Transcript Unit One: AC Electronics - Helderberg Hilltowns Association

1
ET115 DC Electronics
Unit Five:
Series Circuits
John Elberfeld
[email protected]
WWW.J-Elberfeld.com
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Schedule
Unit Topic
Chpt Labs
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
1
2
3
3
4
5
6
6
6
7
Quantities, Units, Safety
Voltage, Current, Resistance
Ohm’s Law
Energy and Power
Series Circuits
Exam I
Parallel Circuits
Series-Parallel Circuits
Thevenin’s, Power Exam 2
Superposition Theorem
Magnetism & Magnetic Devices
Course Review and Final Exam
2 (13)
3 + 16
5 (35)
6 (41)
7 (49)
9 (65)
10 (75)
19 (133)
11 (81)
Lab Final
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Unit 5 Objectives - I
• Identify a series resistive circuit.
• Determine total resistance, current, and
power in a series resistive circuit.
• State Kirchhoff’s voltage law.
• Solve for an unknown voltage in a circuit
using Kirchhoff’s voltage law (KVL).
• State the voltage divider formula.
• Apply the voltage-divider formula to series
circuits to find an unknown quantity
(either a voltage or resistance).
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Unit 5 Objectives – II
• Calculate the range of output voltages
when a potentiometer is used as a
voltage-divider.
• Construct basic DC circuits on a
protoboard.
• Use a digital multimeter (DMM) to measure
a predetermined low voltage on a power
supply.
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Unit 5 Objectives – III
• Measure resistances and voltages in
a DC circuit using a DMM.
• Apply Ohm’s Law, Thevenin’s
theorem, KVL and KCL to practical
circuits.
• Troubleshoot circuits constructed in
Multisim exercises using simulated
instruments.
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Reading Assignment
• Read and study
• Chapter 4: Series Circuits:
Pages 109-147
7
Lab Assignment
• Lab Experiment 7:
“Series Circuits,” page 49
• Complete all measurements, graphs,
and questions and turn in your lab
before leaving the room
8
Written Assignments
• Complete the Unit 5 Homework sheet
• Show all your work!
• Be prepared for a quiz on questions
similar to those on the homework.
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Series Circuit
• A series circuit has one path for
current flow.
• The current is the same at any point
in the circuit.
10
SERIES CIRCUITS
• Total resistance in a series
circuit is the summation of the
individual resistor values:
RT = R1 + R2 + R3 …. + Rn
Where n = the
number of
resistors
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OHM’S LAW
• Ohm’s Law applies to the total circuit
• Total resistance in a series circuit is
equal to the total circuit voltage divided
by the total current
• IT is the same at every point in the
circuit
VT
RT 
IT
VT = ITRT
IT = VT / RT
12
Single Resistor Circuit
• Single resistor example (Simplest)
• V = I R or R = V/I
• R = 10V/2A = 5 
?
13
Find the Current
Voltage = 15 V
Resistance = 1 k
I = _______
V = 15V
R = 1kΩ
I=?
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Find the Current
Voltage = 15 V
Resistance = 1 k
I = _______
V = 15V
R = 1kΩ
I=?
V 15 V
I 
 .015 Amps 15 mA
R 1 k
15
Find the Voltage
Resistance = 10 k
Current = 4 mA
Voltage = ______
V=?V
R = 10kΩ
I = 4mA
16
Find the Voltage
Resistance = 10 k
Current = 4 mA
Voltage = ______
V=?V
R = 10kΩ
I = 4mA
V  IR  4 mA 10 k  40 Volts
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Find the Resistance
Current = 4 mA
Voltage = 60 V
Resistance = _______
V = 60V
R = ?Ω
I = 4mA
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Find the Resistance
Current = 4 mA
Voltage = 60 V
Resistance = _______
V = 60V
R = ?Ω
I = 4mA
V 60 V
R 
 15 k
I
4 mA
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Multiple Resistor Example
• Series circuit comprised of resistors:
• To calculate the total resistance we use
the formula RT = R1 + R2 + R3 + etc.
• The total resistance for the figure is:
• RT= 10Ω + 20Ω + 30Ω = 60 OHMS
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Three Resistors
• Brown Black Black = ___________
• Brown Grey Black = ___________
• Red Red Black = ___________
Br Bk Bk
Br Gy Bk
V = 120V
R R Bk
• Total Resistance = ___________
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Simplify the Circuit
• To make life easier, you can replace many
resistors in series in the circuit with a
single resistor that has the same effect on
the circuit values as all the resistors it
replaced
• The value of the replacement resistor is
equal to the sum of all the series resistors
that it is replacing
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The BIG Idea
• In a pure series circuit, the current in
every resistor is the same – the same
as the current that leaves the battery
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Find the Current
R1
mA
mA
R2
V = 95 V
mA
mA
R3
• If the current in the bottom left Ammeter is
5 mA, what is the current in the:
• Bottom Right = ______________
• Upper Left = ______________
• Upper Right = ______________
24
Find the Current
R1 = 82 Ω
R2
=
56 Ω
V = 120V
Replacement
Resistor
V = 120V
R=
???? Ω
I = ?mA
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Find the Current
R1 = 82 Ω
R2
=
56 Ω
V = 120V
Replacement
Resistor
V = 120V
R=
138 Ω
I = ?mA
RT = 82 Ω + 56 Ω = 138 Ω
IT = VT/RT = 120V / 138 Ω = 870 mA
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Find the Current
R1 = Gn Be Br
V = 120V
R2
= Br Bk R
RT = R1 + R2 = __________________
R1 = ______
R2 = ______
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Find the Current
R1 = Gn Be Br
V = 120V
R1 = 560 Ω
R2
= Br Bk R
R2 = 1 kΩ
Replacement
Resistor
V = 120V
R=
1.56 kΩ
I = ?mA
RT = 560 Ω + 1 kΩ = 1.56 kΩ
IT = VT/RT = 120V / 1.56 kΩ = 76.8 mA
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Find the Current
R1 = O O Bk
V = 120V
R2
= Y V Bk
RT = R1 + R2 = __________________
R1 = ______
R2 = ______
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Find the Current
R1 = O O Bk
V = 120V
R1 = 33 Ω
R2
= Y V Bk
R2 = 47 Ω
Replacement
Resistor
V = 120V
R=
80 Ω
I = ?mA
RT = 33 Ω + 47 Ω = 80 Ω
IT = VT/RT = 120V / 80 Ω = 1.5 A
30
Find the Current
R1 = Br Gn R
R2
=RRR
V = 95V
R1 = ______
R2 = ______
R3 = ______
R3 = Br Bk O
RT = R1 + R2 + R3 = __________________
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Find the Current
R1 = 1.5 kΩ
R2
=
2.2 kΩ
V = 95 V
R3 = 10 kΩ
V = 95 V
R=
13.7 kΩ
Replacement
Resistor
I = ?mA
RT = 1.5 kΩ + 2.2 kΩ + 10 kΩ = 13.7 kΩ
IT = VT/RT = 95V / 13.7 kΩ = 6.93 mA
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Multiple Resistors
• What is the total resistance of twelve
5.6 kΩ resistors in series?
• RT = 12 x 5.6 kΩ = 67.2 kΩ
• What is the total resistance of six 47
Ω resistors, eight 100 Ω resistors and
two 22 Ω resistors in series?
• _________________
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Multiple Resistors
• What is the total resistance of twelve
5.6 kΩ resistors in series?
• RT = 12 x 5.6 kΩ = 67.2 kΩ
• What is the total resistance of six 47
Ω resistors, eight 100 Ω resistors and
two 22 Ω resistors in series?
• 1.126 kΩ
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Missing Resistor
• Five resistors in series have a total
resistance of 20 kΩ.
• The value of four of the resistors are
4.7 kΩ, 1.0 kΩ, 2.2 kΩ, and 3.9 kΩ.
• What is the value of the fifth
resistor?
• R5 = ________________
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Missing Resistor
• Five resistors in series have a total
resistance of 20 kΩ.
• The value of four of the resistors are
4.7 kΩ, 1.0 kΩ, 2.2 kΩ, and 3.9 kΩ.
• What is the value of the fifth resistor?
• 20 kΩ = 4.7kΩ+1.0kΩ+2.2kΩ+3.9kΩ+X
• 20 kΩ = 11.8kΩ+X
• X = 20 kΩ - 11.8kΩ = 8.2kΩ
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Voltage Drop
• V=IR (Ohm’s Law) applies to each
resistor in the circuit
• I = 2 mA = the same current in each
and every resistor in a series circuit
R1 = 10kΩ
+
V = 120V
RT = 60kΩ
R2
=
20kΩ
R3 = 30kΩ
IT=VT/RT
IT = 120 V/ 60kΩ
IT = 2 ma
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Voltage Drop
• I = 2 mA
R1 = 10kΩ
V = 2 ma 10k Ω =20 V
+
+
- +
-
-
+-
V = 120V
R2 = 20kΩ
V = 2ma 20k Ω
V= 40 V
R3 = 30kΩ
V = 2ma 30k Ω = 60 V
• VA = V1 + V2 + V3 = 20V + 40V + 60V = 120 V
• The sum of the voltage drops across all
the resistors adds up to the input voltage
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Voltage Drop in a Series Resistive
Circuit
• Voltage drop in the series circuit is
given by the sum of the individual
voltage drops across all the
resistors:
V  IR1  IR2  IR3  ........ IRn
 V1  V2  V3  ......... Vn
RT = 8.8kΩ
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Voltage Drop
IT = 5.5 V/8.8kΩ
R1 = 2.2 kΩ
V1 = ????
+
+
- +
-
-
+-
V = 5.5 V
IT=VT/RT
IT = 625 µa
R2 = 5.6 kΩ
V2 = ?????
R3 = 1.0 kΩ
V3 = ??????
RT = 8.8kΩ
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Voltage Drop
R1 = 2.2 kΩ
V1 = 625 µa 2.2 kΩ =1.375 V
+
+
- +
-
-
+-
V = 5.5 V
IT=VT/RT
IT = 5.5 V/8.8kΩ
IT = 625 µa
R2 = 5.6 kΩ
V2 = 625 µa 5.6 kΩ =3.5 V
R3 = 1.0 kΩ
V3 = 625 µa 1.0 kΩ = 625 mV
What is the sum of the
voltage drops in the
resistors?
RT = 8.8kΩ
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Voltage Drop
R1 = 2.2 kΩ
V1 = 625 µa 2.2 kΩ =1.375 V
+
+
- +
-
-
+-
V = 5.5 V
IT=VT/RT
IT = 5.5 V/8.8kΩ
IT = 625 µa
R2 = 5.6 kΩ
V2 = 625 µa 5.6 kΩ =3.5 V
R3 = 1.0 kΩ
V3 = 625 µa 1.0 kΩ = 625 mV
• VT = V1 + V2 + V3 = 1.375V + 3.5V + 625m V
= 5.5 V
• The sum of the voltage drops across all
the resistors adds up to the input voltage
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Kirchhoff’s Voltage Law
• The sum of all voltage drops in a
closed loop is equal to the value of
the applied voltage
– This deals with magnitudes
• The algebraic sum of all the voltage
drops and all the voltage sources in
any closed loop equals 0
– This includes the sign of the voltage
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Kirchhoff’s Voltage Law
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Sample Problems - 1
R1 = 1 MΩ
+
R2 = 2.2 MΩ
V = 16V
R3 = 560 kΩ
Comp.
Resistance Voltage
R1
R2
R3
Total
1 MΩ
2.2 MΩ
560 kΩ
16V
Current
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Sample Problems - 1
R1 = 1 MΩ
+
R2 = 2.2 MΩ
V = 16V
R3 = 560 kΩ
Comp.
Resistance Voltage
Current
R1
R2
R3
Total
1 MΩ
2.2 MΩ
560 kΩ
3.76 MΩ
4.26 μA
4.26 μA
4.26 μA
4.26 μA
4.26 V
9.37 V
2.39 V
16V
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Power Calculations
• Power = Voltage x Current = Watts
• P = VI
• The more electric power a motor
uses, the more work it can do per
second
• Combining with Ohm’s Law:
• P = I2R
P = V2/R
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Power in a Series Circuit
• P = I2R
• Because current is the same at every
point in a series circuit, the
resistance with the smallest value
will also dissipate the smallest power
value.
• The largest resistor in the circuit will
dissipate the largest amount of
power.
48
Power in a Series Circuit
• Since the current is the same at any
point in a series circuit, the equation
P = I 2 R is often the
best equation
to use when both
I and R are known.
49
Power in a Series Circuit
• The total power dissipated in a
series circuit is also the amount
of power the Power Source must
deliver.
– The power supplied is equal to the total
power dissipated in the circuit elements
• This may also be expressed as:
PT = PR 1 + PR 2 … + PR n
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Sample Problems - 2
R1 = 4.7 kΩ
+
R2 = 4.7 kΩ
V = 48 V
R3 = = 4.7 kΩ
Comp.
Resistance Voltage
R1
R2
R3
Total
4.7 kΩ
4.7 kΩ
4.7 kΩ
48 V
Current
Power
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Sample Problems - 2
R1 = 4.7 kΩ
+
R2 = 4.7 kΩ
V = 48 V
R3 = = 4.7 kΩ
Comp.
Resistance Voltage
Current
Power
R1
R2
R3
Total
4.7 kΩ
4.7 kΩ
4.7 kΩ
14.1 kΩ
3.40 mA
3.40 mA
3.40 mA
3.40 mA
54.3 mW
54.3 mW
54.3 mW
163.4 mW
16 V
16 V
16 V
48 V
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Why use multiple voltage sources?
• A single voltage source is not able to
supply above its rated power (rated power
= rated voltage X current capacity).
• If more power is needed, multiple sources
are connected in parallel at constant
voltage.
• However, multiple sources may also be
connected in series, where additional
(voltage) power is required, with the
current remaining constant.
53
Series Voltage Sources
• If two voltage sources are in series
and have the same polarity, the
voltages add
• If two voltage sources are in series
and have opposite polarities, the
voltages subtract
54
Same polarity
• Equivalent circuits
+
V = 9V
+
V = 14V
+
-
V = 5V
electron flow
55
Opposite Polarity
• Equivalent circuits
electron flow
+
V = 3V
-
+
+
V = -6V
-
V = -9V
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Complex Sources
• Equivalent source
+
V = 12V
-
-
V = -9V
+
+
V = 3V
-
+
V = +6V
electron flow
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Complex Sources
• Equivalent source
+
V=5V
V = 10 V
V=8V
+
+
-
+
V = + 23V
electron flow
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Complex Sources - 2
• Equivalent source
electron flow
+
+
+
V = -10 V
-
V = -50 V
V = - ??V
+
-
-
V = 25 V
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Complex Sources - 3
• Equivalent source
electron flow ?
+
+
V = -8 V
V = ??
+
-
-
V=8V
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Find the Input Voltage - 1
R1 = 2.75kΩ
V = ?????
I = 2 mA
+
+
- +
-
-
+-
V = ??V
R2 = 4.1kΩ
V = ????
R3 = 6.15kΩ
V = ????
Determine the input voltage
61
Find the Input Voltage - 1
R1 = 2.75kΩ
V = 5.5 V
I = 2 mA
+
+
- +
-
-
+-
V = ??V
R2 = 4.1kΩ
V = 8.2 V
R3 =
6.15kΩ
V = 12.3 V
• VA = V1 + V2 + V3 = 5.5V + 8.2V + 1.23V = 26 V
• The sum of the voltage drops across all
the resistors adds up to the input voltage
62
Find the Missing Voltage
• Five resistors are in series with a 20
V source.
• The voltage drops across four of the
resistors are 1.5 V, 5.5 V, 3 V, and 6V.
• How much is the voltage drop across
the fifth resistor?
63
Find the Missing Voltage
• Five resistors are in series with a 20
V source.
• The voltage drops across four of the
resistors are 1.5 V, 5.5 V, 3 V, and 6V.
• How much is the voltage drop across
the fifth resistor?
20 V = 1.5 V + 5.5 V + 3V + 6 V + X
X = 20 V - 1.5 V - 5.5 V - 3V - 6 V
X=4V
64
Voltage Reference Points
• Voltages are always relative
– That is, one point on a circuit has a
voltage that is higher (+) or lower (-)
with respect to another point
• Often voltages are written with
respect to ground
– Ground is often the negative terminal of
the power supply
65
What is a circuit ground?
• In most electronic equipment, a large
conductive area on a printed circuit board
is used as the common reference point.
• This point is called the ground.
• This ground provides a convenient way of
connecting all common points within the
circuit back to one side of the battery or
the voltage source.
66
Voltages with Respect to Ground
I= 3mA
+
V=
18V -
R1 = 1 kΩ
VR1=3V
+
+
R2 = 3 kΩ
VR2=9V +
R3 = 2 kΩ
VR3=6v V=0V (Ground!)
V=18V =VT
V=15V
V=6V
V=0V
15V = voltage
across 5k
replacement
67
Another Reference Point
+
V=
18V -
R1 = 1 kΩ
VR1=3V
+
+
R2 = 3 kΩ
VR2=9V +
R3 = 2 kΩ
VR3=6v -
V=+3V
V=0V (Black Lead)
V=-9V
V=-15V (Ground!)
68
Voltage Reference Points
69
Decoding the Symbols
• VC = Voltage from point C to ground
– Single letter implies the second point is
ground
– Red lead on the letter, black on ground
• VDA = Voltage from Point D to Point A
– Double letter implies the second point is
another location in the circuit
– Red lead on the first letter, black on the
second letter
70
Voltage Calculations
• I = 794 μA
C
+
V = 15V
G
VG VA
VB
VC
0V 2.62 10.56 15
V
V
V
-
VBA
V
R1 = 5.6 kΩ
V = 4.45 V
+
- +
-
+-
B
R2 = 10 kΩ
V = 7.94 V
A
R3 = 3.3 kΩ
V = 2.62 V
VCB
V
VCA
VAB
VAC
V
-7.94
V
V
71
Voltage Calculations
• I = 794 μA
C
+
V = 15V
G
VG VA
VB
VC
0V 2.62 10.56 15
V
V
V
-
VBA
R1 = 5.6 kΩ
V = 4.45 V
+
- +
-
+-
B
R2 = 10 kΩ
V = 7.94 V
A
R3 = 3.3 kΩ
V = 2.62 V
VCB
7.94V 4.45
V
VCA
VAB
12.39 -7.94
V
V
VAC
-12.39V
72
Voltage Calculations
• I = _____ mA
C
R1 = 470 Ω
VCB = _____ V
B
V = 12 V
G
VG VA VB
VC VBA
R2 = 1.0 kΩ
VBA = _____ V
A
R3 = 680 Ω
VAG = _____ V
VCB VCA VAB
VAC
73
Voltage Calculations
• I = 5.58 mA
C
R1 = 470 Ω
VCB = 2.62 V
V
B
V = 12 V
G
VG VA
0
VB
3.79 9.37
V
V
VC
R3 = 680 Ω
VAG = 3.79 V
VBA
VCB
12V 5.58V 2.62V
R2 = 1.0 kΩ
VBA = 5.58 V
A
VCA
VAB
VAC
8.2V
-5.58V
-8.2V
74
Voltage Calculations
• I = 2 mA
C
R1 = 10kΩ
V = 20 V
B
R2 = 20kΩ
V = 40 V
V = 120V
G
VG VA VB
VC
R3 = 30kΩ
V = 60 V
VBA
A
VCB VCA VAB
VAC
75
Voltage Calculations
• I = 2 mA
R1 = 10kΩ
V = 20 V
C
B
R2 = 20kΩ
V = 40 V
V = 120V
G
R3 = 30kΩ
V = 60 V
VG VA VB
VC
0V
120V 40V
60V 100V
VBA
A
VCB VCA VAB
VAC
20V
-60V
60V
-40V
76
Voltage Divider
• A voltage divider consists of several
resistors connected in series across
a voltage source.
• The voltages developed across each
of these resistors can be used for
different applications.
77
Voltage Divider
• Resistive circuits are used to
obtain some percentage of the
applied voltage source.
RT = 30kΩ
IT=VT/RT
IT = 90 V/ 30kΩ
IT = 3 ma
VR1= 3ma 10kΩ
VR1= 30V
RT = 10kΩ
78
Example
+
V=
30V -
R1 = 2 kΩ
VR1=6V
+
V=30V
IT = 30 V/ 10kΩ
IT = 3 ma
VR3= 3ma 5kΩ
VR3= 15V
+
V=24V
R2 = 3 kΩ
VR2=9V +
V=15V
R3 = 5 kΩ
VR3=15v -
IT=VT/RT
V= 0 (Ground!)
79
Voltage Divider Shortcuts
• Voltage dividers are used in so many
circuits they deserve more study
• Voltage dividers are a way to access
a smaller voltage from a large
voltage source
80
Example
• Find the total current, and then find
the voltage drop across the resistors
V=
24V
R1 = 3 kΩ
VR1=____V
R2 = 5 kΩ
VR2=____V
+
RT = 3 kΩ + 5 kΩ
RT = 8 kΩ
+
IT = VT/RT = ????
-
V1 = ITR1 = ?? 3 kΩ
V= 0 (Ground!)
IT = ????
V1 = ???
V2 = ITR2 = ???? 5 kΩ
V2 = ????
81
Example
• Find the total current, and then find
the voltage drop across the resistors
V=
24V
R1 = 3 kΩ
VR1=____V
R2 = 5 kΩ
VR2=____V
+
RT = 3 kΩ + 5 kΩ
RT = 8 kΩ
+
IT = VT/RT = 24 V/8 kΩ
-
VR1 = ITR1 = 3 mA 3 kΩ
V= 0 (Ground!)
IT = 3 mA
VR1 = 9 V
VR2 = ITR2 = 3 mA 5 kΩ
VR2 = 15 V
82
Example
• Find the total current, and then find
the voltage drop across the resistors
V=
24V
+
V=24V
R1 = 3 kΩ
VR1=9V +
V=15V
R2 = 5 kΩ
VR2=15v -
V= 0 (Ground!)
83
Voltage Divider in 1 Step
•
•
•
•
IT = VT / (R1 + R2)
VR2 = IT R2
VR2 = [ VT / (R1 + R2) ] x R2
VR2 = (VT R2 ) / (R1 + R2)
 R2 
V2  
 Vs
 R1  R2 
VT = VS = VSource
• If you work with lots of voltage dividers,
you will find it easier to memorize the
voltage divider equation
84
Example
I = 16V/(3kΩ+5kΩ)=2mA
 R2 
V2  
Vs
 R1  R2 
5k 


V2  
 16V  10V
 3k   5k  
V=
16V
V2 = 2mA 5kΩ = 10 V
+
R1 = 3 kΩ
R2 = 5 kΩ
+
-
VR2=?
V= 0 (Ground!)
85
Voltage Divider Rule
• The voltage across any resistor
in a series may be determined by
the following equation:
RX
VX 
 VT
RT
• This works with many resistors!
86
Potentiometers
• Rotating the control divides the total
resistance into different proportions
87
Voltage Divider
• The potentiometer
acts like a voltage
divider:
V=
15V
+
R1 = 15 kΩ
+
R2 = 35 kΩ
-
Wiper
V2=?
V= 0 (Ground!)
88
Voltage Divider
• The potentiometer
acts like a voltage
divider:
V=
15V
 R2 
V2  
Vs
 R1  R2 
35k 


V2  
 15V  10.5V
 15k   35k  
+
R1 = 15 kΩ
+
R2 = 35 kΩ
-
Wiper
V2=?
V= 0 (Ground!)
89
Voltage Divider
• If a 60k potentiometer is used to
generate 5 V from a 16 V power
source, what is the resistance
between the wiper and ground?
V=
16V
+
R1 = ?? kΩ
+
R2 = ?? kΩ
-
Wiper
V2=5 V
V= 0 (Ground!)
90
Grind it out!
•
•
•
•
RT = 60 kΩ, V = 16 V,
I = V / R so I = 267 μA
R2 = V2 / I
R2 = 5 V / 267 μA = 18.7 kΩ
• Double Check: The other resistance is
60 kΩ - 18.7 kΩ = 41.25 kΩ
• The voltage drop is 41.25 kΩ • 267 μA
• V1 = 11 V and V1 + V2 = 11V + 5 V = 16 V,
the source voltage
91
Another Approach
• R1+R2= 60 kΩ – the value of the
potentiometer
 R2 
V2  
Vs
 R1  R2 
 X 
5V  
16V
 60k  
5V  60k 
X
 18.75k 
16V
92
Trouble Shooting:
Open in a Series Circuit
• A series circuit is open when no
current flows in the circuit even
though there is a voltage source in
the circuit.
• Why does the meter read V Volts?
93
Trouble Shooting:
Open in a Series Circuit
• An opens blocks all current
• How much power is used in the
circuit if there is no current?
94
Shorts in a Series Circuit
• A short in a series circuit represents
zero resistance between the terminals
that are shorted.
• Why does the meter read 0 V?
Shorted Resistor
95
Shorts in a Series Circuit
• A short reduces the resistance in a series
circuit
• A short INCREASES the current
• A short INCREASES the total power
• P = V I - V stays the same, I increases
96
Think about it…
• A circuit consists of 10 electrical
bulbs in series. If one of the bulbs is
removed from the circuit, what will
happen when the circuit is switched
on? Give reasons to support your
answer.
• What is the voltage across the open in
the circuit?
97
Troubleshooting a Series Circuit
• If an open occurs at any point in
a simple series circuit, current
will decrease to 0 A.
• All voltage drops across all
resistors will decrease to 0 V
• All the applied voltage appears
across the break in the circuit
98
Troubleshooting a Series Circuit
• If a short occurs across
(through) a resistor, current will
increase because resistance
decreases.
• As current increases, the voltage
across the remaining resistors
will increase.
• Total Power will increase
99
Troubleshooting a Series Circuit
100
Troubleshooting a Series Circuit
If a total short occurs, RT = 0 
• Current will attempt to increase
to unacceptable levels
• A protective device reacts to
open the circuit, or ...
101
Ohm
Meter
(Top)
Br Bk R = _____
(Bottom)
0 0 R = _____
(Diagonal)
V R R = _____
(Vertical)
Ohm meter = _____
102
Sample Problems - 1
R1 = 2kΩ
+
R2 = 4kΩ
V = 36V
R3 = 6kΩ
Comp.
Resistance Voltage
R1
R2
R3
Total
2kΩ
4kΩ
6kΩ
36V
Current
Power
103
Sample Problems - 1
R1 = 2kΩ
+
R2 = 4kΩ
V = 36V
R3 = 6kΩ
Comp.
Resistance Voltage
Current Power
R1
R2
R3
Total
2kΩ
4kΩ
6kΩ
12kΩ
3ma
3ma
3ma
3ma
6V
12V
18V
36V
18mW
36mW
54mW
108mW
104
Sample Problems – 2
R1 = ? kΩ
+
R2 = 4 kΩ
V = 10V
R3 = 6 kΩ
Comp.
R1
R2
R3
Total
Resistance Voltage
Current
500μA
4kΩ
6kΩ
10V
Power
105
Sample Problems - 2
R1 = ? kΩ
+
R2 = 4 kΩ
V = 10V
R3 = 6 kΩ
Comp.
Resistance Voltage
Current
Power
R1
R2
R3
Total
10kΩ
4kΩ
6kΩ
20kΩ
500μA
500μA
500μA
500μA
2.5mW
1mW
1.5mW
5mW
5V
2V
3V
10V
106
Sample Problems - 3
R1 = ? kΩ
+
R2 = 5 kΩ
V = 60V
R3 = 7 kΩ
Comp.
Resistance Voltage
R1
R2
R3
Total
5kΩ
7kΩ
Current
Power
112mW
60V
107
Sample Problems - 3
R1 = ? kΩ
+
R2 = 5 kΩ
V = 60V
R3 = 7 kΩ
Comp.
Resistance Voltage
Current
Power
R1
R2
R3
Total
3kΩ
5kΩ
7kΩ
15kΩ
4mA
4mA
4mA
4mA
48mW
80mw
112mW
240mW
12V
20V
28V
60V
108
Summary Unit 5
• Identifying series circuits
• Calculating total resistance in series
• Solving for unknown voltage using
KVL
• Solving problems with the voltagedivider formula
• Solving problems with a
potentiometer as a voltage-divider