bmayer@chabotcollege.edu engr-43_lec-02a_sp_vi-divide_nodemesh.pptx 1 bruce mayer, pe...

BMayer@ChabotCollege.edu • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Engineering 43

Series/Parallel,Dividers,Nodes & Meshes

Series Parallel

Up To Now We Have Studied Circuits That Can Be Analyzed With One Application Of KVL Or KCL

We will see That In Some Situations It Is Advantageous To Combine Resistors To Simplify The Analysis Of A Circuit

Now We Examine Some More Complex Circuits Where We Can Simplify The Analysis Using Techniques:• Combining Resistors• Ohm’s Law

Resistor Equivalents Series

• Resistors Are In Series If TheyCarry Exactly The Same Current

Parallel• Resistors Are In Parallel If

They have Exactly the Same Potential Across Them

NS RRRRR 321

NP RRRRR

Conductance Equivalents

ReCall: G = 1/R For SERIES

Connection

For PARALLELConnection

GS = 1.479 S

GP = 15 S

Combine ResistorsExample:Find RAB

6k||3k = 2k

(10K,2K)SERIES

SERIESk3

kkk 412||6

(4K,2K)SERIES

kkk 36||6 (3K,9K)SERIES

kkk 312||4

More Examples Step-1: Series

Reduction Step-2: Parallel

Reduction

kkk 69||18

kkk 1066

Example w/o Redrawing

Step-2: 12k 12k = 6k Step-3: 3k 6k = 2k

Step-1: 4k↔8k = 12k

Step-4: 6k (4k↔2k) = 3k = RAB

kkk 612||12

kkk 26||3

)24(||6 kkk

Series-Parallel Resistor Circuits

Combing Components Can Reduce The Complexity Of A Circuit And Render It Suitable For Analysis Using The Basic Tools Developed So Far• Combining Resistors In SERIES Eliminates

One NODE From The Circuit• Combining Resistors In PARALLEL

Eliminates One LOOP From The Circuit

S-P Circuit Analysis Strategy

Reduce Complexity Until The Circuit Becomes Simple Enough To Analyze

Use Data From Simplified Circuit To Compute Desired Variables In Original Circuit • Hence Must Keep Track Of Any

Relationship Between Variables

Example – Ladder Network

Find All I’s & V’s in Ladder Network• 1st: S-P Reduction

k12kk 12||4

kk 6||63

• 2nd: S-P Reduction– Also by Ohm’s Law

Ladder Network cont.• Final Reduction; Find Calculation Starting

Points

a 30.13

mAIIIImAk

5.05.06

• Now “Back Substitute” Using KVL, KCL, and Ohm’s Law– e.g.; From Before

The Voltage Divider

tiRRtv

tiRtiRtv

Ohm’s Law in KVL

Find i(t) by

Ohm’s Law

KVL ON THIS LOOP

Voltage Divider cont.

Now Sub i(t) Into Ohm’sLaw to Arrive at TheVoltage Divider Eqns

1 21and

tvRv RR

tvvR RR

1 0and0

1112 21

tvRvR RR

Quick Chk → In Turn, Set R1, R2 to 0

KVL ON THIS LOOP

V-Divider Summary

Governing Equations

• The Larger the R, The Larger the V-drop

Example• Gain/Volume Control

– R1 is a VariableResistor Called aPotentiometer, or “Pot” for Short

Volume Control Example• Case-I → R1 = 90 kΩ

2.25V93090

V4.593020

• Case-II → R1 = 20 kΩ 30kΩ

Practical Example Power Line

8.25% of Pwr Generated is Lost to Line Resistance!* How to Reduce Losses?

Power Dissipated by the Line is a LOSS

Using Voltage Divider

kV 367

kV 400Ω 16.5183.5

Ω 183.5load

V MW 7345.183kA2 2

MW 665.16kA2 2line

loadsrcLOSS-line

Equivalent Circuit

The Equivalent Circuit Concept Can Simplify The Analysis Of Circuits• For Example, Consider A Simple

Voltage Divider

+-Sv 21 RR

SERIES Resistors → 1R 2Rº

– As Far As TheCurrent IsConcerned BothCircuits AreEquivalent The One On The Right Has Only One Resistor

Schematic vs. Physical

Sometimes, For Practical Construction Reasons, Components That Are Electrically Connected May Be Physically Quite Apart• Each Resistor Pair Below Has the SAME

Node-to-Node Series-Equivalent Circuit

COMPONENT SIDE

CONNECTOR SIDE

ILLUSTRATING THE DIFFERENCEBETWEEN PHYSICAL LAYOUT ANDELECTRICAL CONNECTIONS

PHYSICAL NODE

SECTION OF 14.4 KB VOICE/DATA MODEM

CORRESPONDING POINTS

Generalization Multiple v-Sources

Voltage Sources In SeriesCan Be AlgebraicallyAdded To Form AnEquivalent Source• We Select The Reference

Direction To Move AlongThe Path

i(t)+-

01542321 vvvvvvv RR

– Voltage Rises AreSubtracted From Drops

Apply KVL

Multiple v-Source Equivalent

Collect All SOURCES On One Side

The Equivalent Circuit:• V-source in Series

ADD directly

2154321 RR vvvvvvv

21 RReq vvv

Generalization Mult. Resistors

Apply KVL (rise = Σdrops)

tiRv kRik

Now by Ohm’s Law

And Define RS

Then Voltage Division For Multiple Resistors

• [Rk/RS] is the Divider RATIO

Example

Find: I, Vbd, P30kΩ

Apply KVL & Ohm

VVIkV bdbd 10 KVLby So 0][2012

WWARIP 3001030)1030()10( 43242

Solving for I

Now Vbd

Finally, The 30 kΩ Resistor Power Dissipation

APPLY KVLTO THIS LOOP

Examples

Find: I, Vbd

• Use KVL and Ohm’s Law

APPLY KVLTO THIS LOOP

mAIkIkI 05.004012806

VVVkIV bdbd 1001240

VVS 9320

201525

Find VS by V-Divider • The V20k Divider Eqn

201525

• Solving for VS

When In Doubt → ReDraw

From The Last DiagramIt Was Not ImmediatelyObvious That This Wasa V-Divider Situation• UnTangle/Redraw at Right

251520

3251520

Single Node-Pair (SNP) Circuits

SNP Circuits Are Characterized By ALL the Elements Having The SAME VOLTAGE Across Them → They Are In PARALLEL

SNP Example

EXAMPLE OF SINGLE NODE-PAIR

This Element is INACTIVE

• The Inactive Element Has NO Potential Across it → SHORT Circuited

UnTangling Reminder

Nodes Can Take STRANGE Shapes

LowDistortion

PowerAmplifier

NODE → A region of Constant Electrical Potential

e.g.; a group of connected WIRES is ONE Node

LOW VOLTAGE POWER SUPPLY FOR CRT - PARTIAL VIEW

SOME PHYSICAL NODES

COMPONENT SIDE CONNECTION SIDE

The Current Divider

Basic Circuit

The Current i(t) Enters The Top Node then Splits, or DIVIDES, into the the Currents i1(t) and i2(t)

Apply KCL at Top Node

Use Ohm’s Law to Replace Currents

APPLY KCL

The Current Divider cont.

Basic Circuit

By KCL & Ohm The Current Division

Define PARALLEL Resistance

Current Divider Example

By I-Divider

For This Ckt Find: I1, I2, Vo

When in doubt… REDRAW the circuit to Better Visualize the Connections

2-Legged Divider is

more Evident

kΩ802

Real World Example

Car Stereo and Circuit Model

Use I-Divider to Find Current thru the 4Ω Speakers

Thus the Speaker Power

Power Per Speaker by Joule

mA215 mA215

Current & Power Example

Find I2 by I-Divider OR KCL• Choose KCL

For This Ckt Find: • I1, I2,

• P40k Power ABSORBED by 40 kΩ Resistor

By I-Divider

1640120

The 40k Power by RI2

millikmilli

76.55760

Generalization: Multi i-Sources

KCL on Top Node:

Given Single Node-Pair Ckt w/ Multiple Srcs

The Equivalent Ckt Combine Src Terms To Form Equivalent Source

Generalization: Multi i-Sources

By Analysis and Electrical Physics of KCL

Thus CURRENT Sources in PARALLEL ADD directly • Compare to VOLTAGE Sources in SERIES

which also ADD Directly

=Oiiiii 6431

i-Source Example

Combine Srcs to Yield Equivalent Ckt

For This Ckt Find Vo, and the Power Supplied by the I-Srcs

Vo by Ohm’s Law V10mA5k2 Srcpo IRV

mA10 mA15k3

kkRp 2

5mA1V10

mW100mA10V10

Use PASSIVE SIGN Convention for Power

Generalization: Multi Resistors

KCL on Top Node:

Given Single Node-Pair Ckt w/ Multiple R’s

Ohm’s Law atEach Resistor

The EquivalentResistance & v(t)

RtitvRR

Multi-R Example

Find Rp

For This Ckt Find i1, and the Power Supplied by the I-Source

Recall the General Current Divider Eqn

k4 k20 k5

kΩ2kΩ2

Multi-R Example cont.

Find v for Single-Node-Pair by Ohm

Find i1, by Divider • Take Care with

Passive Sign Conv

mA8k4 k20 k5

mA4mA8kΩ4

kΩ21 i

V16kΩ2mA8

psrc Riv

mW128mA8V16 srcsrc viP

• Note: this time For Passive Sign Convention CURRENT Direction assigned as POSITIVE Find Psrc by v•i

Multi-R: Alternative Approach

The Ckt After the R-Combination

Start by Combining R’s NOT associated with i1

Now Have 1:1 Current Divider so

k4 k20 k5

1i20k||5k

k4 k41i

1kΩ4kΩ4

Example: Multi-R, Multi-Isrc SNP

Soln Game Plan: Convert The Problem Into A Basic CURRENT DIVIDER By Combining Sources And Resistors

Given Single Node-Pair Ckt: Find IL

mA2mA4mA1,

Combine Sources• Assume DOWN =

POSITIVEeqSI ,

Multi-R, Multi-Isrc SNP cont.

Next Combine Parallel Resistors

Given Single Node-Pair Ckt: Find IL

IL by 3:1 I-Divider

Then the Equivalent Circuit →

Note MINUS Sign

The SAME Ckt Can Look Quite Different

3mAmA993

UnTangling Utility Redrawing A Circuit May, Sometimes, Help To

Better Visualize The Electrical Connections

k3k3 k6k6

• Be FAITHFUL to the Node-Connections

CBI1 I2

Another Example

Alternatives for P• By vi & passive sign:

For This Ckt Find the I-SrcPower, P20

Use ||-Resistance

mA2020 VIVP

k2 k4 k3

• By Joule and Energy Balance

220 mA20 PR RPPj

suppliedmW13

mA20k1312

Nodal Analysis (based on KCL)

A Systematic Technique To Determine Every Voltage and Current in a Circuit

The variables used to describe the circuit will be “Node Voltages”• The voltages of each node Will Be

Determined With Respect To a Pre-selected REFERENCE Node– The Reference Node is Often Referred to as

Ground (GND) Or

COMMON

Consider Resistor Ladder

Goal: Determine All Currents & Potentials in this “Ladder” Network

Plan• Use Series/Parallel Transformation to Find I1

• Back-Substitute Using KVL, KCL, Ohm to Find Rest

Series-Parallel Transformations Xform1

• Combine 3 Resistors at End of Network

kk 6||6

k12kk 12||4

Xform2• Combine 3 Resistors

at End of Network

Note By Ohm’s Law

Xform cont. Xform3

• To Single-Loop Ckt

VI a 5.0

mAIII 5.0213

Now Back Substitute• Recall

• By KCL

Xform cont.

Recall Xform2

375.04

c 375.03

Law sOhm'by Then

In Summary• I1 = 1 mA

• I2 = I3 = 0.5 mA

• I4 = 0.375 mA

• I5 =0.125 mA

• Va = 3 V

• Vb = 1.5 V

• Vc = 0.375 V

Finally by KCL

Node Analysis Perspective

KVL KVL KVL

REFERENCE

In General: Vx5 = Vx−V5 = Vx−0 = Vx • Then the KVL Eqns

Take Node-5 As the Ref, →V5 = 0, Always

Node Analysis cont

If We Know Va, Vb, and Vc, Then • Can Calc V1, V2, V3 by KVL, Then

– Use Ohm’s Law to Find I1→I5

i.e., If we Know All Node Potentials, Then Can Calc All Branch Currents

Theorem: IF ALL NODE VOLTAGES WITH RESPECT TO A COMMON REFERENCE NODE ARE KNOWN, THEN ONE CAN DETERMINE ANY OTHER ELECTRICAL VARIABLE FOR THE CIRCUIT

ALWAYS Define Reference Node

The Statement V1 = 4V is Meaningless• UNTIL The

Designation of a REFERENCE NODE

VVVVVV 6242112

By Convention The Ground (GND) Symbol Indicates the Reference Point• ALL Node Voltages

are Measured Relative to GND

Strategy for Node Analysis1. Identify All Nodes And

Select A Ref. Node

2. Identify Known Node Voltages

3. at Each Node With Unknown Voltage Write A KCL Equation• e.g., (Sum Of

Current Leaving) =0

4. Replace Currents in Terms Of Node V’s

0:@ 321 IIIVa

VV baaaS

0:@ 543 IIIVb

0:@ 65 IIVc

VV cbbba

VV ccb

Yields Algebraic Eqns In The Node Voltages

Final Desired Eqn Set

REFERENCE

Node Equation Mechanics

When Writing Node Equations• At Each Node We

Can Choose Arbitrary Directions for Currents

• Then select any form of KCL

When The Currents Are Replaced In Terms Of The Node Voltages The Node Eqns That Result Are The Same Or Equiv.

2R2I3I1I

0LEAVING CURRENTS

321321

VVIII cbdbba

0 NODE INTO CURRENTS

321321

VVIII cbdbba

Node Eqn Mechanics cont. When Writing The Node Equations

• Use Ohm’s Law to Write The Equation Directly In Terms Of The Node Voltages

• BY Default Use KCL In The Form Sum-of-currents-leaving = 0– But The Reference Direction For The Currents

Does NOT Affect The Node Equation

2R '2I

0LEAVING CURRENTS

VVIII bcdbab

0 NODE INTO CURRENTS

VVIII bcdbab

Ckts w/ Independent Sources

At Node-1• Using Resistances

Using Conductances Eliminates Tedious Division Operations

Replacing R’s w/ G’s• At Node-1

0)( 21211 vvGvGiA

• At Node-2

Node Analysis of Indep Src Ckts

ReOrder Terms in Eqns for iA & iB

The Manipulation Of Systems Of Algebraic Equations Can Be Efficiently Done Using Matrix Analysis• c.f., MTH-6 or ENGR-25 (MATLAB)

The Model For The Circuit is a System Of Algebraic Equations

Example Write the KCL Eqns

• @ Node-1 We Visualize The Currents Leaving And Write the KCL Eqn

Similarly at Node-2

• Could Use (i Entering Node) Just as well

KCL Eqn Example Write KCL At

Each Node In Terms Of Node Voltages• 3 Nodes Implies 2

KCL Equations

k8 k8k2 k2AV

Mark the nodes (to insure that None is missing)

Select C asReference

Solving by Algebra, Find:

Two simple eqns in Two Unknowns

Linear Algebra Analysis

mAimAi BA

Recall R=1/G, Then Insert Numerical Values, and Change to Time Independent Notation (All CAPS)

The Math Model

The Node Eqns in Conductance Form

Linear Algebra Analysis cont. The Numerical Model

Multiply the 1st Eqn by 4kΩ to Find V1 in Terms of V2

Back Sub into 2nd eqn

Then V2

And V1

Alternatively, Multiply Both Sides of Math Model by LCD in kΩ

• R.H.S. of Eqn Now in Volts

• V1, V2 CoEffs are No.s

Linear Algebra Analysis cont. The “Clean” Eqns

Proceed with Gaussian Elimination• Add Eqns to Eliminate V2

• Back Substitute to Find V2

Use Matrix Algebra Recall The Math Model

From MTH-6 the Form of Matrix Multiplication

IGV • In this Case

The Matrix Eqn Soln

IGV 1• In this Case

Calculating the Matrix Inverse, G-1, is NOT Trivial• Use Matrix Manipulation

– Adjoint Matrix– Determinant Calculation

• Or use MATLAB

GV = I By MATLAB

Construct the Coefficient Matrix G >> G = [1/4e3 -1/6e3;

-1/6e3 1/3e3]

1.0e-003 *

0.2500 -0.1667 -0.1667 0.3333

Construct the Constraint Vector, I

>> I = [1e-3; -4e-3]I = 0.0010 -0.0040

Matrix Inversion by “Left” Division for V

>> V = G\IV = -6 -15

Example Ckt w/ V-controlled Isrc

Write Node Equations

Treat Dependent Source as a Normal Current-Source• Node Eqns

Express Controlling Variable In Terms Of Node Voltages

Example Ckt w/ V-controlled Isrc

4 Eqns in 4 Unknowns• Solve Using Most

Convenient Method– Choose SUB &

GAUSSIAN ELIM

Sub for vxin •vx Isrc

Continue w/ Gaussian Elim OR UseMatrix Algebra

Solve Using MATLAB

Define Components (m-file Node_Anal_0602.m)

,4,2,4

mAimAikR

R1 = 1000; R2 = 2000; R3 = 2000;R4 = 4000; %resistances in OhmsiA = 0.002; iB = 0.004; %sources in AmpsAlpha = 2; %gain of dependent source in Siemens

Solve Using MATLAB cont

Define Coefficient Matrix

,4,2,4

mAimAikR

G=[(1/R1+1/R2), -1/R1, 0; % first Matrix row-1/R1,(1/R1+alpha+1/R2),-(alpha+1/R2); % 2nd row0, -1/R2,(1/R2+1/R4)] %third row.

G = 0.0015 -0.0010 0 -0.0010 2.0015 -2.0005 0 -0.0005 0.0008

Solve Using MATLAB cont

Define Constraint Vector

,4,2,4

mAimAikR

I=[iA;-iA;iB]; Solve by Left/Back Division; V in voltsV=G\I % end with carriage return and get the ReadBackV = 11.9940 15.9910 15.9940

V 994.11 V 991.15V 994.15

Loop Analysis (Based on KVL)

Loop Analysis is The 2nd Systematic Technique To Determine All Currents And Voltages In A Circuit• IT Is the DUAL To Node Analysis

– It First Determines All Currents In A Circuit And Then It Uses Ohm’s Law To Compute Voltages

• There Are Situations Where Node Analysis Is Not An Efficient Technique And Where The Number Of Equations Required By Loop Analysis Is Significantly Smaller

Loop Analysis Illustration

Apply Node Analysis Observe

Need 3 Eqns to Find All Node Potentials; 24

Notice There is Only ONE Current Flowing Thru All Components• A Single Loop Ckt• Can Use Ohm’s Law to

Determine Voltages

Apply KVL for Clockwise Loop Starting at GND

2RV 1RV

3V2V1V

0][18][12 321 RRR VVVVV

• Have 4 Non-Ref Nodes• One SuperNode• One Node Connected to

GND Thru a V-src

Loop Analysis Illustration cont

Now Use Ohm’s Law to Express V’s In Terms of the Loop Current

By KVL

Note:• Recalling that V=IR Allows

Writing the Ohm Eqn “by Inspection” for a Single Loop Ckt

The Loop Generates a SINGLE Eqn to Yield the Loop CURRENT

2RV 1RV

3V2V1V

0][18][12 321 IRVIRIRV

Loops, Meshes, Loop-Currents

Each Component Is Characterized By The • VOLTAGE

ACROSS It• CURRENT THRU It

A Loop Is A Closed Path That Does Not Go Twice Over Any Node

This Circuit Has 3 Loops1. fabef

A BASIC CIRCUIT

2. ebcde

3. fabcdef

Loops, Meshes cont.

A MESH is a LOOP That Does Not Enclose Any OTHER Loop• This Ckt Has

Meshes– fabef

A Loop Current is a Fictitious or Virtual Current That is Assumed to Flow Around a Loop• The Loop Currents

of This Ckt– I1, I2, I3

Mesh Current = Current Within a Mesh Loop• e.g.: I1, I2

A BASIC CIRCUIT

– ebcde

Loops, Meshes cont.

Claim• In a Circuit, the

Current Through Any Component Can Be Expressed In Terms of the (perhaps multiple) Loop Currents

Ckt Examples

A BASIC CIRCUIT

• The DIRECTION Of The Loop Currents is SIGNIFICANT

• FACT– Not ALL Loop

Currents are Required To Compute All The Currents Through Components

Loops, Meshes cont.

For Every Circuit There is a MINIMUM Number of Loop Currents Needed to Find Every Current in the System

• Such A Collection is Called the “MINIMAL SET” (of Loop Currents)

For a Given Circuit Let• B Number of

BRANCHES• N Number of NODES

The Minimum Number of Loop Currents is

A BASIC CIRCUIT

)1( NBL

Illustration For This Ckt

• B = 7• N = 6• L = 7-(6-1) = 2

Need Two Loop Currents• The Currents Shown

are MESH Currents– Hence They are

Independent and form a Minimal Set

Determination of Loop Currents• KVL on Left Mesh

• KVL on Right Mesh

• Using Ohm’s Law

Illustration cont. Substituting and

Rearranging

We Obtain in MATRIX FORM the Loop Equations for This Circuit

Mesh Practice Write The Mesh

Equations Some Bookkeeping

• 8 BRANCHES• 7 NODES• L = 8-(7-1) = 2

– Need TWO Mesh Currents

This is MESH Current Practice• Choose as the

Two Loops Meshes– i1– i2

Identify All Voltage Drops

KVL on Bottom Mesh

02211 RSRS vvvv

033452 RSRRR vvvvv

KVL on Top Mesh

Mesh Practice cont. Now Use Ohm’s Law To

Find The Mesh Current Equation Set

221 )( Rii

Develop a ShortCut Whenever An Element

Has More Than One Loop Current Flowing Through It We Calculate NET Current In The DIRECTION of TRAVEL

Draw The Mesh Currents• Orientation can be

arbitrary, But Conventionally Defined as CLOCKWISE

NOW • Write KVL For Each Mesh • Apply Ohm’s Law To Every

Resistor

V 2R 1

R 2 R 3

R 4R 5

WRITE THE MESH EQUATIONS

Develop a ShortCut cont At Each Loop must Follow

The Passive Sign Convention Using Loop Current REFERENCE DIRECTION• This Defines the Polarity of

the Voltage Drops

Then KVL for Meshes 1 & 2

Note The NET Currents

V 2R 1

R 2 R 3

R 4R 5

0)( 51221111 RIRIIRIV

0)( 21242322 RIIRIRIV

)( 21 II

)( 12 II

Numerical Example Use Loop Analysis

to Find Io

SHORTCUT: • POLARITIES ARE

NOT NEEDED.• Apply Ohm’s Law To Each

Element As KVL Is being written

KVL for Meshes 1 & 2

Collect Like Terms & Solve

mAIVkI

-----------------------

Add and2396

mAIkIkI4

561212 121

mAIIIo 4

Numerical Example - Alternate Alternative Loop Current

Selection KVL for Mesh1 & Loop2

Collect Like Terms & Solve In This Case one mesh

and one loop• Io = I1

– This Selection is More Efficient than 2 small meshes; Only Need to Find l1

oImAIkI

-----------------

Substract and2996

312612

Circuits w/ Indep. I-Sources Find Both Vo & V1

There is NO Relationship Between V1 and the 2 mA Source Current However ...• The Mesh-1 Current is

CONSTRAINED by the 2mA Source– Thus the Mesh-1 Eqn

mAI 21

In General• Current Sources That Are

NOT SHARED By Other Meshes (Or Loops) Serve To DEFINE a Mesh (Loop) Current And Reduce The Number Of Required Equations

Circuits w/ Indep. I-Sources cont The Mesh-2 KVL Eqn

ReArranging Find Equivalent Eqn

mAkVmAkV

75.06224

To Obtain V1 Apply KVL To Any Closed Path That Includes V1

VkIkI 282 21 Use I1 Constraint

to Calculate I2

211 6240 IkVIkV

Numerical Example

Two Mesh Currents Are Defined By Current Sources

Only Need Eqn for Mesh-3

mAImAI 24 21

Numerical Example cont

Collect Terms for eqn-3 Finally Use KVL to Calculate VoVkIkIkI 31242 321

mAkmAkVI

)2(4)4(233

Then I3

KVL FOR Vo

036 3 oVVkI

STOP Here if Short on Time

Eqns by “Inspection” If The Circuit Contains

Only INDEPENDENT VOLTAGE Sources Then The Mesh Equations Can Be Written “By Inspection”• MUST HAVE All Mesh

Currents With The Same Orientation

In loop “k”• The Coefficient Of Ik Is The

Sum Of Resistances Around The Loop

The Right Hand Side Is The Algebraic Sum Of Voltage Sources Around The Loop• VoltageRise = POSITIVE

on R.H.S. of eqn

Eqns by “Inspection” cont The Coefficient Of Ij Is

The Sum Of Resistances COMMON To Both k and j and With a NEGATIVE Sign

In This Example• Loop1: k = 1, j = 2

VkIIk 12612 21

VIkIk 396 21

• Loop2: k = 2, j = 1

Equation Practice Loop-1

• Coefficient of I1

Similarly The Coeffs for Loop-2

3coeff

39coeff

0coeff

kkI 641

• Right Hand Side (RHS)

In Summary for Loop1

Equation Practice cont.

In Summary for Loop-2

Applying the Method to Loop-3 Yields

Solve 3-Eqns in 3-Unknowns Using Normal Linear Algebra, or MATLAB, Techniques

Numerical Example Use Mesh Eqns to

Determine Vo

1. Draw the Mesh Currents

2. Write the Mesh Eqns for Mesh-1 & Mesh-2

][32)242( 21 VkIIkkk

)36()62(2 21 VVIkkkI

Divide Both Sides of Both Eqns by 1kΩ• Units on RHS become

V/kΩ, or mA

Solve System of Eqns

mAImAI

336then

113330

-----------------

Add and4982

Example 1. Draw the Mesh

Currents

2. Write Mesh Eqns by KVL

Or Eqns by Inspection

0)(61212 :1 MESH 311 IIkVkI

0)(4)(412 :2 MESH 3242 IIkIIkV

0)(4)(69 :3 MESH 2313 IIkIIkV

02)(49 :4 MESH 424 kIIIkV

VkIkI 12618 31

VkIkIkI 12448 432

VkIkIkI 91046 321

VkIkI 964 42

Calculate Currents Using Multi-Eqn Solver Tools• 4 Eqns in 4 unknowns

– Solve using Standard Linear Algebra Techniques

– Perfect for MATLABVI

All Done for Today

A DifferentType ofNODE

BRANCHES Connect to NODES

WhiteBoard Work

Let’s Use KCL to Derive the Req for N Parallel Resistors

Done Previously

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Engineering 43

AppendixΔ↔WYE

& others

Recall Passive Sign Convention

Rv 'Rv

vvi Nm LAW SOHM'

vvi mN 'LAW SOHM'

iivv RR ' Thus '

If V’ Drops R←L• i’ by Passive Sign

Convention

If V Drops L→R• i by Passive Sign

Convention

Single Loop Ckts - Background

Using KVL And KCL We Can Write Enough Equations To Analyze ANY Linear Circuit

Begin The Study Of Systematic And Efficient Ways Of Using The Fundamental KCL & KVL Circuit Laws• This Time →

Single LOOP Circuits

The Power of Loop Analysis Consider this

Circuit

6 branches6 nodes1 loop

ALL ELEMEN TS I N SER I ESON LY ON E CUR R EN T

Alternative Analyses• Write N-1 (5) KCL Equations• Determine Only the SINGLE Loop Current

– An Easy Choice

The Plan for Loop Analysis• Begin With The Simplest One-Loop Circuit• Extend Results To Multiple-Source

and Multiple-Resistor Circuits

“Inverse” Voltage Divider

The Std V-Divider1R

+- 2RSV

SO VRR

Inverse

Divider OS VR

kV 500 458.3kV220

220202

• Use Inverse Divider

Example Find VS

Example

Find• Vx, Vab

• P3Vx

– The Power Absorbedor Supplied By theDependent Source

Apply KVL VVVV XXX 203412

xV3-+ V4

VVS 12=+ -

VVVV abXab 10034

100212

IVP XVX3)3(

mWmAVPXV 6][1][23)3(

Find DS Power• Passive Sign Conven.

• Ohm’s Law

• Then Pwr ABSORBED

Example

Find: VDA, VCD, I• Apply KVL & Ohm

k*Ik*Ik*I-

05.060

0103092012

VVIkV DADA 5.110*1012

VmAkIkVCD 5.105.030*30

• Ohm’s Law

WhiteBoard Work

Let’s Work This Problem

1 2 0 m A 4 k 4 k

Inverse Series Parallel Combo

Find R: Simple Case• Constraints

– VR = 600 mV

– I = 3A– Only 0.1Ω R’s Available

Recall R = V/I 2.0

1.01.02 availRR

Since R>Ravail, Then Need to Run in Series

More Complex Case Find R for Constraints

• VR = 600 mV

• I = 9A• Only 0.1Ω Resistors Available

m 67.660667.09

R 0.1║(0.1↔0.1)

Either of These 0.1Ω R-Networks Will Work

33.33 mΩ

Effect Of Resistor Tolerance The R Spec: 2.7k, ±10% What Are the Ranges for

Current & Power?• I = V/R

VInom 115.4

7.29.0

10367.3

7.21.1

10704.3

10maxmin

VPnom 15.41

7.29.0

1067.33

7.21.1

1004.37

• P = V2/R

• For Both I & P the Tolerance.: -9.1%, +11.1% – Asymmetry Due to Inverse Dependence on R

Final Example

Find VS

• Straight-Forward

VmAkVIkVV

mAmAmAImAI

915.020620

15.005.01.01.0

05.0120

61.060

• Or, Recognize As Inverse V-Divider

Final Example cont

Inverse DividerCalculation

40120||60

Before As61.060

Wye↔ Transformations

This Circuit Has NoSeries or ParallelResistors

If We Could MakeThe Change Below Would Have Series-Parallel Case

Y↔ Xforms cont Then the Circuit Would Appear as Below and

We Could Apply the Previous Techniques

Wye↔ Transform Eqns

→Y (pg. 58)

←YR ac =

R 1||(

R 2↔

R 3) R

ab = Ra + R

accbba

RRRRRRR

←Y (pg. 58)

↔Y Application Example

Find IS

Connection

Calc IS

kkkkkkR

10)62(||936

• Use the →Y Eqns toArrive at The ReducedDiagram Below

Another ↔Y Example

For this Ckt Find Vo

Convert this Y to Delta

Keep This Node-Pair

Another ↔Y Example cont Notice for Y→Δ in

this Case Ra = Rb = Rc = 12 kΩ• Only need to Calc

ONE Conversion

RRRRRRRRR

accbba 3612

12123321

The Xformed Ckt

4mA 36k

• ||-R’s Form a Current Divider

Another ↔Y Example cont

The Ckt After ||-Reductions

4mA 36k

36 ||12 9k k k

Can Easily Calc the Current That Produces Vo

36 18 3O

kI mA mA

Then Finally Vo by Ohm’s Law

89 9 24

3O OV k I k mA V

“BackSubbing” Example

Given I4 = 0.5 mA, Find VO

BackSubbing Strategy

Always ask: What More Can I Calculate?• In the Previous Example Using Ohm’s Law,

KVL, KCL, S-P Combinations - Calc:

IkVIkV

bmayer@chabotcollege.edu engr-43_lec-02a_sp_vi-divide_nodemesh.pptx 1 bruce mayer, pe...

pe engineering

circuit slide

simplified circuit

original circuit

kseries slide

series reduction step

parallel reduction

kseries series

Documents

bmayer@chabotcollege.edu...

bmayer@chabotcollege.edu...

bmayer@chabotcollege.edu...

bmayer@chabotcollege.edu...

mth55_lec-54_sec_8-5a_polynom_inequal.ppt 1 bruce mayer, pe...

bmayer@chabotcollege.edu...

bmayer@chabotcollege.edu...

bmayer@chabotcollege.edu...

bmayer@chabotcollege.edu...

engr-25_hw-01_ 1 bruce mayer, pe engineering/math/physics...

bmayer@chabotcollege.edu...

mth55_lec-41_sec_7-3a_radical_product_rule.ppt.ppt 1 bruce...

bmayer@chabotcollege.edu...

e engr36_tutorial_areal_moment_of_ 1 bruce mayer, pe...

bmayer@chabotcollege.edu...

bmayer@chabotcollege.edu...

bmayer@chabotcollege.edu...

bmayer@chabotcollege.edu...

bmayer@chabotcollege.edu...

bmayer@chabotcollege.edu...