Eq Ckts Voltage Division Current Division

8/13/2019 Eq Ckts Voltage Division Current Division

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1

Equivalent Circuits

Originalcircuit

I

V+

-

1

2

The Current Voltage (I-V) Relationshipdefines any element. It can also be used todefine any circuit in the box. We assumethe original circuit is complicated.

Equivalentcircuit

I

V

+

-

1

2

Replace original circuit with simplerequivalent circuit that exhibits the same I-Vrelationship (at nodes 1 & 2) !


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3

Circuit Simplification for Parallel Current Sources & ResistancesIS

VS R 1 R 2 R 3

IS1 IS2 IS3

IR1 IR2 IR3

IS

VS R 1

R 2 R 3IS1 IS2 IS3

IR1 IR2 IR3

Apply arbitrary voltage source Vs

Since all elements are in parallel, they alsohave Vs across them.We assign currents in red

KCL at top node I in = I out

IS + I S2 = I S1 + I R1 +I R2 +I R3 +I S3

Is = (I S1 -I S2+ I S3 )+I R1 +I R2 +I R3

IS= I STOT +I R1 + I R2 + I R3

ISTOT -


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4

Continued..

IS

VS R 1 R 2 R 3

ISTOT

IR1 IR2 IR3

Therefore, the ckt can be redrawn asshow on the left where:

IS= I STOT +V S(1/R 1 + 1/R 2 + 1/R 3)

IS= I STOT +V S(1/R TOTAL )

321 /1/1/11

R R R R TOTAL

N

i i

TOTAL

R

R

1

11

IS

VSRTOTALISTOT

+

-

VS

Resistors in parallel

1. So we can simply add curr ent sources in paral lel bu t consideri ng theirorientation* For th is problem, Curr ent sour ces point in g down are posit ive (I s1 and I s3) thosepoint in g up are negative (Is2)2. And we can combine paral lel r esistors


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5

Numeric ExampleIS

VS 6K 3K1A 3A 10A

2K

IS

VS RTOTALISTOT

+

-

VS

Find Circuit 2 that is equivalent toCircuit 1

ISTOT = 1 + 10 3 = 8 A K

K K K R TOTAL 1

)3/1()2/1()6/1(1

IS = 8A +V S/1K=IS = (1/1K)V S + 8A

I S

V S

1/1K=slopey = mx + b

8 A

- 8000 V

Circuit 1

Circuit 2


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Similar Numeric ExampleIS

VS 6K 3K1A 3A 10A

2K

IS

VS RTOTISTOT

+

-

VS

Find Circuit 2 that is equivalent toCircuit 1 ( changing the orientation ofIsTOT ):ISTOT = -1 - 10 + 3 = -8 A

K K K K RTOTAL 1)3/1()2/1()6/1(

1

Circuit 1

Circuit 2


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Last Numeric ExampleIS

VS 12K 6K

-3A 4A -5A12K

IS

VS RTOTISTOT

+

-

VS

Find Circuit 2 that is equivalent to Circuit 1:ISTOT = -(-3A) (4A) + (-5A)= -6A

K K K K RTOTAL 3)6/1()12/1()12/1(

1

Circuit 1

Circuit 2


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Circuit Simplification for Series Voltage Sources & Resistances

IS R 1VS1

+

-

VS

R 2

R 3

VS2

VS3IS

Is there a way to combine series resistances and voltage sources in the gray box (circuit on the

left)? The resulting simplified circuit will be in the form of the circuit on the right.

Apply arbitrary current source I S. I S is the only current in the circuit; it enters and leaves allof these elements in series.

KVL gives us:-V S + V S1 + R 1IS VS2 + R 2I S +V S3 + R 3I S = 0

-V S + (V S1 VS2 +V S3) + R 1I S + R 2IS + R 3I S = 0

Algebraic properties allows us to change the order of the items in the circuit and in theequation, so we can rewrite it as

-V S + V STOTAL + I S(R 1+R 2+R 3) =0

VSTOTAL

IS

RTOTAL

VSTOTAL

+

-

VS


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Continued..-V S + V STOTAL + I S(R 1+R 2+R 3) =0

Now solving the previous equation for I S:

321 R R R R TOTAL

N

i

iTOTAL R R1

IS

RTOTALV

TOTAL

+

-

VS Resistors in series

* * * 1. So we can simply add voltage sources in series but consider in g their ori entation* * * 2. Simply add series resistors together

TOTAL

STOTALS

TOTAL

STOTALS S R

V V

R R R RV

V R R R

I

1)(

1

321321

This gives us Linear relationship such asy = mx + b


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Numeric Example

Find V TOTAL and R TOTAL so that the circuit on the right has the same I-V

characteristic as the circuit on the left??

We add R TOTAL = 2K + 5K + 7K = 14KVSTOTAL = -6V -3V + 15V = 6V

IS 2K6V

+

-

VS

5K

7K

3V

15V

IS

RTOTAL

VSTOTAL

+

-

VS

IS

14K

6V

+

-

VS


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Numeric Example

How to Find Vs and Rs so that the circuit on the right has the same I-V

characteristic as the circuit on the left??

VS2

VS1R 1

VS3R 2

VSTOTAL

R TOTAL

A

B

A

B

For V S1 = 3V, R 1 = 5K, V S2 = 7V, R 2 = 2K and V S3 = 11VWe add R TOT = 5K + 2K =7KVTOT = 3V -7V -11V = -15V

Pay attention to the sign of the final voltage source in the final circuit!


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Series/Parallel Reduction of Resistance

R 1

R 2

R 3

R 4

A

B

I3

I3

R 1

R 2

R 3

R 4

A

B

R TOTAL

A

B

R 1

R 2 R 34

Find R TOTAL

Step 1: Working with only 2 resistors at a time, start

opposite ports A and B R 3 and R 4 are in series R 34 = R 3 + R 4

Step 2: Next we see that R 2 and R 34 have the sameendpoints (ie share the same voltage)

R 34 // R 2 = R 234

Step 3: R 1 is in series with R 234 .Therefore R TOT = R 1 + R 234 .

R TOTAL = R 1 + (R 2//(R 3+R 4))

R1

R234


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Example Series/Parallel Reduction of Resistance

R 3=1

A

B

R TOTAL

A

B

R 4=5R 2=3

R 1=8

R 34 =6

R 1=8

R 2=3

Find R TOTAL

Step 1:R 3 and R 4 are in series R 34 = 1 + 5 =6New equivalent resistance gets connected across same 2 nodes

Step 2: R 2 and R 34 have the same endpoints (ie share the samevoltage)

R 34 // R 2 = R 234R 234 =6//3= 1/{(1/6)+(1/3)}=2

New equivalent resistance gets connected across same 2 nodes

Step 3: R 1 is in series with R 234 . Therefore, R TOT = R 1 + R 234 .R TOTAL = R 1 + (R 2//(R 3+R 4))= 8+2 = 10 ohms

R1=8

R234 =2

Step 1:

Step 2:

Step 3:


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14

Voltage Division

Consider two resistors in series with a voltageacross them:

R 1

R 2

v 1

+

+

v 2

v s 21

11 R R

Rvv s

21

22 R R

Rvv s

+

KVL: - Vs + V1 +V

2 = 0


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Voltage DivisionChanging the orientation of the element voltages:

R1

R 2

v1

+

+

v 2

v s

21

11 R R

Rvv s

21

22 R R R

vv s+

+

_

+

vX

v Y

21

1

R R

Rvv s X

21

2

R R R

vv sY If we instead use v X, we make a sign change,since KVL still applies

KVL: - Vs + V 1 +V 2 = 0KVL: - Vs - V X -V Y = 0


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In General: Voltage Division

Consider N resistors in series :

Source voltage(s) are divided between the resistors in direct proportion to their resistances

N

j j

iS

i R

R

RV V

1


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Voltage Division- example 1

5

10

v 1

+

+

v 2

Vs=9V V V V

R R R

vv s 331

9105

5)9(

21

11

+

+

_

+

v X

v Y

V V V R R

Rvv s 6

32

9105

10)9(

21

22

V V R R

Rvv s X 3105

5)9(

21

1

V V R R

Rvv sY 6105

10)9(

21

2

KVL: -Vs + V 1 +V 2 = 0.KVL: -Vs - V X -V Y = 0. For the total voltages around the loop to equal zero, V X and V Y must be negative for this circuit! So apply negative sign in voltage dividers for V X and V Y .


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Voltage Division- example 2

10

30

v 1

+

+

v 2

Vs=12V V V V

R R R

vv s 341

123010

10)12(

21

11

+ +

_

+

v X

v Y

V V V R R

Rvv s 9

43

123010

30)12(

21

22

V V R R

Rvv s X 33010

10)12(

21

1

V V R R

Rvv sY 93010

30)12(

21

2

Note: The orientation of the source voltage has been flipped from previous example!

KVL: Vs + V 1 +V 2 = 0. For the total voltages around the loop to equal zero, V 1 and V 2 mustbe negative for this circuit! So apply negative sign in voltage dividers for V 1 and V 2 .KVL: Vs - V X -V Y = 0


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19

10

V s

5

+- Find I x , V S

I X

+ -10 V

20 30

Where to start?1) Try Ohms Law 2) Use the values found to write KVL or KCL3) Alternative: use voltage division

Example 1 similar to HW


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20

10

V s

5

+- Find I x , V S

I X +

++

+

--

-

-10 V

20 30

V 20

V 10

V 30

I 5 I 30

I 10

I S

Making assignments shown in blue

1) Ohms Lawassign I 5 so I 5 = - (10V/5 ) = - 2A (note not PSC)2)KCL at B : 0=I 5 + I 10 therefore I 10 = -(-2A) = 2A3) V 10 = I 10(10 ) = (2A)(10 )=20 V4) Cannot apply KCL yet to get other unknown currents

Example 1B


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21

10

V s

5

+- Find I x , V S

I X +

++

+

--

-

-10 V

20 30

V 20

V 10

V 30

I 5 I 30

I 10

I S

5) Instead apply KVL

+10+V 10-V 20=0 +10+20-V 20=0 V 20 = 30V6) I x=V 20 /20 = 30/20 = 1.5 A note: follows PSC7) KCL at A: I 5 = I S +I X IS=I 5 I X = (-2A)-1.5A = - 3.5A =I 30 8) V 30=I 30(30 ) = (-3.5A)(30 ) = -105 V9) KVL: +V S +V 30-V 20 =0 V S= V 20-V 30 =30-(-105) = 135 V

AExample 1


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22

10

V s

5

+- Find I x , V S

I X +

++

+

--

-

-10 V

20 30

V 20

V 10

V 30

I 5 I 30

I 10

I S

Alternative: Applying Voltage Division would have saved us steps

205105

5 V V V V V 30)5105(1020

Example 1


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Review: Voltage DivisionConsider two resistors in series with a voltage v

across them:

R1

R 2

v1

+

+

v 2

v s

21

11 R R

Rvv s

21

22 R R R

vv s+

+

_

+

vX

v Y

21

1

R R

Rvv s X

21

2

R R R

vv sY


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24

How do I know if I need to apply a negative signwhen using voltage division?

Answer: Pay attention to the orientation of the element voltages withrespect to the source voltages, knowing that the sum of the voltagesaround a closed loop must be zero (they must cancel each other out KVL still applies!)

For the circuit on the previous page:

KVL: -V S + V 1 + V 2 =0 so V S = V 1 + V 2

These voltages have the same signs as the source voltage

KVL: -V S + V X + V Y =0 so V S = -V X - V Y

These voltages have the signs opposite the source voltage


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Example: Voltage Division

RX

R Z

v

X

+v s

3 R R R

Rvv

Z X

X s

X

+

R4 R 5

R 3

Resistors in Series indicates voltage division.

Do NOT include R 4 and R 5 which arent inseriesnot part of the divider, they also havevoltage v S

Find VX

v S

+


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Example: Double Voltage Divider

VS

R 1

R 234 +

- V 2

FIND V 4

First assign element voltages V 2 and V 3 as shown in blue in themiddle diagram.

We will first find an expression for V 2 in terms of V S, and thenwe will find an expression for V 4 in terms of V 2 which we willthen combine.

To find our first expression, we combine resistances obtain thelast circuit where we apply voltage division:

)//(432234

R R R R

S V R R R

V 123 4

23 42

Returning to the middle circuit:

234

44 V R R

RV

Substituting we get our double divider:

S V R R R R R R R

R R R

V 1432

432

34

44 )]//([

)//(

+

- V 4 VS

R 1

R 2

R 3

R 4

+

- V 2

+- V 3

+

- V 4 VS

R 1

R 2

R 3

R 4


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27

Example: Numeric Double Voltage DividerFIND V 3

First assign element voltage V 30 as shown in blue in the middlediagram.

We will first find an expression for V 30 in terms of V S, and thenwe will find an expression for V 3 in terms of V 30 which we willthen combine.

To find our first expression, we combine resistances obtain thelast circuit where we apply voltage division:

5.7)532//(30 NEW

R

V V V 30405.25.7

5.730

Returning to the middle circuit we write another voltage

division equation where we substitute :

V V V V 9)30(103

5233

303

+

-

V 3

40 V

2.5

30

5

3

2

+

-

V 3

40 V

2.5

30

5

3

2

+

- V 30

40 V

2.5

7.5

+

- V 30


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Review: Current Divider

I s R

2 R 1

I 1 I

2

21

11

11

1

R R

R I I S

Consider tworesistors in parallel that divide current I S:

21

21 R R

R I I S

Current divider presented last week: Equivalent to:

WHY?

Multiply fraction by R 1R 2:

21

2

2

21

1

21

1

21

21

21

211

1

*11

*1

R R R

I

R

R R

R

R R R R R

I

R R

R R

R R R I I S S S


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29

More on Current Divider

R X R 2

I 2

I X

2

2

211

1 R R

R I R R

R I I X

S

X

X S X

I s

Resistors in Parallel indicate currentdivision.

Do NOT include R 1 which isnt in parallel with Isnot part of thedivider

In general DO NOT include

resistors in series with IS

R1

Find IX


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More on Current Divider

R X R 2

I 2

I X

52111

1

R R R

R I I X

X S X

I s

R1

Find IX

R 5

The addition of R 5 tothe circuit

R 5 is in parallel with I Sas well!

I 5

52

55 111

1

R R R

R I I X

S


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Review: Current Divider

R2 R 1

I 1 I

2

V S+

21

21 R R

R I I S

I s

21

12 R R

R I I S

I X I Y

21

2

R R R I I S X

21

1

R R R I I S Y

Pay attention to the orientation

of the element currents withrespect to the source currents.For currents directed oppositethe source, use a negative signin the current divisionequation


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Double Current Divider Example 1

R 2 IS

R 1 R3

I5

R 4 R 5

R2

IS

R1R3'

IyIx

R 2 IS

R 1 R 3

I5

R 4 R 5

Iy

Ix

Original Problem: Find I 5

Disregard R1 since it is in series withthe current source and doesnt dividecurrent. Assigning Currents I x andIy, we can write the first currentdivider:

y I R R

R I

45

55

/1/1

/1

Obtain new resistanceR3 and write thesecond current divider:

)//(' 5433 R R R R

S y I R R R

I 23

3

/1'/1'/1

Finally substituting (2) into (1) we get:

(1) (2)

S I R R R

R R R

I 23

3

45

55 /1'/1

'/1/1/1

/1


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Double Current Divider Example 2

Original Problem: Find i 1 thru i 4

SECOND CURRENT DIVIDERReturning to the original circuit, we can nowutilize the values we just found for ia and ib tofind the currents through each resistor:

FIRST CURRENT DIVIDERFirst combine resistors on the right in parallel, thencombine resistors on the left in parallel:20//30=(20*30)/(20+30)=600/50=1210//40=(10*40)/(10+40)=400/50=8

Define new currents i a and i b and apply current division:

30A

20

3040

10

i1

i2i4

i3

30A

128

ibia

A Aia 18)30(12812

A Aib 12)30(1288

A Aii b 2.7)12(203030

)(2030

301

A Aii b 8.4)12(203020

)(2030

202

A Aii a 4.14)18(401040

)(4010

403

A Aii a 6.3)18(401010

)(4010

104

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Eq Ckts Voltage Division Current Division