Chapter 6 – Series and Parallel Combination Circuits electronics fundamentals circuits, devices,...
Click here to load reader
-
Upload
madeline-wade -
Category
Documents
-
view
305 -
download
17
Transcript of Chapter 6 – Series and Parallel Combination Circuits electronics fundamentals circuits, devices,...
Chapter 6 – Series and Parallel Combination Circuits
electronics fundamentalscircuits, devices, and applications
THOMAS L. FLOYDDAVID M. BUCHLA
Series and Parallel CircuitsSeries and Parallel Circuits
Most practical circuits have combinations of series and parallel components.
Identifying series-parallel relationships
Components that are connected in series will share a common path.
Components that are connected in parallel will be connected across the same two nodes.
1 2
From Chapters 4 and 5
Series and Parallel CircuitsSeries and Parallel Circuits
You can frequently simplify analysis by combining series and parallel components.
Combination circuits
Solve by forming the simplest equivalent circuit possible.
1. characteristics that are electrically the same as another circuit but
2. is generally simpler.
Circuits containing both series and parallel circuits are called COMBINATION circuits
An equivalent circuit is one that has:
Series and Parallel CircuitsSeries and Parallel Circuits Equivalent circuits
R 1
R 2R 2
1
1 .0 k
1 .0 k
is equivalent to R 11
2 .0 k
There are no electrical measurements that can distinguish the boxes.
Series and Parallel CircuitsSeries and Parallel Circuits
Another example:
There are no electrical measurements that can distinguish the boxes.
R R1 2
1 .0 k 1 .0 kR
1 ,2
5 0 0
is equivalent to
Equivalent circuits
Series and Parallel CircuitsSeries and Parallel Circuits
R
R1
R 2R 2
31 .0 k
4 .7 k2 .7 k
is equivalent to
R R1 ,2 3
4 .7 k3 .7 k
R 1 ,2 ,3
2 .0 7 k
is equivalent to
There are no electrical measurements that can distinguish between the three boxes.
Equivalent circuits
Series and Parallel CircuitsSeries and Parallel Circuits
What Do We Know1.For Series Circuits:
a. Current at all points is the IT = IR1 = IR2
b. Voltage across each resistorVT = V1 + V2
2.For Parallel Circuita. Current
IT = IR1 + IR2
b. Voltage at all nodes is the VT = V1 = V2
same
same
drops
divides across each resistor in the branch
Series and Parallel CircuitsSeries and Parallel CircuitsSeven Step Process for Solving a Combination Circuit
1.Simplify the circuit to a series circuit by finding the effective equivalent resistance (REQ) of each parallel section in the circuit. Redraw the simplified circuit.
2.Calculate the total resistance (RT) of the circuit by adding all REQ’s to the other series resistances.
3.Calculate the total current (IT) using RT in Ohm’s law.
4.Calculate the voltage drop across any series resistances or REQ’s using Ohm’s law.
5.Calculate the branch currents in all parallel sections of the circuit using the voltage drop across REQ and Ohm’s law.
6.Use the branch currents and resistance values to calculate the voltage of the parallel resistances.
7.Make a summary of the voltage drops and currents for each resistance to make sure they total correctly.
Series and Parallel CircuitsSeries and Parallel Circuits
A simple series-parallel resistive circuit.
Series and Parallel CircuitsSeries and Parallel Circuits
R4 is added to the circuit in series with R1.
Series and Parallel CircuitsSeries and Parallel Circuits
R5 is added to the circuit in series with R2.
Series and Parallel CircuitsSeries and Parallel Circuits
R6 is added to the circuit in parallel with the series combination of R1 and R4.
Series and Parallel CircuitsSeries and Parallel Circuits Sketch the circuits for nodesA to BA to CB to C
Series and Parallel CircuitsSeries and Parallel Circuits Sketch the circuits for nodesA to B
)( 321 RRRRT
Series and Parallel CircuitsSeries and Parallel CircuitsSketch the circuits for nodes
A to C
)( 213 RRRRT
Series and Parallel CircuitsSeries and Parallel Circuits Sketch the circuits for nodesB to C
)3( 12 RRRRT
Series and Parallel CircuitsSeries and Parallel Circuits
7412365)( )())(( RRRRRRRR AB
Series and Parallel CircuitsSeries and Parallel Circuits
Reduce the following circuit to REQ
Series and Parallel CircuitsSeries and Parallel Circuits
)( 43 RRREQ
6 Ω
Series and Parallel CircuitsSeries and Parallel Circuits
643 )( RRRREQ
10 Ω
Series and Parallel CircuitsSeries and Parallel Circuits
5643 ))(( RRRRREQ 5 Ω
Series and Parallel CircuitsSeries and Parallel Circuits
215643 )))((( RRRRRRREQ 50 ΩIT = 2
amps
IT
Series and Parallel CircuitsSeries and Parallel Circuits
Review of voltage relationships.
Series and Parallel CircuitsSeries and Parallel Circuits
I
+
+
26.5 mA
I
+18.5 mA
I
+8.0 mA
R5100
R3330
R2470
R1270
VS5.0 V
R4
100
R6
100
A- -
-
Kirchoff’s current law
What are the readings for node A?
Series and Parallel CircuitsSeries and Parallel Circuits Loaded voltage divider
1. A voltage-divider with a resistive load forms a combination (parallel) circuit.
The voltage-divider equation was developed for a series circuit. Recall that the output voltage is given by
A
22 S
T
RV V
R
R1
R2 R3
+
2. The voltage divider is said to be LOADED.
3. The loading reduces the total resistance from node A to ground.
Series and Parallel CircuitsSeries and Parallel Circuits Loaded voltage divider
A
What is the voltage across R3?
Form an equivalent series circuit by combining R2 and R3; then apply the voltage-divider formula to the equivalent circuit:
2,33 2,3 S
1 2,3
387 15 V
330 387
RV V V
R R
+15 V R1
R2 R3
330
470 2.2 k
VS =
2,3 2 3 470 2.2 k = 387 R R R
8.10 V
Series and Parallel CircuitsSeries and Parallel Circuits The loading effect of a voltmeter.
Series and Parallel CircuitsSeries and Parallel Circuits Loading effect of a voltmeter
1. A voltmeter has internal resistance
2. This RINT can change the resistance of the circuit under test.
3. A 1 M internal resistance of the meter accounts for the readings.
Given: VS = 10 V and R1 and R2 are not defective but the meter reads only 4.04 V when it is across either R1 or R2.
R1
470 k
R2
47 k0
VS +10 V
+ 10 V
What is a possible explanation of the meter not displaying 10 volts?
+4.04 V
+4.04 V
Series and Parallel CircuitsSeries and Parallel CircuitsWheatstone bridge
• The Wheatstone bridge consists of:1. a dc voltage source and 2. four resistive arms
forming two voltage dividers.
• The output is taken between the dividers.
• Frequently, one of the bridge resistors is adjustable. (R2)
-
+
R 1R 3
R 4R 2
V SO u tp u t
.
Series and Parallel CircuitsSeries and Parallel CircuitsBalanced Wheatstone bridge
When the bridge is balanced, the output voltage is
-
+
R 1R 3
R 4R 2
V SO u tp u t
zero
The products of resistances in the opposite diagonal arms are equal.
Series and Parallel CircuitsSeries and Parallel Circuits Wheatstone bridge.
Voltage Divider 1
Voltage Divider 2
Series and Parallel CircuitsSeries and Parallel Circuits Wheatstone bridge.
V1=V2
I1=I3
V3=V4
I2=I4
4
231
R
RRR
Series and Parallel CircuitsSeries and Parallel Circuits Wheatstone bridge
-
+
R 1R 3
R 4R 2
V SO u tp u t
What is the value of R2 if the bridge is balanced?
470 330
270
12 V
384
4
231
R
RRR
Series and Parallel CircuitsSeries and Parallel CircuitsFinding an Unknown Resistance
4
2
R
RRR VX
Scale Factor
Series and Parallel CircuitsSeries and Parallel Circuits
Measuring a physical parameter using a transducer.
Unbalanced Wheatstone Bridge
• Unbalance occurs when VOUT ≠ 0
• Used to measure1. Mechanical Strain2. Temperature3. Pressure
• VOUT is converted to a digital output indicating the value of the reading.
Series and Parallel CircuitsSeries and Parallel Circuits
ST
XX V
R
RV
VOUT = VA-VB = 0
Series and Parallel CircuitsSeries and Parallel Circuits
10
10
10
10
10
Using the Thermistor chart, what is VOUT when the temperature is 50o C
VA=8.8v VB=6.0v VA-B=2.8v
Series and Parallel CircuitsSeries and Parallel Circuits Example of a load cell.
Series and Parallel CircuitsSeries and Parallel Circuits
1. Remove RL to make an open circuit between A & B
2. Calculate R1||R3 =
3. Calculate R2||R4 =
4. Calculate the voltage from A to ground
5. Calculate the voltage from B to ground
165
179
7.5 V
6.87 V
R3 R4
R2
RL
R1VS
-
+A B
330 390
330 330
+15 V150
Wheatstone Bridge
Series and Parallel CircuitsSeries and Parallel Circuits
The maximum power is transferred from a source to a load when: RL = RS(resistance of the voltage source)
The maximum power transfer theorem assumes the source voltage and resistance are fixed.
RS
RL
VS +
Maximum power transfer Theorem
Series and Parallel CircuitsSeries and Parallel CircuitsMaximum power transfer Theorem
What is the power delivered to the load?
The voltage to the load is
RS
RL
VS + 50
50 10 V
22
LL
5.0 V= 0.5 W
50
VP
R
The power delivered is:
5.0 V
Series and Parallel CircuitsSeries and Parallel CircuitsSuperposition theorem
A way to determine currents and voltages in a linear circuit that has multiple sources 1.Take one source at a time and 2.Algebraically summing the results.
+-
-
+
-
+
R 1 R 3
R 2
I 2
V S 2V S 1
1 2 V
2 .7 k 6 .8 k
6 .8 k
1 8 V
Series and Parallel CircuitsSeries and Parallel CircuitsSuperposition theorem
Four Step Process1.Leave one voltage or current source at a time in the circuit (circuit 1) and replace the other (circuit 2) with a short.2.Calculate REQ/Total and then calculate the voltage or current for the resistor(s).3.Repeat steps 1 and 2 for the other circuit (circuit 2).4.Algebraically add the results for all sources.
Series and Parallel CircuitsSeries and Parallel Circuits
Series and Parallel CircuitsSeries and Parallel Circuits
SummarySummary
6.10 k
What does the ammeter read for I2?
1.97 mA 0.98 mA
8.73 k 2.06 mA
+-
-
+
-
+
R 1 R 3
R 2
I 2
V S 2V S 1
1 2 V
2 .7 k 6 .8 k
6 .8 k
1 8 V
0.58 mA
1.56 mA
Source 1: RT(S1)= I1= I2= Source 2: RT(S2)= I3= I2= Both sources I2=
Set up a table of pertinent information and solve for each quantity listed:
The total current is the algebraic sum.
+ -
-
+
R1 R3
R2
I2VS1
12 V
2.7 k 6.8 k
6.8 k
+ -
-
+
R1 R3
R2
I2
VS2
2.7 k 6.8 k
6.8 k
18 V+-
-
+
-
+
R 1 R 3
R 2
I 2
V S 2V S 1
1 2 V
2 .7 k 6 .8 k
6 .8 k
1 8 V1.56 mA
Series and Parallel CircuitsSeries and Parallel Circuits
A B
VTH VTH
RTH RTH'
'165 179
7.5 V 6.87 V
RLA B
150 VTH VTH
RTH RTH'
'165 179
7.5 V 6.87 V
Putting the load on the Thevenin circuits and applying the superposition theorem allows you to calculate the load current. The load current is: 1.27 mA
Thevenin’s theorem and Wheatstone Bridge
.0152-.0191
Series and Parallel CircuitsSeries and Parallel CircuitsTroubleshooting
The effective troubleshooter must think logically about circuit operation.
Understand normal circuit operation and find out the symptoms of the failure.
Decide on a logical set of steps to find the fault.
Following the steps in the plan, make measurements to isolate the problem. Modify the plan if necessary.
Series and Parallel CircuitsSeries and Parallel Circuits Troubleshooting
The output of the voltage-divider is 6.0 V. Describe how you would use analysis and planning in finding the fault.
From an earlier calculation, V3 should equal 8.10 V. A low voltage is most likely caused by a low source voltage or incorrect resistors (possibly R1 and R2 reversed). If the circuit is new, incorrect components are possible.
Decide on a logical set of steps to locate the fault. You could decide to 1) check the source voltage, 2) disconnect the load and check the output voltage, and if it is correct, 3) check the load resistance. If R3 is correct, check other resistors.
A+15 V R1
R2 R3
330
470 2.2 k
VS =
Series and Parallel CircuitsSeries and Parallel Circuits FIGURE 6–97
Series and Parallel CircuitsSeries and Parallel Circuits
Loading
Load current
Bleeder current
Wheatstone bridge
The current left after the load current is subtracted from the total current into the circuit.
The output current supplied to a load.
The effect on a circuit when an element that draws current from the circuit is connected across the output terminals.
A 4-legged type of bridge circuit with which an unknown resistance can be accurately measured using the balanced state. Deviations in resistance can be measured using the unbalanced state.
Selected Key Terms
Series and Parallel CircuitsSeries and Parallel Circuits
Thevenin’s theorem
Maximum power transfer
Superposition
The condition, when the load resistance equals the source resistance, under which maximum power is transferred to the load.
A method for analyzing circuits with two or more sources by examining the effects of each source by itself and then combining the effects.
A circuit theorem that provides for reducing any two-terminal resistive circuit to a single equivalent voltage source in series with an equivalent resistance.
Selected Key Terms
Series and Parallel CircuitsSeries and Parallel Circuits
1. Two circuits that are equivalent have the same
a. number of components
b. response to an electrical stimulus
c. internal power dissipation
d. all of the above
Series and Parallel CircuitsSeries and Parallel Circuits
2. If a series equivalent circuit is drawn for a complex circuit, the equivalent circuit can be analyzed with
a. the voltage divider theorem
b. Kirchhoff’s voltage law
c. both of the above
d. none of the above
Series and Parallel CircuitsSeries and Parallel Circuits
3. For the circuit shown,
a. R1 is in series with R2
b. R1 is in parallel with R2
c. R2 is in series with R3
d. R2 is in parallel with R3
-
+
R 1
R 3
R 2
V S
Series and Parallel CircuitsSeries and Parallel Circuits
4. For the circuit shown,
a. R1 is in series with R2
b. R4 is in parallel with R1
c. R2 is in parallel with R3
d. none of the above
-
+R 1
R 4
R 3
R 2
V S
Series and Parallel CircuitsSeries and Parallel Circuits
5. A signal generator has an output voltage of 2.0 V with no load. When a 600 load is connected to it, the output drops to 1.0 V. The Thevenin resistance of the generator is
a. 300
b. 600
c. 900
d. 1200 .
Series and Parallel CircuitsSeries and Parallel Circuits
6. For the circuit shown, Kirchhoff's voltage law
a. applies only to the outside loop
b. applies only to the A junction.
c. can be applied to any closed path.
d. does not apply. R1
R3
470
270
R2
330
VS +10 V A
Series and Parallel CircuitsSeries and Parallel Circuits
7. The effect of changing a measured quantity due to connecting an instrument to a circuit is called
a. loading
b. clipping
c. distortion
d. loss of precision
Series and Parallel CircuitsSeries and Parallel Circuits
8. An unbalanced Wheatstone bridge has the voltages shown. The voltage across R4 is
a. 4.0 V
b. 5.0 V
c. 6.0 V
d. 7.0 V
-
+
R L
R 1R 3
R 4R 2
V S
12 V
1 .0 V
7 .0 V+ -
Series and Parallel CircuitsSeries and Parallel Circuits
9. Assume R2 is adjusted until the Wheatstone bridge is balanced. At this point, the voltage across R4 is measured and found to be 5.0 V. The voltage across R1 will be
a. 4.0 V
b. 5.0 V
c. 6.0 V
d. 7.0 V
-
+
RL
R1R3
R4R2
VS
12 V
5.0 V
+ -
Series and Parallel CircuitsSeries and Parallel Circuits
10. Maximum power is transferred from a fixed source when
a. the load resistor is ½ the source resistance
b. the load resistor is equal to the source resistance
c. the load resistor is twice the source resistance
d. none of the above
Series and Parallel CircuitsSeries and Parallel Circuits
Quiz
Answers:
1. b
2. c
3. d
4. d
5. b
6. c
7. a
8. a
9. d
10. b