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ECE2262 Electric Circuits

Chapter 2: Resistive Circuits ! ENG 1450

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2.1 Ohm’s Law

R1

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The voltage across a resistance is directly proportional to the current flowing through it.

v = Ri

R = vi

- 1 ! (ohm) =1V1A

Conductance: G = 1R

- 1 S (siemens) =1A1V

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The power absorbed by a resistor

p = vi = Ri2 ! v = Ri

= v2

R ! v

R= i

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R = vi

!

R = 0 ! “Short” Circuit ! v = 0

R = ! ! “Open” Circuit ! i = 0

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Example

(a) determine the current and the power absorbed by the resistor (b) determine the voltage and the current in the resistor(c) determine the voltage of the source and the power absorbed by the resistor(d) determine the values of R and VS

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!(d) determine the values of R and VS

I = 4 !10"3A , P = 80 !10"3W

P = RI 2 ! R = PI 2

! R = 5 !103" = 5k!

VS = RI ! VS = 20V

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2.2 Kirchhoff’s Laws

nodeloop/closed path

loop: closed path such that no node is visited more than once

loop: R2 ! v1! v2 ! R4 ! i1 ! 1" 4" 3" 5" 2"1

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Kirchhoff’s Current Law - KCL

The algebraic sum of all currents entering any node is zero ! KCL is based on the principle of the conservation of charge !

Convention*: In writing KCL we consider the current that enters a node with a positive sign and the current that leaves a node with negative sign.

i1 > 0

i2 < 0

i3 > 0 i1 ! i2 + i3 = 0

Convention**: !i1 + i2 ! i3 = 0

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iin! = iout!

The sum of all the currents flowing into a node equals the sum of all the currents flowing out

In applying this rule we just need to pay attention to the arrows indicating the flow directions of the element currents present at each node

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Example

• Node 1: !i2 ! i1 + i5 = 0 ; • Node 2: i2 ! i3 + 50i2 = 0

• Node 3: i1 ! i4 ! 50i2 = 0 ; • Node 4: i3 + i4 ! i5 = 0 Convention*

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In writing these equations, a positive sign is used for a current leaving a node Convention**

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SUPERNODE: we can make supernodes by aggregating nodes

node 2 : !i1 ! i6 + i4 = 0 node 3 : i2 ! i4 + i5 ! i7 = 0

Adding 2 and 3 : !i1 ! i6 + i2 + i5 ! i7 = 0 ! Supernode 2&3

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Example: Write all KCL equations

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Kirchhoff’s Voltage Law - KVL

The algebraic sum of all voltages in every loop of the circuit is zero ! KVL is based on the principle of the conservation of energy !

Convention: In writing KVL when we cross a component from positive to negative terminal we consider its voltage with a positive sign. When we cross a component from negative to positive terminal we consider its voltage with a negative sign.

+ !V > 0

+! V < 0

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vdrop! = vrise!

The sum of all voltage rises encountered around any closed loop of elements in a circuit equals the sum of all voltage drops encountered around the same loop voltage rise ! loop moves from ! to + voltage drop ! loop moves from + to !

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The voltage of a component can be labeled in different ways

The arrow notation is particularly useful since we may want to label the voltage between two points that are far apart in a circuit.

a

b

+

!

Vab

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Example

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Example Find Vad , Veb

• loop a!b!c!d!a: 24 ! 4 + 6 !Vad = 0 ! Vad = 26V

• loop a!b!e! f!a: 24 !Veb ! 8 ! 6 = 0 ! Veb = 10V

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Example R1 , R2 , RC , RE , VCC , V0 , ! - known; (a) find the equations needed to determine the current in each element of this circuit; 6 unknown currents (b) Devise a formula for computing iB

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• KCL for nodes a, b, c, d

1. Node a: i1 + iC ! iCC = 0 ; 2. Node b: iB + i2 ! i1 = 0

3. Node c: iE ! iB ! iC = 0 ; • Node d: xxxxxx

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• Dependent current source: 4. iC = !iB

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• KVL loop b!c!d!b: 5. V0 + iERE ! i2R2 = 0

loop b!a!d!b: 6. !i1R1 +VCC ! i2R2 = 0

These six equations allow to determine the current in each element of the circuit and obtain

iB =VCCR2 / R1 + R2( )!V0

R1R2 / R1 + R2( ) + 1+ "( )RE

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Example Find Vx

b!

6 ! 24 + I " (2 "103) + I " (5 "103) + I " (2"103 ) = 0

I

9 "103( ) " I =18

I = 2mA

Vx = 5 "103( )" I = 10V

KVL

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2.3 Series and Parallel Combinations of Resistors

A. Series Connection - Single Loop Network

KVL: !vs + isR1 + ....+ isR7 = 0

vsis= Req = R1 + ....+ R7

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Equivalent Circuit

Req = Rii!

Note: Req ! Ri

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Multiple-Source/Resistor Networks

KVL: R1i + R2i ! v1 + v2 ! v3 + v4 + v5 = 0 ! R1 + R2( )i = v

v = v1 + v3 ! v2 ! v4 ! v5

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B. Parallel Connection - Single Node Network

KCL: is = i1 + ...+ i4

The voltage across each resistor is the same: i j =vsRj

is =vsR1

+ ...+ vsR4

! isvs

= 1R1

+ ...+ 1R4

Req =1

1R1

+ ...+ 1R4

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Req =1

1Rii

! Combining Resistors in Parallel

Note: Req !mini Ri

Note: Two resistors in parallel Req =1

1R1

+ 1R2

= R1R2R1 + R2

! R1 || R2

R || R = ?

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Multiple-Source/Resistor Networks

KCL: i1 ! i2 ! i3 + i4 ! i5 ! i6 = 0 ! i1 ! i3 + i4 ! i6sources

! "## $## = i2 + i5

i0 = i2 + i5

i0 = i1 ! i3 + i4 ! i6

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Example Find is ,i1,i2

• Equivalent circuit

Req =18 ! 6 + 3( )18 + 6 + 3( ) = 6"

is =1204 + Req

= 12A! vxy = 12 " Req = 72V ! i1 =vxy18

= 4A , i2 =vxy9

= 8A

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Example (E2.14)

RAB = 22k!

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Example Find v and the power delivered to the circuit by the current source.

P5A = 300W - delivered

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Example (Mathematical Inequality)

a b

ab

A B

•Switch open ! RAB =a + b2

• Switch closed (short circuit) ! RAB =aba + b

+ aba + b

= 2aba + b

is smaller than before

2aba + b

! a + b2

2aba + b

= 121a+ 1b

!"#

$%&

'()

*+,

-1

! The arithmetic mean is greater than the harmonic mean

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C. ! ! Y Equivalent Circuits

RaRc Rb

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• Terminal c, a: Rca= R1 || R2 + R3( ) Rca = Ra + Rc

R2

R1 R3

Ra Rb

Rc

! Y

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• Terminal b, c: Rbc= R3 || R1 + R2( ) Rbc = Rb + Rc

R2

R1 R3

Ra Rb

Rc

! Y

a

c

b b

c

aR3 R1

R2 Rb Ra

Rc

• Terminal a, b: Rab= R2 || R1 + R3( ) Rab = Ra + Rb

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For a given ! network the equivalent Y network is obtained by setting

R2 R1 + R3( )R1 + R2 + R3

= Ra + Rb

R3 R1 + R2( )R1 + R2 + R3

= Rb + Rc

R1 R2 + R3( )R1 + R2 + R3

= Ra + Rc

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Ra =R1R2

R1 + R2 + R3 ; Rb =

R2R3R1 + R2 + R3

; Rc =R1R3

R1 + R2 + R3

Note: R1 = R2 = R3 = R ! Ra ,Rb ,Rc = ?

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Example Find the current and power supplied by the 40 V source.

We wish to replace the upper ! network by the equivalent Y network

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Rc

Ra

Rb

Ra = 100 !125100 +125 + 25

= 50"

Rb = 125 ! 25100 +125 + 25

= 12.5"

Rc = 100 ! 25100 +125 + 25

= 10"

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The equivalent circuits

Req = 55 +

10 + 40( )! 12.5 + 37.5( )10 + 40( ) + 12.5 + 37.5( ) = 55 + 50 ! 50

100 = 80!

! i40V = 4080

= 0.5A P40V = 40 ! 0.5 = 20W - supplied

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2.5 Single-Loop (Voltage Division) and Single-Node (Circuit Division) Circuits

A. Voltage Division / Single-Loop Circuit

In electronic circuits it is often necessary to generate multiple voltage levels from a single voltage supply

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KVL + Ohm’s L vs = iR1 + iR2 ! i = vs

R1 + R2

v1 = iR1 , v2 = iR2

v1 =R1

R1 + R2vs , v2 =

R2R1 + R2

vs

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v2 =R2

R1 + R2vs

v1 =R1

R1 + R2vs

Given vs , we wish to specify v2 ! there is a large number of combinations of R1 and R2 yielding the proper ratio

R2R1 + R2

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Example The resistors used in the voltage-divider circuit have a tolerance of ±10% . Find the maximum and minimum value of v0 .

v0 = 80V

v0 max( ) = R2 +10%R2

R1 !10%R1( ) + R2 +10%R2( ) vs = 11022.5 +110

!100 = 83.02V

v0 min( ) = R2 !10%R2R1 +10%R1( ) + R2 !10%R2( ) vs = 90

27.5 + 90!100 = 76.60V

The output voltage will lie in the interval 76.60,83.02[ ]

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Example (Practical Power Application)

16.45367.09

= 65.8

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Multiple-Resistor Networks

i = vR1 + ....+ Rn

= vReq

vj = iRj =Rj

Reqv Voltage Division Equation

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B. Current Division/ Single Node Circuit

v = i1R1 = i2R2 and v = R1R2

R1 + R2is

i1 =R2

R1 + R2

larger R2 larger i1

!"#$is , i2 =

R1R1 + R2

is

ibranch =RtotalRbranch

is ! Rtotal =R1R2R1 + R2

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Rtotal = 3.75k!

I1 =RtotalR1

Is = 30mA

I2 =RtotalR2

Is = 10mA

I1 =Vs5k

= 2mA

I2 =Vs20k

= 0.5mA

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Example Find the power dissipated in the 6! resistor

P6! = i6! " v6! = 6 " i6!2

10 R4||6 = 2.4

R1.6!4||6 = 410

i4

R1.6+4||6 = 4"

CD : i4 =16

16 + 4!10 = 8A

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8Ai6

CD : i6 =44 + 6

! 8 = 3.2A

P6! = 3.22 " 6 = 61.44W

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Multiple-Resistor Networks

v = i ! Req ; Req =

11Rll

!

i j =vRj

! ibranch =

ReqRbranch

! i Current Division Equation

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Example Use CD to find i0 and use VD to find v0

Req = 36 + 44( ) ||10 || (40 +10 + 30) || 24 = 6 !

CD: i0 =Req24

! 8 = 2A

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The voltage drop across 40!10! 30

2A

v40!10!30 = 24 " 2v! = 48V

VD: v0 =

3040 +10 + 30

! v40"10"30 = 3080

! 48 = 18V

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Example

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+!

!

+

KCL: 16mA+I2 ! I1 = 0

CD: I1 =120

120 + 40!16 = 12 mA

! I2 = 12 !16 = !4 mA

! P40k! = I12 " 40k! = 5.76W

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Example

P6k! = 6k " I2

2

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+

I1

I2

I3

4 ||12 = 3 k!

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!2mA 3k"I2

CD: I2 =33+ 6

! "2mA( ) = " 23mA

P6k! = 6 "103( )" I22 = 6 !103( )! " 23

#$%

&'(2

!10"6 = 2.67mW

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2.5. Measuring Resistance: The Wheatstone Bridge

Galvanometer µA( )

R1,R2 ,R3 - known resistors; Rx - unknown resistor

To find Rx we adjust R3 until there is no current in Galvanometer, i.e., ig = 0 .

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Let ig = 0 , i.e., the bridge is balanced ! KCL: i1 = i3 , i2 = ix

By KVL (points a and b are at the same potential):

i3R3 = ixRx , i1R1 = i2R2

i1 = i3 , i2 = ix

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i3R3 = ixRx ! i1 = i3 , i2 = ix

i1R1 = i2R2 ! i1R3 = i2Rx

i1R1 = i2R2 !

R3R1

= Rx

R2

!

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Rx =R2R1

! R3

Rx =R2R1R3

• The range of Rx : 1! - 1 M! ! with the accuracy ±0.1%

• The range of R3 : 1! - 11 k!

• The ratio R2 / R1 range: 0.001 - 1000

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Example: R3 - varies from 10! to 2k!

1k! 4k!

Rx =R2R1

! R3

• R3 = 10! ! Rx =40001000

10( ) = 40! • R3 = 2k! ! Rx =

40001000

2000( ) = 8k! ! 40! " Rx " 8k!

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2.6 ! ! Y ( ! ! T ) Transformations (Delta ! Wye)

v Req!

A resistive network generated by a Wheatstone bridge circuit

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! (Delta) Resistance Network

R2

R1 R3

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Y (Wye) Resistance Network

Ra Rb

Rc

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! ! Y Equivalent Circuits

R2

R1 R3

Ra Rb

Rc

! Y

• Terminal c, a: Rca= R1 || R2 + R3( ) Rca = Ra + Rc

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• Terminal b, c: Rbc= R3 || R1 + R2( ) Rbc = Rb + Rc

R2

R1 R3

Ra Rb

Rc

! Y

a

c

b b

c

aR3 R1

R2 Rb Ra

Rc

• Terminal a, b: Rab= R2 || R1 + R3( ) Rab = Ra + Rb

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For a given ! network the equivalent Y network is obtained by setting

R2 R1 + R3( )R1 + R2 + R3

= Ra + Rb

R3 R1 + R2( )R1 + R2 + R3

= Rb + Rc

R1 R2 + R3( )R1 + R2 + R3

= Ra + Rc

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Ra =R1R2

R1 + R2 + R3 ; Rb =

R2R3R1 + R2 + R3

; Rc =R1R3

R1 + R2 + R3

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For a given Y network the equivalent ! network is obtained by setting

! Y

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Example* Find the current and power supplied by the 40 V source.

We wish to replace the upper ! network by the equivalent Y network

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Rc

Ra

Rb

Ra = 100 !125100 +125 + 25

= 50"

Rb = 125 ! 25100 +125 + 25

= 12.5"

Rc = 100 ! 25100 +125 + 25

= 10"

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The equivalent circuits

Req = 55 +

10 + 40( )! 12.5 + 37.5( )10 + 40( ) + 12.5 + 37.5( ) = 55 + 50 ! 50

100 = 80!

! i40V = 4080

= 0.5A P40V = 40 ! 0.5 = 20W

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Example

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!

Ra =54 ! 36108

=18k"

Rb = 6k"Rc = 9k"

Ra = 18k"

Rb =18 ! 36108

= 6k"

Rc =18! 54108

= 9k"

RT = 6 +18 + 9 + 3( ) || 6 +18( ) + 2 = 34 k!

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Example

!

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4

4 4

1212

V0

V0 = 12k!" 4mA2

symmetry!

= 24V

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Example

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6

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12 6I1

I1

CD : I1 =6 + 6( )

12 + 6( ) + 6 + 6( ) ! 3A = 1230

! 3A = 1.2A

I1 = !I1 = !1.2A

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Electrical Safety

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2.7 Circuits with Dependent Sources

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Example

KVL: !12 + 3kI1 !Va + 5kI1 = 0 and Va = 2000I1 ! I1 = 2mA

V0 = 5k ! I1 = 10V

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Example

KCL + OL: 10 !10"3 + Vs

2 + 4+ Vs3" 4I0 = 0 and I0 =

Vs3

! Vs = 12V

VD: V0 =44 + 2

Vs = 8V

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Example

KVL: !6 + 4kI ! 2VA + 8kI = 0 and VA = 4I ! I = 1.5mA

V0 = I ! 8k = 1.5m ! 8k = 12V

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Example

I2

V0 = 2k ! I2 The equivalent circuit source: I0 =

V02!10"3 + 2 !10"3A ------> !

CD: I2 =63+ 6

! I0 =23I0 and V0 = 2k!" I2 ! V0 =

43!103 ! I0

! V0 =43!103 ! V0

2!10"3 + 2 !10"3#

$%

&'(

! V0 = 8V

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Example V0 = 12k ! I3

I2I3

• I3 =6

6 +12! I2 ! I3 =

13! I2

• The equivalent circuit source: I0 = 6 !10

"3 + 0.5Ix A ! • I2 =

6 ||126 ||12 + 4

! I0 = 44 + 4

! I0 ! Ix =4

4 + 4! I0 ! Ix = I2

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• Ix = 4mA ! I2 = 4mA

• I3 =13! I2 = 4

3mA

• V0 = 12k ! I3 = 16V

I2I3

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2.8. Analyzing Circuits Containing V/C Sources and Interconnection of Resistors: Examples

Example

V0 = 2V

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I I1

Req=60 ! 3060 + 30

= 20k" V0 = 20k ! I1

• I = 12V20k + 20k

= 0.3mA

• CD : I1 =30

30 + 40 + 20( ) I = 0.1mA ! V0 = 20k!" I1 = 2V

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Example

I1

• I1 =3V30k!

= 0.1mA

• CD : I1 =60

60 + 90 + 30( ) IS =13IS ! IS = 3I1 = 0.3mA

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Example V1 = 3k ! I1

I1 I2

• The equivalent circuit source: I0 = 10mA

• CD: I1 =ReqRbr

I0 =9k || 6k9k

!10mA = 4mA

• V1 = 3k ! I1 = 12V

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Example

I1

I2

I1 =63+ 6

I2

• CD : I2 =12

3 || 6 + 4( )+12 ! 9mA = 6mA

• CD : I1 =63+ 6

I2 = 4mA ! I0 = !I1 = !4mA

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Example

I1

I2 I3

• V0 = 5k ! I5k"

• V1 = 4k ! "I3( ) • V2 = 8k ! I4 k"8k

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• Equivalent Circuits

I3I2

I1

I1 Req = 10 || 20 + 5 ||10{ } || 4 +12 || 4 + 8( )

6! "# $#

!"#

$#

%&#

'#= 5

I1 =16V3k + 5k

= 2mA

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I3I2

I1

10 || 20 + 5 ||10{ }= 10

4 +12 || 4 + 8( )6

! "# $#

= 10

By symmetry I2 = I3 =12I1 = 1mA

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I3I2

I1 2mA

1mA1mA

CD: I5k! = 105 +10

I2 =23mA ! V0 = 5k ! I5k" = 3.33V

OL: V1 = 4k ! "I3( ) = "4V

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I3I2

I1 2mA

1mA1mA

I4

CD : I4 =12

12 + 4 + 8( ) ! I3 = 0.5mA

V2 = 8k ! I4 = 4V

Answer: V0 = 3.33V , V1 = !4V , V2 = 4V

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2.9 Summary of Chapter 2

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105

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2.10 Problems ! Tutorials 2 + 3

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Georg Simon Ohm

On March 16, 1787, German physicist was born in Erlanger, the son of a master mechanic. After achieving his Ph.D., Ohm was only able to secure a high school teaching position. Needing impressive research publications to get a university post, he began experiments in the new field of electric currents. Lack of funds forced him to make his own wires, which led to experiments on wires of different thicknesses, and his discovery of Ohm’s law [electric current is inversely proportional to resistance and directly to proportional to voltage]. His resistance lead to discovering resistance.