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


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.


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


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


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.


A simple series-parallel resistive circuit.


R4 is added to the circuit in series with R1.


R5 is added to the circuit in series with R2.


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


7412365)( )())(( RRRRRRRR AB


Reduce the following circuit to REQ


)( 43 RRREQ

6 Ω


643 )( RRRREQ

10 Ω


5643 ))(( RRRRREQ 5 Ω


215643 )))((( RRRRRRREQ 50 ΩIT = 2

amps

IT


Review of voltage relationships.


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


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.


ST

XX V

R

RV

VOUT = VA-VB = 0


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.


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


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.


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


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


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


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


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


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


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


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


-

+R 1

R 4

R 3

R 2

V S


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 .


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


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


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+ -


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

+ -


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



Quiz

Answers:

1. b

2. c

3. d

4. d

5. b

6. c

7. a

8. a

9. d

10. b

Chapter 6 – Series and Parallel Combination Circuits electronics fundamentals circuits, devices,...

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Transcript of Chapter 6 – Series and Parallel Combination Circuits electronics fundamentals circuits, devices,...