Chapter 6 Series and Parallel Circuits 1 of 81 MECH1100 1 of 81 Chapter 6 Series and Parallel Circuits MECH1100 Topics Identifying Series- Parallel Relationships

Chapter 6 Series and Parallel Circuits

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MECH1100

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Chapter 6 Series and Parallel Circuits MECH1100

TopicsIdentifying Series- Parallel Relationships

Analysis of Series-Parallel Resistive Circuits

Voltage Dividers with Resistive Loads

Loading Effect of a Voltmeter

Wheatstone Bridge

Thevenin’s Theorem

Maximum Power Transfer

Superposition


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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


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You can frequently simplify analysis by combining series and parallel components.

Combination circuits

Solve by forming the simplest equivalent circuit possible.

1. has characteristics that are electrically the same as another circuit

2. is generally simpler.

Circuits containing both series and parallel circuits are called COMBINATION circuits

An equivalent circuit is one that :


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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 between the boxes.


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Another example:

There are no electrical measurements that can distinguish between the boxes.

R R1 2

1 .0 k 1 .0 kR

1 ,2

5 0 0

is equivalent to

Equivalent circuits


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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


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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


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Seven 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 circuit each time it is simplified.

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.


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A simple series-parallel resistive circuit.


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R4 is added to the circuit in series with R1.


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R5 is added to the circuit in series with R2.


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R6 is added to the circuit in parallel with the series combination of R1 and R4.


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Sketch the circuit and write the equation for

nodes A to B

)( 321 RRRRT


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)( 213 RRRRT


nodes A to C


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)3( 12 RRRRT


nodes B to C


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Draw the schematic for this circuit board and give the

equation for the relationship of the resistor

branches


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7412365)( )())(( RRRRRRRR ABEQ


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Reduce the following circuit to REQ


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)( 43 RRREQ

6 Ω


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643 )( RRRREQ

10 Ω6 4


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5643 ))(( RRRRREQ 5 Ω10 10


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215643 )))((( RRRRRRREQ 50 ΩIT = 2

amps

IT

5 15 30


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Review of voltage relationships

688ABR

647BCR

1334)( ACEQR

15.510*1334

687)( ABV

85.410*1334

647)( BCV

107.49

1334TI mA


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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?


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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.


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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 R

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


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Stiff voltage divider

• A stiff voltage-divider is one in which the loaded voltage is nearly the same as the no-load voltage.

• To accomplish this, the load current must be small compared to the bleeder current (or RL is large compared to the divider resistors).

R1

R2 RL

VS

21 & RRRL


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Stiff voltage divider

R1

R2 RL

VS

Given: R1 = R2 = 1.0 kW, 1. What value of RL will

make the divider a stiff voltage divider?

2. What fraction of the unloaded voltage is the loaded voltage?

Rule of Thumb: RL = R2*10; Thus: RL should be 10 kW or greater.

2 LRL S S S

1 2 L

|| 0.91 k 0.476

|| 1.0 k 0.91 k

R RV V V V

R R R

This is 95% of the unloaded voltage.

SSSloadnoR VVkk

kV

RR

RV 5.

11

1

21

2)(2

-


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Wheatstone bridge

Voltage Divider 1

Voltage Divider 2


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Wheatstone Bridge

V1=V2

I1=I3

V3=V4

I2=I4

4

231

4

2

3

1

R

RRR

R

R

R

RL R

L R

U U

L L


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Wheatstone 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 tpu t

.

+4.04 V


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Balanced Wheatstone Bridge

When the bridge is balanced, the output voltage is

-

+

R 1R 3

R 4R 2

V SO u tpu t

zero

The products of resistances in the opposite diagonal arms are equal.


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Wheatstone bridge

-

+

R 1R 3

R 4R 2

V SO u tpu t

What is the value of R2 if the bridge is balanced?

470 W 330 W

270 W

12 V

385 W

4

321

R

RRR


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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.


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ST

XX V

R

RV

VOUT = VA-VB = 0


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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

3

3A S

RV V

R Rtherm

SB VRR

RV

42

4


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Example of a load cell


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Load Cell or Strain Gauge


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Typical load cell (strain gauge) schematic


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Troubleshooting

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.


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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 =


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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.

Key Terms


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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.

Key Terms


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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

Quiz


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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

Quiz


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3. For the circuit shown,

a. R1 is in series with R2∥R3b. R1 is in parallel with R2

c. R2 is in series with R3

d. R2 is in parallel with R1 + R3

-

+

R 1

R 3

R 2

V S

Quiz


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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

Quiz


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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

Quiz


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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

Quiz


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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+ -

VR2 = 5V

VR2 – VRL = 4V

VR3 = 8V

Quiz


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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

+ -

Quiz


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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

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Chapter 6 Series and Parallel Circuits 1 of 81 MECH1100 1 of 81 Chapter 6 Series and Parallel Circuits MECH1100 Topics Identifying Series- Parallel Relationships