temperature of a PTC (Positive Temperature Coefficient) thermistor increases, its resistance increases.
6
B. A thermistor could be used as an input transducer in a circuit to control the temperature inside a refrigerator
because its resistance varies with temperature.
7
A. As the amount of light falling on an LDR increases, the resistance of the LDR decreases.
8
At minimum operating level
VL = Vin − VLED = 9 − 2 = 7 V
RL=
VL
I
=
7
20 × 10−3
= 350 Ω
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ISBN 978 14 8861 12​6 ​1
Heinemann Physics 1 4
​e
9
a Circuit (i): In parallel, required current = 3 × 20 mA = 60 mA
VL = Vin − VLED = 9 − 2 = 7 V
RL =
VL
I
=
7
60 × 10−3
= 117 Ω
Circuit (ii): In series, required current = 20 mA
VL = Vin − 3 × VLED = 9 − 6 = 3 V
RL =
VL
I
=
3
20 × 10−3
= 150 Ω
b Both circuits will emit the same light—all are operating at the same V and I.
c Circuit (ii) requires only 20 mA, so will run longer.
10 a From the graph, RT = 500 Ω at 20°C
b VT = 10 − 2.5 = 7.5 V, IT =
7.5
500
= 0.015 A
since ILED = 0.011, IR = IT − ILED = 0.004 A and RR =
c VR = VLED = 2 V since the two are in parallel
Hence IR =
2
625
2.5
0.004
= 625 Ω
= 0.0032A
And IT = 0.0032 + 0.0048 = 0.008 A
VT = 10 − 2.0 = 8.0 V
Hence RT =
8.0
0.008
= 1000 Ω
Using this value on the graph supplied, T = 10°C
Section 4.3 Electrical safety
Worked example: Try yourself 4.3.1
CALCULATING THE COST OF ELECTRICITY
A 2500 W iron is used for 2.5 hours. Assume the price for household electricity is 26 cents per kW h. How much
would it cost (to the nearest cent) to use this iron for 2.5 hours?
Thinking
Working
Convert the power consumption of the appliance to kW.
2500 W = 2.5 kW
Use the appropriate equation to multiply the power of the
appliance in kW by the number of hours it operates.
E = Pt
= 2.5 kW × 2.5 h
= 6.25 kWh
Multiply the number of kWh by the cost per kWh. Give
your answer correct to two decimal places.
Cost = 6.25 × 0.26
(26 cents = $0.26)
= $1.63
Section 4.3 Review
KEY QUESTIONS SOLUTIONS
1
A. In the event of an electrical fault the current will rapidly increase through the zero resistance path offered by the
earth connection. Once this current exceeds the fuse’s rating the fuse will blow, shutting off power to the appliance.
2
D. Double insulated appliances usually do not have an earth connection.
3. 1 kWh = 1000 W × 3600 s
= 3 600 000 J
10 kWh = 10 × 3 600 000
= 3.6 × 107 J
4
This air conditioner would cost 0.75 × 5 × 0.27 = approximately $1 to run for 5 hours. Therefore, the figure $10 in the
statement is incorrect.
5
The neutral and earth are common.
6
It is much safer to place the fuse in the active circuit because then it cuts off the supply to the circuit.
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Heinemann Physics 1 4
​e
7
The earth stake ensures that the neutral and earth conductors are at zero potential.
8
The toaster will work normally, but the connection is very unsafe because it will remain live even when switched off.
9
The outer casing of the appliance could become live.
10 I =
240 V
= 2.4 mA
1.0 × 105 Ω
Chapter 4 Review
1
A. The equivalent resistance of series resistors is the sum of their individual resistances.
2
a RT = Rparallel pair + R3
∴ R3 = RT − Rparallel pair = 8.5 − 5 = 3.5 Ω
b I3 = IT =
VT
RT
3
=
8.5
= 0.35(3) A
c V3 = I3 × R3 = 0.35 × 3.5 = 1.2 V
Vparallel pair = 3 − 1.2 = 1.8 V
V2
d I2 =
1.8
=
R2
15
= 0.12 A
e I1 = IT − I2 = 0.35 − 0.12 = 0.24 A
f R1 =
3
V1
1.76
=
I1
= 7.46 Ω
0.24
a Ammeter. The meter is connected in series so it must be an ammeter.
1
b
=
Rtop parallel group
1
40
+
1
40
=
2
40
Rtop parallel group = 20 Ω
1
Rbottom parallel group
=
1
20
+
1
60
=
3
60
+
1
60
=
4
60
Rbottom parallel group = 15 Ω
1
Rtotal
=
3
=
60
1
20
+
Rtotal =
+
4
1
15
=
60
60
7
60
= 8.57 Ω
7
4
The earth wire is usually connected to the metal casing of an electrical appliance. If the insulation around the wire
inside the appliance becomes degraded, the casing of the appliance could become ‘live’ and dangerous to touch. In
this situation, the earth wire provides an alternative low-resistance path to earth, protecting any users of the appliance
from electrocution.
5
The circuit will need to have either two pairs of series resistors connected in parallel or two pairs of parallel resistors
connected in series.
6
RT = R1 + R2 = 600 + 1200 = 1800 Ω
I=
V
R
=
9
1800
= 0.005 A
V600 = IR
= 0.005 × 600
=3V
7
B
Vout = IR400
8 = I × 400
I = 0.02 A
VR = IRR
12 = 0.02 × RR
R = 600 Ω
Alternatively,
R: 400 Ω = 12 V : 8 V
∴ R = 400 ×
= 600 Ω
12
8
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Heinemann Physics 1 4
​e
8
C. An LDR is a type of input transducer.
9
a RT = R1 + R2 + R3 = 20 + 20 + 20 = 60 Ω
V = IR
V
∴I=
=
R
12
= 0.2 A
60
P = VI = 12 × 0.2 = 2.4 W
b
1
1
=
RT
20
RT =
I=
+
R1
=
R
+
1
R3
=
1
20
+
1
20
+
1
20
=
3
20
= 6.67 Ω
3
V
1
R2
12
= 1.8 A
6.67
P = VI = 12 × 1.8 = 21.6 W
c
Pparallel
Pseries
=
21.6
2.4
= 9 The parallel circuit draws 9 times more power.
10 D. A 50 mA current for over 4.5 s is likely to cause severe shock and possible death.
11
E = Pt
=3×4
= 12 kWh
Cost = 12 × 0.30
= $3.60
12 D: Power is how quickly energy is consumed/supplied/transformed.
1 watt = 1 joule per second
13 a
1
RT
=
=
600
10
=
RT =
b
1
100
6
600
600
10
1
+
+
200
3
600
+
+
1
600
1
600
= 60 Ω
V = IR
120 = I × 60
I = 2.0 A
c Each branch has 120 V across it
I1 =
I2 =
I3 =
V
RT
V
RT
V
RT
=
=
=
120
100
120
200
120
600
= 1.2 A
= 0.6 A
= 0.2 A
I1 = 1.20 A, I2 = 0.60 A, I3 = 0.20 A
d P = VI
= 120 × (1.2 + 0.6 + 0.2)
= 240 W
e 240 W
14 a 4 V. This is calculated using Ohm’s Law, or alternatively, by recognising that the voltage drop at Vout is half the
voltage drop across the LDR.
V = IR
12 = I × 300
I = 0.04 A
V = IR
= 0.04 × 100
=4V
b above; as light increases, the resistance of the LDR decreases, hence Vout rises
c Vout approaches zero, as the LDR has increased resistance and therefore the voltage drop across the LDR
approaches 12 V.
Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd)
ISBN 978 14 8861 12​6 ​1
Heinemann Physics 1 4
​e
15 a A thermistor is a temperature-sensitive resistor whose resistance decreases as its temperature increases. It is
usually constructed from Ge, Si or a mixture of various oxides. The resistance of a typical thermistor may range
from 10 kΩ at 0°C to 100 Ω at 100°C.
b
20
R (Ω)
15
10
5
0
10
20
30
40
Temperature (°C)
50
(not linear)
16 VR = IR
1 = I × 1000
I = 0.001 A
VT = IR
5 = 0.001 × R
R = 5000 Ω
From the graph, this corresponds to T = 20°C
17 a Combining Ohm’s law, V = IR, and the equation for power:
P = VI = I2R =
2
25 = IX RX
2
IX =
25
100
V2
R
= 25 ×10−2
IX = 5 × 10−1 = 0.5 A
VX = IRX = 0.5 × 100 = 50 V
VY = RY ×
1
2
= 100 × 0.25 = 25 V
Vtotal = VX + VY = 75 V
b P = VI = 75 × 0.5 = 37.5 W
18 a 3
b 1
c 2
19 The finger provides less contact with the live wire and hence more resistance.
20 A fuse will melt when a high current flows in a circuit. Without the fuse the heat generated from a high current could
be enough to start a fire and burn the house down. A safety switch switches off a circuit when the current in the active
and neutral wires are not equal, thus preventing possible electrocution.
Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd)
ISBN 978 14 8861 12​6 ​1
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