ECE 3183 – EE Systems Chapter 2 – Part A Parallel, Series and General Resistive Circuits.

ECE 3183 – EE Systems

Chapter 2 – Part A

Parallel, Series and General Resistive Circuits

ECE 3183 – Chapter 2 – Part A CHAPTER 2 A-2

Chapter 2 Resistive Circuits

1. Solve circuits (i.e., find currents and voltages of interest) by combining resistances in series and parallel.

2. Apply the voltage-division and current-division principles.

3. Solve circuits by the node-voltage technique.

4. Solve circuits by the mesh-current technique.

5. Find Thévenin and Norton equivalents.6. Apply the superposition principle.


+-

+-

+ -

+-

+ -

FIRST GENERALIZATION: MULTIPLE SOURCES

i(t)

KVL

Voltage sources in series can be algebraically added to form an equivalent source.

We select the reference direction to move along the path.Voltage drops are subtracted from rises

1R

2R

1Rv

2Rv

1v

2v

3v

4v

5v


+-

FIRST GENERALIZATION: MULTIPLE SOURCES

eqv

1R

2R

veq = ?


SECOND GENERALIZATION: MULTIPLE RESISTORS

APPLY KVLTO THIS LOOP

VOLTAGE DIVISION FOR MULTIPLE RESISTORS

iRv iRi



)30(,, kPVI bd FIND

APPLY KVLTO THIS LOOP

bdV FOR LOOP


THE CONCEPT OF EQUIVALENT CIRCUITS

THIS CONCEPT WILL OFTEN BE USED TO SIMPLFYTHE ANALYSIS OF CIRCUITS. WE INTRODUCE ITHERE WITH A VERY SIMPLE VOLTAGE DIVIDER

+-

1R

2R

Sv

i

21 RRv

i S

+-Sv 21 RR

i

AS FAR AS THE CURRENT IS CONCERNED BOTHCIRCUITS ARE EQUIVALENT. THE ONE ON THERIGHT HAS ONLY ONE RESISTOR

1R 2R

21 RR

SERIES COMBINATION OF RESISTORS


THE DIFFERENCE BETWEEN ELECTRICCONNECTION AND PHYSICAL LAYOUT

SOMETIMES, FOR PRACTICAL CONSTRUCTION REASONS, COMPONENTS THAT ARE ELECTRICALLYCONNECTED MAY BE PHYSICALLY FAR APART

IN ALL CASES THE RESISTORS ARE CONNECTED IN SERIES


SUMMARY OF BASIC VOLTAGE DIVIDER

VR1 = ?

VR2 = ?



kRkRVVS 30,90,9 21 :EXAMPLE

kR 151

VOLUMECONTROL?



A “PRACTICAL” POWER APPLICATION

HOW CAN ONE REDUCE THE LOSSES?


THE “INVERSE” VOLTAGE DIVIDER

+-

1R

2RSV

OV

SO VRR

RV

21

2

VOLTAGE DIVIDER

OS VR

RRV

2

21

"INVERSE" DIVIDER


THE “INVERSE” VOLTAGE DIVIDER

SV COMPUTE


Current Division

total21

2

11 i

RR

R

R

vi

total21

1

22 i

RR

R

R

vi


FIND Vx, i3


Find i1, i2, and i3


SERIES AND PARALLEL RESISTOR COMBINATIONS

UP TO NOW WE HAVE STUDIED CIRCUITS THATCAN BE ANALYZED WITH ONE APPLICATION OFKVL(SINGLE LOOP) OR KCL(SINGLE NODE-PAIR)

WE HAVE ALSO SEEN THAT IN SOME SITUATIONSIT IS ADVANTAGEOUS TO COMBINE RESISTORS TO SIMPLIFY THE ANALYSIS OF A CIRCUIT

NOW WE EXAMINE SOME MORE COMPLEX CIRCUITSWHERE WE CAN SIMPLIFY THE ANALYSIS USINGTHE TECHNIQUE OF COMBINING RESISTORS…

… PLUS THE USE OF OHM’S LAW


SERIES AND PARALLEL RESISTOR COMBINATIONS

SERIES COMBINATIONS

PARALLEL COMBINATION

N21p G...GGG

N21s G...

GGG

1111


FIRST WE PRACTICE COMBINING RESISTORS

FIND RAB


FIRST WE PRACTICE COMBINING RESISTORS

6k||3k

(10K,2K)SERIES = 12K

SERIESk3

k12k3

k5

RAB = 5K

(12K||6K) = 4K

(4K,2K)SERIES = 6KWe need to re-draw!


EXAMPLES COMBINATION SERIES-PARALLEL

k9

kkk 69||18

kkk 1066

AN EXAMPLE WITHOUT REDRAWING

kkk 612||12 kkk 26||3

)24(||6 kkk

kRAB 3

kRAB 22

k12


EXAMPLES COMBINATION SERIES-PARALLEL

RESISTORS ARE IN SERIES IF THEY CARRYEXACTLY THE SAME CURRENT (SHARE ONE COMMON NODE)

RESISTORS ARE IN PARALLEL IF THEY HAVE THE SAME VOLTAGE ACROSS THEM AND ARE CONNECTED EXACTLY BETWEEN THE SAME TWO NODES


Strategy for analyzing circuits with series and parallel combinations of resistors:

• Systematically reduce the resistive network so that the resistance seen by the source is represented by a single resistor.

• Determine the source current for a voltage source or the source voltage for a current source.

• Expand the network, apply Ohm’s law, KVL, KCL, voltage division, and current division to determine all currents and voltages in the network


Circuit analysis example

6 k

6 k

2 k 4 k

1 2 V

I o

3 k

Find Io in the circuit shown.


SV FIND

THIS IS AN INVERSE PROBLEMWHAT CAN BE COMPUTED?

V9

mA05.0

V6

OV FIND

DIVIDER VOLTAGEUSE

FIND :STRATEGY 1V

1V k60

kkk 2060||30

+-

1Vk20

k20

V12V

kk

k6)12(

2020

20

V6

DIVIDER VOLTAGE

14020

20V

kk

kV

O

V2

VmAkVS

615.0*20

mA15.0

mAkV 1.0*601

k

VI

120

61


Circuits with dependent sources

• When writing the KVL and/or KCL equations for the network, treat the dependent source as though it were an independent source.

• Write the equations that specify the relationship of the dependent source to the controlling parameters.

• Solve the equations for the unknowns. Be sure that the number of linearly independent equations matches the number of the unknowns.


A PLAN:IF V_s IS KNOWN V_0 CAN BE DETERMINED USING VOLTAGE DIVIDER.TO FIND V_s WE HAVE A SINGLE NODE-PAIR CIRCUIT

ALGEBRAICALLY, THERE ARE TWO UNKNOWNSAND JUST ONE EQUATION

THE EQUATION FOR THE CONTROLLINGVARIABLE PROVIDES THE ADDITIONAL EQUATION

SUBSTITUTION OF I_0 YIELDS

6056/* SVkVOLTAGE DIVIDER

VVkk

kV SO )12(

3

2

24

4

OV FINDKCL TO THIS NODE. THEDEPENDENT SOURCE IS JUSTANOTHER SOURCE


Problem solving strategy: Circuit with dependent sources

V S

R S

R in

R o

R L V in

+

V in

_

+

V o

_

For the network shown, what is the resulting ratio , Vo /Vs ?



V S

R S

R in

R o

R L V in

+

V in

_

+

V o

_


inin S

S in

RV V

R R

(voltage divider for resistors in series)



V S

R S

R in

R o

R L V in

+

V in

_

+

V o

_


inin S

S in

RV V

R R

inL Lo in S

o L S in o L

RR RV V V

R R R R R R





V S

R S

R in

R o

R L V in

+

V in

_

+

V o

_


inin S

S in

RV V

R R

inL Lo in S

o L S in o L

RR RV V V

R R R R R R



o in L

S S in o L

V R R

V R R R R

ECE 3183 – EE Systems Chapter 2 – Part A Parallel, Series and General Resistive Circuits.

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Transcript of ECE 3183 – EE Systems Chapter 2 – Part A Parallel, Series and General Resistive Circuits.