Transcript Unit One: AC Electronics - Helderberg Hilltowns Association

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ET115 DC Electronics
Unit Seven
Series-Parallel 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 7 Objectives - I
• Identify series-parallel relationships in a
combinational resistive circuit.
• Determine total resistance, current, and
power in a series-parallel resistivecircuit.
• Draw the schematic for the Wheatstone
bridge.
• Explain how to determine if the
Wheatstone bridge is balanced.
• Determine the effect of one or more
resistive loads on a voltage divider
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Unit 7 Objectives – II
• Determine the bleeder current in a
resistive voltage divider.
• Determine the loading effect of a voltmeter
on a circuit given the internal resistance of
the voltmeter.
• Define bipolar voltage divider
• Use proper prototype board wiring and
test procedures for DC resistive circuit
components including using the digital
multimeter..
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Unit 7 Objectives – III
• Construct basic DC circuits on a
protoboard.
• Use a digital multimeter (DMM) to measure
a predetermined low voltage on a power
supply.
• Measure resistances and voltages in a DC
circuit using a DMM.
• Apply Ohm’s Law, Thevenin’s theorem,
KVL and KCL to practical circuits.
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Reading Assignment
• Read and study
• Chapter 6: Series-Parallel Circuits:
Pages 212-237 (First half of chapter)
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Lab Assignment
• Experiment 10, “Series-Parallel
Combination Circuits,” beginning on
page 75 of DC Electronics: Lab
Manual and MultiSim Guide.
• Complete all measurements, graphs,
and questions and turn in your lab
before leaving the room
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Written Assignments
• Complete the Unit 7 Homework sheet
• Show all your work!
• Be prepared for a your second
MAJOR EXAM on questions similar
to those on the homework.
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Ohms Law
• MEMORIZE: V = I R
• Ohms Law
• If you increase the voltage, you
increase the current proportionally
– 3 times the voltage gives you three
times the current
– Resistance (ohms) is the proportionality
constant and depends on the atomic
structure of the material conducting the
current
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Power Formula
• Power = Work / time
• P=VI
• Voltage is the work done per
coulomb, and current is the number
of coulombs per second passing by
a point.
• The product of voltage and current
gives the work done per second, or
power.
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Series-Parallel Combination of
Resistances
• This type of circuit consists of a
combination of resistances in series
and parallel.
• Series-parallel combination of
resistances can be classified as:
– Simple series-parallel circuits
– Complex series-parallel circuits
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Parallel Resistors
• For resistors to be in parallel, both
ends of each resistor must be in
direct electrical contact with each
other – with no resistance between
the ends
• Parallel
Not Parallel (blue)
R1
R1||R2
R2
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Series-Parallel Combination of
Resistances
• To differentiate between a simple seriesparallel circuit and a complex circuit:
– observe simple series and parallel
combinations of resistances in the given
circuit.
– replace the simple series and simple parallel
combinations by their single equivalent
resistor.
– repeat the simplification process to verify
whether the circuit is reduced to a single
resistance circuit.
– if a single resistor can replace ALL the
resistors, it is a simple series-parallel circuit
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Recognizing Series Components
RT= R1+R2+R3||R4||(R5+R6)
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Analysis
• Analysis of the circuit also
requires one to recognize the
various paths for current flow.
• The ability to recognize the
points where current branches
out and where current converges
(sums) is vital.
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EXAMPLE
Electron Flow →
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Series-Parallel Circuits
• Pure series-parallel circuits can be
reduced to a single equivalent resistor:
• Example
R1
R1
R2
R3
R2||R3
RT= R1+R2||R3
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Example
• Another simple series-parallel circuit
R1
R2
R4
R5
R3
R1
R2||R3
R4||R5
RT= R1+R2||R3+R4||R5
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RT=R2||[R1+R3||(R4+R5)]
R1
R2
R1
R3||(R4+R5)
R2
R4
R3
R1
R5
R2
R1+R3||(R4+R5)
R2
R3 R4+R5
RT
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Complex Circuit
• Unable to simplify to a single resistor
– Where the battery connects to the
circuit is important!!!!
• Complex
Simple
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Complex Circuit
• Unable to simplify to a single resistor
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Practice
• Classify this circuit as series-parallel
or complex
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Practice
• Classify this circuit as simple seriesparallel or complex
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Practice
• Classify this circuit as series-parallel or
complex
• It is the electrical connections, not the
drawing, that determines the type of
circuit
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Practice
• Classify this circuit as series-parallel
or complex
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Voltage Drop in Series-Parallel
Resistive Circuit
• To find the equivalent resistance of a
series-parallel resistive circuit:
– Identify the resistance combinations as either
series or parallel.
– Simplify, combine and recombine series and
parallel combinations
– Find the value of the total resistance of the
circuit.
• Once you know the equivalent resistance,
you can calculate voltage or current
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Total Resistance
• One of the common approaches
is called Outside Toward the
Source approach.
• To implement this method, one
begins farthest from the source
and works back toward the
source.
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Equivalent Resistance
• The analysis of the circuit uses
equivalent resistance as circuit
reductions are performed.
• For instance, if a 6-k and a 3-k
resistor are in parallel, their
equivalent series resistance is 2
k.
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Current In a Series-Parallel Circuit
• The analysis of current in this
circuit is a fundamental step.
• Kirchhoff’s Current Law must be
followed to see how current
divides and sums together.
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Circuit Analysis Tools
• The circuit tools used to
determine circuit parameters
have already been covered.
They include:
-Ohm’s law
-KCL and KVL
rule
-Power equations
-Voltage divider
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Voltage In a Series-parallel Circuit
• Voltage distribution throughout
the circuit follows the laws
appropriate to series and parallel
connections.
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Circuit Analysis
• Combine the two parallel resistors to
one equivalent resistor
• 1/R23 = 1/20Ω + 1/30Ω = 3/60Ω + 2/60Ω
• 1/R23 = 5/60Ω
• R23= 12 Ω
R1 = 10Ω
V=
25V
R2 =
20Ω
R3 =
30Ω
V=
25V
R1 = 10Ω
R23 =
12Ω
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Circuit Analysis
• Combine the two parallel resistors to
one equivalent resistor using POS
• R23 = 20Ωx30Ω / (20Ω+30Ω) = 12 Ω
– Same results!
R1 = 10Ω
V=
25V
R2 =
20Ω
R3 =
30Ω
V=
25V
R1 = 10Ω
R23 =
12Ω
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Analysis
• Combine the series resistors
• R123 = 10 Ω + 12 Ω = 22 Ω
R123 = 22Ω
R1 = 10Ω
V=
25V
R23 =
12Ω
V=
25V
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Analysis
•
•
•
•
•
•
V = 25 V, RT = 22Ω, IT = 1.14 A
VR1 = 11.4V, VR2 = VR3 = 25V-11.4V=13.6V
I R2= VR2/20 Ω = .68A
I R3= VR3/30 Ω = .45A
Check: IT = IR1 = IR2 + IR3 = 1.14 A
VT = VR1+VR2 = 25V
R1 = 10Ω
V=
25V
R2 =
20Ω
R3 =
30Ω
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Current in Parallel Resistive Circuit
• In the circuit shown, the current flowing
through R1, R2, R3, and R4 has two
separate paths called branches.
• Applying KCL, the current, I, is equal to
the sum of the branch currents, I1 and I2.
←Conventional Flow
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Class Activity
• What is the equivalent resistance of
the following circuit?
R1 = 5.6kΩ
V=
30V
R2 =
10kΩ
R3 =
4.7kΩ
R4 = 2.2kΩ
R5 =
3.3kΩ
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Description
• What is the equivalent resistance of
the following circuit?
R1 = 5.6kΩ
V=
30V
R2 =
10kΩ
R3 =
4.7kΩ
R4 = 2.2kΩ
R5 =
3.3kΩ
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Class Activity
• R2 and R3 are….
– In series with each other AND the series
combination is in parallel with R5
• R1 and R4 are in series with the
combination of R5 in parallel with R2
and R3
• RT= R1+R4+R5||(R2+R3)
V=
30V
R1 = 5.6kΩ
R2 =
10kΩ
R3 =
4.7kΩ
R4 = 2.2kΩ
R5 =
3.3kΩ
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Predict Total Resistance
• The total resistance must be
between:
• 5.6k Ω + 2.2 kΩ = 7.8kΩ Minimum
• 5.6k Ω + 2.2 kΩ + 3.3 kΩ = 11.1 kΩ
Maximum
• WHY???
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Current Analysis
• Combine resistors in series
• R23 = 10kΩ + 4.7kΩ = 14.7kΩ
R1 = 5.6kΩ
V=
30V
R5 =
3.3kΩ
R23 =
14.7kΩ
R4 = 2.2kΩ
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Current Analysis
• Combine parallel resistors (POS)
• R235 = 14.7 kΩ x 3.3kΩ / (14.7kΩ + 3.3kΩ)
• R235 = 2.70 kΩ
R1 = 5.6kΩ
V=
30V
R235 =
2.7 kΩ
R4 = 2.2kΩ
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Current Analysis
• Combine series resistors
• R12345 = 5.6 kΩ + 2.7kΩ +2.2 kΩ =10.5 kΩ
R12345 = 10.5kΩ
V=
30V
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More Analysis
• IT = VT/RT = 30V / 10.5 KΩ = 2.85 mA
• VR1 = 16V, VR235 = 7.7V, VR4 = 6.3V
R1 = 5.6kΩ
V=
30V
R235 =
2.7 kΩ
R4 = 2.2kΩ
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Current Analysis
• VR235 = 7.7V
• IR23 = 523μA, IR5 = 2.33mA
R1 = 5.6kΩ
V=
30V
R5 =
3.3kΩ
R23 =
14.7kΩ
R4 = 2.2kΩ
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Current
• VR235 = 7.7V, IR23 = 523μA
• VR2 = 5.23V, VR3 = 2.46V,
R1 = 5.6kΩ
V=
30V
R2 =
10kΩ
R3 =
4.7kΩ
R4 = 2.2kΩ
R5 =
3.3kΩ
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Check You Results
• 30V =VR1+VR2+VR3+VR4 = VR1+VR5+VR4
• IT = IR23 + IR5
R1 = 5.6kΩ
V=
30V
R2 =
10kΩ
R3 =
4.7kΩ
R4 = 2.2kΩ
R5 =
3.3kΩ
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More Analysis
• Which resistor has the largest
voltage drop?
– Is it necessarily the biggest resistor?
• Definition: I1 is the current through
R1, I2 the current through R2, etc.
• What must I1 be equal to?
• What must I1 be great than?
• Which currents equal IT – the total
current output of the power source?
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Power In A Series-parallel Circuit
• Analysis of power distribution in
the circuit follows the same rules
from pure series and pure
parallel circuits.
• Total power is a sum of all
individual power levels in the
circuit components.
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Power in Series-Parallel Resistive Circuit
• Total power generated by the source of a
series parallel circuit is dissipated in
various branches of the circuit.
Pd  ( Pc )  P1  P2  P3  ..........
... Pn
where Pd = power dissipated in the circuit
•
Pc = power consumed by the circuit,
•
P1, P2 …..P= are the powers dissipated
in the components of the circuit
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Effects Of Opens
• As noted earlier, the location of
the open has a great influence
on how the electrical parameters
change with the open.
• Total current decreases as a
result.
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Troubleshooting An Open
• As the open becomes more
dramatic, the technician will note
that current and power levels
greatly decrease.
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Current Measurements
• Measuring current in the parallel
portion will indicate if an open
exists.
• Measuring voltage in the series
portion will indicate opens via
the absence of voltage drops.
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Effects Of Shorts
• The location of the short
determines how dramatic the
impact will be to the circuit.
• A general rule is that the closer
the short is to the source, the
greater the impact will be to
circuit operation.
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What IF
• What is the current IF R2 is OPEN?
• RT = R1+R5+R4 = 11.1kΩ
• IT= 2.70mA (was 2.85 mA - DEcrease)
R1 = 5.6kΩ
V=
30V
R2 =
10kΩ
R3 =
4.7kΩ
R4 = 2.2kΩ
R5 =
3.3kΩ
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What IF
• What is the current IF R1 is OPEN?
• RT = infinite
• IT = 0 A (was 2.85 mA - DEcrease)
R1 = 5.6kΩ
V=
30V
R2 =
10kΩ
R3 =
4.7kΩ
R4 = 2.2kΩ
R5 =
3.3kΩ
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What IF
• What is the current IF R5 is SHORTED?
• RT = R1+R4 = 7.8kΩ
• IT = 3.9mA (was 2.85 mA - INcrease)
R1 = 5.6kΩ
V=
30V
R2 =
10kΩ
R3 =
4.7kΩ
R4 = 2.2kΩ
R5 =
0
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Application of Kirchhoff’s Current Law
(KCL)
• The following circuit consists of four
resistances and one voltage source,
along with the total and branch
currents.
• Applying KCL
I  I1  I 2
at node B:
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A Completed Example
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Class Activity
• Find the current I4 flowing through
the resistance R4 in the following
circuit.
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Plan of Attack
• How would you tackle this problem?
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Plan of Attack
• How would you tackle this problem?
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Plan of Attack
• How would you tackle this problem?
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Loaded Voltage Dividers
• The Voltage Divider is a common
series-parallel type circuit.
• As loads are placed on the
circuit, the analysis becomes a
bit more difficult.
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Voltage Divider Review
VCC
•
•
•
•
•
•
Under ideal conditions, I1 = I2 = I
I1
V=IR (Ohm’s Law!!!)
R1
VCC = I (R1 + R2)
VR2
I = VCC / (R1 + R2)
VR2 = R2 I
R2
VR2 = R2 VCC / (R1 + R2)
I2
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Voltage Divider Review
VCC
•
•
•
•
•
Under ideal conditions, I1 = I2 = I3 = I
VCC = I (R1 + R2 + R3)
R1
I = VCC / (R1 + R2 + R3)
VR3 = R3 I
VR3 = R3 VCC / (R1 + R2 + R3)
R2
VRr
R3
I1
I2
I3
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Loaded Voltage Dividers
• An unloaded voltage divider circuit has no
other components connected across it.
• The circuit shows a loaded voltage divider
circuit, where resistor R3 is connected
across the points, A and ground.
• The resistors R2 and R3 are
in parallel. Therefore, their
voltages will be same and
currents through them will
be inversely proportional
to their resistances.
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Analysis
• If the applied voltage is
30V, and R1= 4.7kΩ and R2
= 5.6kΩ, find VR2.
VCC
I1
R1
VR2
R2
I2
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Calculations
•
•
•
•
•
•
VCC
VR2 = 30V 5.6kΩ / ( 4.7kΩ + 5.6kΩ)
VR2 = 16.3 V
I1
IR2 = 16.3 V/ 5.6kΩ = 2.91 mA
R1
Check:
VR2
IT = 30V / ( 4.7kΩ + 5.6kΩ)
IT = 2.91 mA
R2
I2
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Analysis
• If the applied voltage is
30V, and R1= 4.7kΩ, R2 =
5.6kΩ, and R3 = 8.7kΩ,
find all currents and find
VR2
VCC
I1
R1
VR2
R2
I2
R3
I3
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Calculations
•
•
•
•
•
•
•
•
•
•
VCC
R2 is parallel to R3 (POS!)
R23 = 5.6k Ω 8.7kΩ / (5.6kΩ + 8.7kΩ)
R23 = 3.4kΩ
I1
V2 = 30V3.4k Ω /(3.4KΩ + 4.7kΩ) R1
V2 = 12.6 V
VR2
I2 = 12.6V/5.6k Ω = 2.25mA
I3 = 12.6V/8.7k = 1.45 mA
R2
IT = 30V / (3.4kΩ + 4.7kΩ)
I2
IT = 3.70 mA
CHECK: IT = I2 + I3
R3
I3
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Voltmeters
• You measure voltage by placing a
voltmeter across a resistor in a
active circuit.
• A perfect voltmeter has infinite
resistance and draws no current, but
in reality, a voltmeter might have a
resistance of 10 MΩ
• Any measuring device affects the
value being measured
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Analysis
VCC
• If the applied voltage is 30V,
and R1= 200 kΩ, R2 = 100 kΩ,
I1
and R3 = 10 MΩ (a voltmeter),
R1
find all currents and find VR2
VR2
• Ideally,
• VR2 = 30V(100 kΩ)/(300 kΩ)
• VR2 = 10 V
R2
I2
R3
I3
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Calculations
VCC
R2 is parallel to R3 (POS!)
R23 = 100kΩ 10 MΩ / (100kΩ + 10 MΩ )
R23 = 99.01 kΩ
I1
V2 = 30V 99.01 kΩ
=9.93 V
R1
(99.01kΩ+200kΩ)
VR2
• V2 = 9.93 V
• The meter causes less than 1%
R2
drop in measured voltage, but
this increases as the size of the
I2
measured resistors increase
•
•
•
•
R3
I3
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Special Conditions
• If two equal resistors are in parallel,
the current AND voltage is the same
for both resistors
• If two equal resistors are in parallel,
the equivalent resistance is equal to
just ½ of one of the resistors
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Example
•
•
•
•
•
•
•
IR1= 25V/4.7kΩ = 5.32 mA
IR2= 25V/4.7kΩ = 5.32 mA
IT = IR1 + IR2 = 5.32 mA + 5.32 mA = 10.6mA
RT = VT/IT = 25V / 10.6mA = 2.36 k Ω
Check: RT = 4.7kΩ 4.7kΩ / (4.7kΩ 4.7kΩ)
RT = 2.35 kΩ
Currents same, RT = R1/2
V=
25V
R1 =
4.7kΩ
R2 =
4.7kΩ
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Example
• All resistors are
20Ω.
• The current in R3
is 10mA.
• Find the voltage
drop across and
the current
through all the
other resistors.
R1
RL1
R2
RL2
R3
RL3
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Analysis
•
•
•
•
•
•
•
•
•
•
VR3 = 10mA 20 Ω = .2 V
VL3 = VR3 so IL3 = .2 V/ 20 Ω = 10 mA
IR2 = IR3 + IL3 = 10 mA + 10 mA = 20 mA
VR2 = 20mA 20 Ω = .4 V
VL2 = VR2 + VR3 so IL2 = .6 V/ 20 Ω = 30 mA
IR1 = IR2 + IL2 = 20 mA + 30 mA = 50 mA
VR1 = 50mA 20 Ω = 1 V
VL1 = VR1+ VR2 + VR3 so
IL1 = 1.6 V/ 20 Ω = 80 mA
IT = IR1 + IL1 = 50 mA + 80 mA = 130 mA
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Checks
• VT = VR1 + VR2 + VR3 = .2V+.4V+1.0V = 1.6V
• RT=VT/IT = 1.6V / 130mA = 12.3 Ω
80
Check your work – Find RT
20Ω
??V
20Ω
18V
20Ω
20Ω
20Ω
20Ω
20Ω
20Ω
20Ω
20Ω
10Ω
81
Simplify
20k
??V
20k
20Ω
20Ω
10 Ω
20Ω
??V
20Ω
20Ω
30 Ω
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Simplify
20Ω
??V
20Ω
20Ω
30 Ω
20Ω
??V
20Ω
12Ω
83
Simplify
20Ω
18V
20Ω
12Ω
20Ω
18V
32Ω
84
Checks!!!!
= 12.3 Ω
20Ω
??V
??V
32Ω
85
Bridge Circuits
• In a balanced bridge circuit, the current
through and the voltage drop across R5 = 0:
• If any one of the outer resistances is
increased, the voltage drop across that
resistance will also increase. Therefore, it is
an unbalanced bridge circuit.
R1
R
 2
R3
R4
A
B
86
The Wheatstone Bridge
• This circuit is a series-parallel
circuit that is very popular in
controls and industrial
applications.
• There are two states for the
bridge:
Balanced
VA = VB
Unbalanced
VA ≠ VB
87
Why??
• In a balanced bridge, the voltage
drop across R3 = voltage drop across
R4
• There is NO current across the
resistor connecting R3 to R4
• VR3 = VInR3/(R3+R1)
• VR4 = VInR4/(R4+R2)
A
• VR3 = VR4
B
88
Calculations
• VInR3/(R3+R1) = VInR4/(R4+R2)
• Multiply both sides by (R3+R1)(R4+R2)
and divide both by Vin
• R3 (R4+R2) =R4 (R3+R1)
• R3 R4+ R3 R2 =R4 R3+ R4 R1
• R3 R2 =R4 R1
• R1 / R3 =R2 / R4
• In most problems, given 3, you find
the 4th value
89
Class Activity
• For the following balanced bridge
circuit, calculate the value of the
resistance, R.
90
Lab Experiment
• With a wire, resistance is
proportional to the length of the wire
– The longer the wire, the more resistance
it has
• A wire exactly 1 meter long is
connected to a battery
– A slider that moves up and down the
wire makes an electrical connection to it
91
Lab Setup
73 cm
XXX Ω
0 mA
35 V
L
A
R
27 cm
1kΩ
92
Suppose you forgot?
• VL = 35V 27 cm / (27cm + 73 cm) = 9.45
V = VR
• I1k = 9.45 V / 1 kΩ = 9.45 mA
• VXXX = 35 V – 9.45 V = 25.55 V
• RXXX = V/I = 25.55 V / 9.45 ma = 2.70 kΩ
• R1 / R3 =R2 / R4
• 73/27 = X/1k Ω
• X = 1k 73/ 27 = 2.70 kΩ
93
Practice
33 cm
XXX Ω
0 mA
20 V
L
A
R
67 cm
5kΩ
94
Unit 7 Summary
1. Identifying series/parallel relationships in
combinational circuits
2. Calculating resistance, current or voltage
in combinational resistive circuits
3. Applying KVL and KCL to combinational
resistive circuits
4. Wheatstone bridge
5. Loaded voltage-dividers
6. Meter loading