Electric Circuits

UAE University Department of Electrical Engineering Dr.Hazem N.Nounou

ELEC 3202

Electric Circuits I

Dr. Hazem N. Nounou

Department of Electrical EngineeringUnited Arab Emirates University


Chapter OneBasic Concepts

(1) Electron :electron is a mobile charge carrier.

•The electron is measured in coulumb [ C ]

• e = 1.6*10-19 C

• Multiple of electrons constitute charge (q).

• This course basically deals with the analysis of electric circuits.• The most basic quantity used in the analysis of electrical circuits

is the electric charge (electron).Basic Quantities


•The movement of charge (q) over time causes current.

(2) Current :the time rate of change of charge produces an electrical current

• the electric current is measured in Amper [A]

1 A = 1 C / 1 sec

•.current convention.

dtdq(t)i(t) = Or ∫

−∞==

t

τd τ)i( τq(t)

e e e

i

---


i(t)

time

There are 2 types of currents 1. Direct current (DC)

2. Alternating current (AC)

i (t)

time


(3) Voltage :The voltage is defined as the work or energy (in Joules) required per unit charge to move a test charge though an element

qWV = And

C 1J 11V =

• Since we are dealing with a changing charge and energy, we have

dqdwv =

(4) Power :

Power is the time rate of change of energy.


dtdw(t)P(t) =

dtdq

dqdw(t)

dtdw(t))(P ⋅==t

i(t)V(t))(P =t

•The unit of power is Watt [W].

• 1 W = 1 V * 1A

(5) Energy: energy can be expressed as

∫=

=∫=

=2

1

2

1

t

ttdti(t)v(t)

t

ttdtp(t)w(t)


Passive sign conventionCurrent flow from the positive to the negative terminal.

i(t) R

(+)

(-)

• Power can be absorbed or supplied by an element.

• Power is absorbed (or dissipated) by an element if the sign ofpower is (+)

• Power is supplied (delivered or generated) by an element if the sign of power is (-)


Circuit Active Elements:

There are 4 types of active elements (sources):

1. Independent voltage source: It is a 2-terminal sources that maintains a specific voltage across its terminals regardless of the current through it

+-

2. Independent current source:It is a 2-terminal sources that maintains a specific current through it regardless of the voltage across it terminals.

3. Dependent voltage source:It is a 2-terminal sources that generates a voltage that is determined by a voltage or current at a specified location in the circuit.


4. Dependent current source:It is a 2-terminal sources that generates a current that is determined by voltage or current at a specified location in the circuit.Example :Compute the power that is absorbed or supplied by each of the elements in the following circuit

R2

-

+-Vs = 36 V

Ix = 4 A R1

+ 12 V

++

--24 V 28 V

1 Ix

I R2 I R3=2 A

R3


)(supplies144W4)(36)(IVP xsvs −=−==

)(absorbs48W(12)(4)IVP xR1R1 ===

(absorbs)48W2)-(24)(4)I-(IVIVP R3xR2R2R2R2

====

(supplies)W8-(4)(-2)))(II(1IVP R3xR3DsDs ====

(absorbs) W56(28)(2)IVP R3R3R3 ===


Important Units

QUANTITY SYMBOL

Length l Current I, i Temperature T Mass m Time t

UNIT ABBREV. meter m ampere A kelvin K kilogram kg second s


Important Units

Voltage V, v, Charge Q, qResistance RPower P, pCapacitance CInductance LFrequency fMagnetic Flux ΦMag. Flux Density B

volt Vcoulomb Cohm Ωwatt Wfarad Fhenry Hhertz Hzweber Wbtesla T


Unit Conversions

UNIT MULTIPLY BY TO GET

in 0.0254 mft 0.3048 mmi 1.609 kmlb 4.448 Nhp 746 WkWh 3.6 x 106 Jft-lb 1.356 J


Prefixes For Engineering Notation

POWER OF 10 PREFIX SYMBOL1012 tera T109 giga G106 mega M103 kilo k10-3 milli m10-6 micro µ10-9 nano n10-12 pico p


Chapter 2

Resistive CircuitsOhm’s law :The voltage across a resistor is directly proportional to the current flowing through it.

V (t) = R i(t) R ≥ 0 R1

v (t)

i (t)The symbol of ohm is Ω( )

A1V1Ω1 =


R(t)v(t)iRi(t)v(t)P(t)

R(t)v

Rv(t)v(t)(t)iR

i(t)i(t)Ri(t)v(t)P(t)

22

22

===∴

===

==

Note: Last equation says that the power at a resistor is always positive

Resistors always absorb power.

The instantaneous power P (t):

The inverse of resistance is conductance

R1G =


The unit of conductance is Siemens (S)

1V1AS1 =

The current can be also expressed as

V(t)Gi(t)=

And the instantaneous power is

G(t)ii(t)

Gi(t)i(t)v(t)P(t)

2

===


(t)G vG(t)ii(t)v(t)

(t)vGv(t)Gv(t)i(t)v(t)P(t)

22

2

===⇒

===

Open and short CircuitsOpen circuit ( R = ) ∞ G = 0

R=∞circuit circuit Open circuit

0v(t)R

v(t)i(t) =∞

==


R= 0circuit circuit Short circuit

0)(*0)(v(t) === titRi

Short circuit ( R = 0) G = ∞


Example :Consider the circuit:

Find the current and power absorbed by the resistor

+-

I

Ωvs=12 v R = 2 k

Am6Ωk2v12

Rv

I s ===

wm72m)(6(12)IvP R ===


Example:The power absorbed by a 10 k resistor in the circuit is 3.6 mW. Find voltage and current in the resistor.

RIIVP 2s ==

Ω

( )

mA0.610*3.6I

)10*(1010*3.6RPI

RPI

7

33

2

==

==

=

−

−

V6V)k(10A)m(0.6RIV

=Ω==

+-

I

vsΩk10R=


Example :

Find the value of the voltage source and the power absorbed by the resistance

G = 50 Sµ R=1/G =2*104

Wm5mA)(0.5V)(10IVPV10Ω)10*(20A)m(0.5RIV

R

4s

======

+-

I=0.5 m A

VsSµ50G =


Example :

Find R and the voltage acrossThe resistor?

Ω===

==

=

==

=

−

−

−

−

k

V

5A10*4

V. 20IVR

RA)10*(4RIV.20V

A10*4W10*80

IsPV

Is VP

3

3

3

3

P=80mW RIs=4mA


Kirchoff’s Laws:

(1) kirchoff’s current law (KCL) :the sum of all currents entering any node is zero.

∑=

=N

1kk 0(t)i

Where N= number of currents.


Here we have (4) nodes:

At node (1) :

At node (2) :

At node (3) :

At node (4) :

(t)i(t)i(t)i 521 =+

(t)i(t)i50(t)i 322 =+

(t)i(t)i(t)i50 142 =+

(t)i(t)i(t)i 543 =+

Example:Write the KCL equation

50 i2

i4 R4

i1 R1

i3

R3

i2R2

+

-

i5

Vs

(1)(2)

(3)

(4)


(2) Kirchoff’s voltage Law (KVL):

The sum of the voltage around any loop is zero.

0(t)vN

1kk∑

=

= N = # of voltage

Example:

Find VR3 ? using KVL

-30+18-5+12-15+ VR3 = 0

VR3 = 20 V

-- VR3 +

R3

VR2= 12 V R2

VR1=18 V

R1

+ -

Vs3= 15 V

+-

Vs2= 5 V

+

-Vs1=30 V


Example :Find the KVL equation for the two paths abda and bcdb

0vvv sR2R1 =−+

0vvv20 R2R3R1 =−+

Path abda:

Path bcdb:

+-

a b c

d

R1

R2 R3

VR1

VR2 VR3

20 VR1

--

-

+

+

+

Vs


Single Loop circuitsWe will discuss (2) issues :

1. Voltage divider rule:Voltage is divided between resistor in direct proportion to their resistance

v(t)RR

R(t)v

v(t)RR

R(t)v

21

22

21

11

+=

+

vRR

Rv

)RR

v(RiRv

21

11

21111

+=

+==

+-

+

-R1

R2

V1

V2-

+V(t)

How?


Multi Sources / resistors :

•Source can be added v=v1+v2+……•Resistors can be added R= R1+R2+…..

Where: v = v1 + v2

R = R1 + R2 + R3

+-

+-

R1 R2

R3v1

v2

+- RV


Single Node-Pair circuits :

We will discuss (2) issues:

1. Current-divider Rule .

(t)iRR

R(t)i

(t)iRR

R(t)i

21

12

21

21

+=

+=

1221

21

21

2211

iiiiii

iRR

i

RiRiv

−=⇒+=

=∴

==Why ??

i (t) R1 R2

i1(t) i2(t)


iRR

Ri

iRR

)R

RR(i

iRR

)RR

1(i

)i(iRR

i

21

21

1

2

1

211

1

2

1

21

11

21

+=

=+

=+

−=

2. Multiple sources/resistors :

•Current source can be added.

•Resistors can added as reciprocals


R1 R2i1(t) i2(t) R3

321

21

R1

R1

R1

R1

(t)i(t)ii(t)

++=

+=

Ri (t)


Series and parallel resistors :

Series :

Parallel

∑=

=⇒+++=N

1kksN21 RRRRRR K

∑=

=

+++=

N

1k kP

N21P

R1

R1

R1

R1

R1

R1

K


Example :Find equivalent resistance

Ωk10

Ωk1Ωk6

Ωk6Ωk6

Ωk2

Ωk4

Ωk2

Ωk2

Ωk9

( ) ( ) Ωk10Ωk6||]Ωk2Ωk1[R1 ++=

Ωk6

Ωk2

Ωk4

Ωk2

Ωk9

Ωk6 Ωk12R 1 =


Ωk2

Ωk4

Ωk9

Ωk6 k6R2=

[ ] 6kk2k6//k12R2 =+=

( ) k3k6//k6R3 ==

Ω

Ω


Ωk2

Ωk4

Ωk9

k21R3 =

( ) k5k2k4||k12Req =+=

Ωk5R eq =

3 kΩ


Example :Find all currents and voltages

The equivalent circuit is :

Ωk9

Ωk4Ωk6+- Ωk3

I2

I1 I5

I4

I3

Va Vb Vc

Ωk9

+ + +

-- -12 V

Ωk3

eqRVa+

-

12 V

Ωk9I1+

-


Am21

k63

Ωk6V

I

V3IRVAm1k3k9

V12I

a2

ieqa1

===∴

==⇒=+

=

Am21Am

21Am1IIII 3213 =−=⇒−=

( )[ ][ ]kΩ3

k6//k3k4//k9k3R eq

=

++=


V1.5V01.53V

0VVV

bb

Ωk3ab

=⇒=+−

=+−∴

Am81

k121.5

k3k9V

I b5 ==

+=

V83)(3kΩAm

81)(3kΩIV 5c ===

V1.5)Ωk(3A)m21()Ωk(3IV 3Ωk3 ===


Example :Find the source voltage Vo if I4=1/2 m A ?

Am1.5Am21Am1III

Am1k3

3k3

VI

V3Ω)k(6A)m21((6k)IV

432

b3

4b

=+=+=

===∴

===

k6

k1k2

+-

I2

I1 I5

Va+

-

Ωk3

Vo

k3 k6I3 I4

Vb

k4

+

-


k6

k1k2

+-

I2

I1 I5

Va+

-

Ωk3

Vo

k3 k6I3 I4

Vb

k4

+

-

V36V36630V

33m)k(310)(Ik4VVI)Ωk(6V

0

0

1ba10

==+=

++=+++=

Am3Am1.5Am1.5III

Am1.54k

331k3kVV

I

V3m)(1.5k)(2IΩk2V

521

ba5

2a

=+=+=∴

=+

=++

=

===∴


Example :Find V0 ? Using KVL

V10Vm)(2k)(5)(IΩ)k(5V

Am2I126kI0)(Ik5I2000)(Ik312

0

10

11

111

==

=⇒==+−+−

+-

2000 I1

R25k ohm

R13k ohm

+

-

Vs112 V

I1 +

--

Vo


Example:Find V0 using KCL:

V8(12)32V

k4k2k4V

V12V

Am10k3

4k3

1k6

1V

0)k3

V(4

k3V

k6V

m10

s0

s

s

sss

==+

=

=

−=

−+

=−++

VS

+

-

4 I0

I0

3 k

2 k

4 k V0

+

-

10 m A


Example:Find V0 using KVL

V6V12V3V

121kV(3k)V

k1VI

012I(3k)V0VV2Ik)(312

0

00

00

0

0

00

==+−

=

+−

=

=−+−=+−+−

+-

12 V

I 3 k 2 V0

1 k V0

+

-


Example:Find V0 in the network

( )

21

21

23211

210

i21i

i)Ωk(3i)Ωk(6AlsoiRRiR

0Am2ii2000V

=∴

=+=

=+−−

i1 i2

R1 R2

R3

Ωk1Ωk6Am2

V0

+

-Ωk2

2000V0


V8VAm2k2

V21

0Am2k2

V23

k2V

00

00

=⇒=

=+

−∴

k2V

iiRV

0Am2i23

2000V

0Am2ii21

2000V

02230

20

220

=⇒=

=+−

=+−−∴

Q


Chapter (3)Nodal and loop analysis

Consider the following circuit:

+-

+-

R2

R4

R5

R7

R6

R3

R1

V1

V2

I

i1

i5 i4

i3

i6i2

a b

c d e

f g


Definitions:Node :A point where two or more circuit elements joinEx. a,b,c,d,e,f,g

Essential Node:A node where three or more circuit element joinEx. b,c,e,g

Path:A trace of adjoining elements with no elements included more than once

1. V1-R1-R5-R62. R5-R6-R4-V2 ,etc


Branch:A path that connects any two nodes.Ex. R1 , V1 , R1-R5 , etc

Essential Branch:A path that connects two essential nodes without passing throughan essential node.Ex. V1-R1 , R5 , R2-R3 , V2-R4 , …

Loop :

A path whose last node is the same as its starting node

Ex. (1) V1-R1-R5-R3-R2

(2) V1-R1-R5-R6-R4-V2


Mesh :A loop that doesn’t enclose any other loops.Ex. V1-R1-R5-R3-R2Ex. V2-R2-R3-R6-R4 , …

• In chapter (2) we studied circuits containing a single loop or asingle node-pair

• Such circuits can be solved easily by one algebraic equation.• Here , we will study circuit containing multiple node and

multiple loops• Hence we will introduce (2) analysis techniques :1. Nodal analysis 2. Loop analysis


(1) Nodal Analysis :

Nodal analysis : is a technique in which KCL is used to determine the nodes’ voltages at all essential nodes with respect to the reference node.

•Here , node voltage is defined as the voltage of a given node with respect to a reference node

+-+

-

i5i3i1 i4i2Vm = 10

R1= 2 R3= 2 R5= 1

R4 = 2R2= 1

1 2

3

Vn = 5

Example:


Essential nodes : 1,2,3Consider (3) to be the reference node (ground)At (1) apply KCL:

( )110V4V

02

V2V5

02

V2

VV

2V

5

02

VV1V

2V10

0R

VVRV

RVV

0iii

21

21

211

1

2111

3

21

2

1

1

1m

321

KK=−

=+−

=+−−−

=−

−−−

=−

−−−

=−−


( ) ( )

( )210V4V

05V22

V

01

5V2

V2

VV

21

21

2221

KK=−

=−−

=+

−−−

0R

VVRV

RVV

0iii:KCLApply,(2)At

5

n2

4

2

3

21

543

=−

−−−

=−−

2V2VBAVBVA

1010

VV

4114

2

11

2

1

−==⇒=⇒=

=

−−

−


Note :Number of equations = N-1Where :N is the number of essential nodes

Example :Circuit with only independent current source

iR1

i1

R3= 5

R1= 60

R4 = 2R2= 15

1 2

3iR2

5 A = I S2iR4

V 1V 2

+

-

+

-

15 A = I S1


Find V1,V2 and i

# of essential node = N=3Select (3) as ground (reference node)# of KCL equations = N-1 = 2

At node (1) apply KCL

05

V5

V15V

60V15

0R

VVRV

RV

15

0iiiI

2111

3

21

2

1

1

1

1R2R11s

=+−−−

=−

−−−

=−−−


(1)15V51V

6017

15V51V

601412

15V51V

601

151

51

21

21

21

KK=−

=−++

=−

++

(2)50V7V2050V5V2V2

052

V5

VV

05RV

RVV

0Iii

21

221

221

4

2

3

21

s2R41

KK=−=−−−

=−−−

=−−−

=−−At node (2)


A105

1060R

VVi

V10VV60V

5015

VV

7251

6017

211

2

1

2

1

=−

=−

=

==

=

−

−

Example :Circuit with dependent current source

i21 2

3

I0

V 2

I S = 2 mA Ωk10R3 =

V 1 Ωk10R2 =

2 Ix

i1

Ix

Ωk10R1 =

Find I0


# of essential nodes = N=3Choose (3) to be reference node ( ground )We have N-1 = 2 KCL equation

at node(1) and( 2)

At node (1) , apply KCL

(1)A2V0.1V0.210kV

5kV

10kVV

10kVAm2

0R

VVRVAm2

0iiI

21

21211

2

21

1

1

21s

KK=−

−=−

+=

=−

−−

=−−


0II2i 0x2 =−−

At node (2) apply KCL,

0II2R

VV0I2Ii

0x2

21

0x2

=−−−

=−−

(2)0V2V1

010kV

10kV2

10kVV

RVI

RViIwhere

21

2121

3

20

1

11x

KK=+

=−−−

⇒

=

==


Am0.4k104

RVI

V4VV8V

02

V2V1

210.10.2

3

20

2

1

−=−

==⇒

−==

=

−


Example:(circuit with independent voltage source)

Find ia ,ib ,ic ,id ,ie

N=3 N-1 = 2

Choose (3) to be ground

ic1 2

3ib

V 2V 1 Ω18

ia

id

+-

+- 70128

ie

Ω10Ω8

Ω48 Ω02


(1)16V181V0.201388

018V

18V

48V

8V

16

018

VV48V

8V128

1

2111

2111

KK=−

=+−−−

=−

−−−

KCL at (1)

(2)7V0.20555V181

010

70V20V

18VV

21

2221

KK−=−

=−

−−−

KCL at (2)


V60VV96V

716

VV

0.2055181

1810.201338

2

1

2

1

==

−

=

−

−

A110

70Vi

A42060

20Vi

A218

609618

VVi

A24896

48Vi

A48

961288

V128i

2e

2d

21c

1b

1a

−=−

=

===

=−

=−

=

===

=−

=−

=∴


A110

70Vi

A42060

20Vi

A218

609618

VVi

A24896

48Vi

A48

961288

V128i

2e

2d

21c

1b

1a

−=−

=

===

=−

=−

=

===

=−

=−

=∴

Special case:What if a branch between two essential non-reference node contain a voltage source ?This case is called “super node" case.


Find I0• # of essential nodes = N = 3• “Super node: is the voltage source and the two connecting nodes• # of equations = N-1-1 = 3-1-1=1

Reference node Super node

But we need (2) equations to find the two unknowns V1 and V2There is an equation that describe the super node.

1 2

3i2

V 2V 1

i1i3

+ -6 V

Ωk6

Io

Ωk6 Ωk6Ωk21Ωk21

Ωk6

supernode


Apply KCL at the super node :

(1)06kV

6kV

012kV

12kV

k12V

k12V

0Iiii

21

2211

0321

KK=+

=+++

=+++

Am0.2512k

312kV2I

V3VV3V

60

V2V1

116k1

6k1

)2(V6V

0

2

1

12

−=−

==⇒

−==

=

−

=+ KK

The super node is described by:


Example:Circuits with dependent voltage sources

Find I0? N = 3 N-1 = 2 equations

Io1 2

3

i2

V 2V 1 Ω5

i1

i3

+-20 V

i4

Ω2Ω2

Ω02 Ω01

8 Io


(1)10V51V

43

10V51V

51

201

21

21

21

KK=−

=−

++

5V

5V

20V

2V10

5VV

20V

2V20

Iii

2111

2111

021

−+=−

−+=

−+=

KCL at node (1):


2V

10V

5VV

5

5VV

42

V10V

5V

5V

2I8V

10V

5VV

iiI

2221

212221

02221

430

+=

−

−

−+=−

−+=

−

+=

(2)0V58V

0V21

1011V

2V

10VVV

21

21

2221

KK=−

=

++−

+=−


A1.256

5VVI

V10VV16V

010

V2V1

581

51

43

210

2

1

==−

=

==

=

−

−

Loop Analysis (Mesh)

Mesh analysis : It is a technique in which KVL is used to determine the current in all meshes# of equations needed = # of meshes = be-(ne-1)Where :be : # of essential branchesne: # of essential nodes


Example :

(1)40i0i8i100)ii(82i40

321

211

KK=+−=−++−

KVL left loop :

Ω6

+-

+- 20 V

Ω4Ω2

Ω840 V

Ω6

+

-

Voi1i3

i2

Find V0?We have (3) meshes


A0.8i,A2i,A5.6i20040

iii

10606208

0810

321

3

2

1

−===

−=

−−−

−

V28.8V(3.6)82)(5.68)ii(8V

0

210

==−=−=

0i6i20i80)i(i6i6)i(i8

321

32212

=−+−=−++−

20i10i6i004i)i(i620

321

323

−=+−=+−+

KVL middle loop:

KVL right loop :


Example:Mesh with dependent source

# of meshes = 3 so 3 equations are needed

(1)50i20i5i25)i(i20)i(i550

321

311

KK=−−−+−+−

KVL around mesh 1:

i2 Ω4

i1 i3

+-

50 V

Ω5

Ω02Φi15

Ω1

Φi


(2)0i4i10i50)i(i5)i(i4i1

321

12322

KK=−+−=−+−+

i3iiwhere0)i(i4)i(i20i15

1Φ

2313Φ

−==−+−+

KVL around mesh 2:

KVL around mesh 3:

0i9i4i50)i(i4)i(i5

0)i(i4)i(i20)i(i15

321

2331

231331

=+−−=−+−−

=−+−+−∴

A28iA,26iA,29.6i00

50

iii

945410520525

321

3

2

1

===

=

−−−−−−


Special Case :What happens if a current source is located between two meshes

“Super mesh”

i2 Ω2

i1 i3

+- 50 V

Ω3

Ω10

+-

100 V

5 A

Ω4Ω6

You don’t know the voltage across the current source !!Remove the whole branch that includes the current source


i2 Ω2

i1i3

+- 50 V

Ω3

Ω10

+-

100 V

Ω4Ω6

Apply KVL around the super mesh

(1)50i6i5i90i6i450)i(i2)i(i3100

321

132321

KK=+−=+++−+−+−


(2)0i2i15i30)i(i3)i(i2i10

321

12322

KK=−+−=−+−+

(3)5ii0iA5ii

321

13

KK=++−=−

A6.75iA1.25iA1.75i

3

2

1

===

KVL around the upper loop:

We also know


Example:

Nodal Analysis :

Ω001

+-

128 V

Ω501

256 V

+-

Ω005Ω502

Ω200 Ω400

Ω300

Φi

Φi50

Use nodal analysis and loop analysis to find power in the 300 (Ω) resistor ?


150256V

3001V

1001

1001

2001

3001

1501V

0100

VV200V

300VV

150V256

0iiii

321

211131

32Φ1

=

−

−

+++

=−

−−−

+−

=−−+KCL at node (1):

Ω001

+-

128 V

Ω501

256 V

+-

Ω005Ω502

Ω200 Ω400

Ω300

Φi

Φi50

V1 V3V2

i1 i6

i5

i4

i2

i3

1 2 3


(1)1.7067V0.00333V0.01V0.0250 321 KK=−−

( )

(2)0.256V0.0118V0.004V0.0033500128

3001

5001

4001

2501V

250V

300V

0300

V1V500

128V400V

150VV

0iiii

321

321

33332

Φ654

KK=−+

=

−−−−

++

=−

−−−

−−−

=−−−KCL at node (3):

You can notice that

(3)0V0.1667V1V0.166

V61V

61

300VV

50V

i50V

321

1313

2

Φ2

KK=−+

−=

−

=

=


V8VV11.75V

V62.5V0

0.2561.7067

V3V2V1

0.166710.16670.01180.0040.00330.00330.010.0383

3

2

1

−=−=

=

=

−−−−

( ) ( ) W16.56753000.235RiP

A0.235i50V

i

22Φ300Ω

Φ2

Φ

=−==

−=⇒=


Using loop analysis:

5 meshes 5 equations

KVL around loop (1):

(1)256i200i350256)i(i200i150

21

211

KK=−=−+

Ω001

+-

128 V

Ω501

256 V

+-

Ω005Ω502

Ω200Ω400

Ω300

Φi

Φi50i1

i5

i4i3

i2


(2)0i150i300i2000)i(i200i50)i(i100

ii0)i(i200i50)i(i100

521

12552

5Φ

12Φ52

KK=−+−=−+−−

−==−++−


5Φ

543

Φ4353

iisince(3)0i200i400i6500i50)i(i400)i(i250

−==−−

=−−+−KK

(4)128i900i4000)i(i400128i500

43

344

KK=+−=−+−

KVL around loop (3)

KVL around loop (4)


(5)0i650i250i1000)i(i100)i(i250i300

532

25355

KK=+−−=−+−+

KVL around loop (5)

=

−−−

−−−−

−

0128

00

256

iiiii

650025010000900400002004006500015000300200000200350

5

4

3

2

1


W16.5675(300)(0.235)R)(iP 22

5300Ω

===

A0.235iA0.24iA0.22i

A0.9775iA1.29i

5

4

3

2

1

=====


Example:Use the loop analysis to find ΦV

10Ai3 −=

(3) Meshes (3) equations

i3 Ω1

i1 i2

+-

Ω2

Ω5 5V2 Φ

10 A

+

-ΦV

Ω57 V


(1)55i5i707520i5i7

075)i(i5)i(i2

21

21

2131

KK=−=−+−

=−−+−


( ) ( )

(2)0i3i2

i2i2ii552i

)i(i5V5V2

i

21

21212

21Φ

Φ2

KK=−

−=−

=

−=

=

Equation of dependent source:


A10iA15i

055

ii

3257

2

1

2

1

==

=

−−

V25VV2510)(155

)i(i5V

Φ

21Φ

==−=

−=


Chapter (4)Additional Analysis Techniques

Here we will study four additional Techniques• Superposition• Source transformation• Thevenin and Norton Theorems• Maximum power principle1. Superposition :Definition :Whenever a linear circuit is excited by more than one independentsource, the total response is the algebraic sum of individual responsesThe idea is to activate one independent source at a time toget individual response.Then add the individual response to get total response


Note:1. Dependent source are Never deactivated (always active)2. When an independent voltage source is deactivated, it is set to zero.

replaced by short circuit3. When an independent current source is deactivated, it is set to zero.

replaced by open circuit

Example:Use superposition to find i1,i2,i3,i4 ?

+- Ω4

Ω2Ω6

Ω3 A12120 V

i1

i2

i3

i4


+- Ω4

Ω2Ω6

Ω3120 V

'i1

'i2

'i3

'i4

V1

open circuit forcurrent source

•Activate independent voltage source 120 V only

•Using KCL at V1 (nodal analysis)

0i'i'i' 321 =−−


V30V

061

31

61V20

042

V3

V6

V120

1

1

111

=⇒

=

++−

=+

−−−

A56

Vii'

A103

303

Vi'

A156

906

V120i'

143

12

11

===

===

==−

=


Ω4

Ω2Ω6

Ω3

''i1

''i2

''i3

''i4

V3

short circuit forvoltage source 12 A

V 4

* Activate the independent current source only

(1)0V3V6-0)V(V3V2V

02

VV3

V6V

0i"i"i"

43

333

4333

321

KK=+=−−−−

=−

−−−

=−−KCL at V3:


0124

V2

VV012i"i"

443

43

=−−−

=−−

(2)48V3V248VV2V2

43

443

KK=−=−−

KCL at V4:

V24VV12V

4

3

−=−=


A1-65i"i'iA1165i"i'iA6410i"i'iA17215i"i'i

444

333

222

111

=−=+==+=+=

=−=+==+=+=

A6424

4Vi"

A62

24122

VVi"

A4312

3V

i"

A26

126V

i"

44

433

32

31

−=−

==

=+−

=−

=

−=−

==

==−

=


Use super position to find V0 ?Activate voltage source only:

Example :

Ω01Ω02 A510 V +

-

Ω5

Φi

ΦV0.4

ΦV

+

-

+

-oV

Φi2

Ω01Ω0210 V +

-

Ω5

Φi'

ΦV'0.4

ΦV'

+

-

+

-oV'

Φi'2

X


0V'V'4V'V'4)V'(0.410V'

ΦΦΦ

ΦΦΦ

=⇒===

V8V'252010V'

0

0

=

=

Dependent current source is open

Ω5

+-

oV'

+

-

Ω2010 V


Activate independent current source only:

Ω01Ω02

Ω5

Φi''

ΦV''0.4

ΦV''

+

-

+

-oV''

Φi''2

y

5 A

Z

KCL at node (y):

(1)0V"8V"5-0V"8V"V"4-

0V"0.420V"

5V-"

Φ0

Φ00

Φ00

KK=+=+−

=+−


V8-168V"V'V

V16580-V"

58V"

V10V"5V" 0.5

0V"0.410V"5

000

Φ0

ΦΦ

ΦΦ

=−=+=

−===∴

−=⇒−=

=++


Example:

Consider the independent source only

Use superposition to find V ?

100 V+-

4 A

Ω12

Ω2Ω5

Ω10+

-V


V59.23V'22141

101

51V'

014V'

10V'

5V'22

014V'

10V'

5V'100

0iii 321

=⇒=

++

=−−−

=−−−

=−−

Apply KCL at node (x) :

100 V+- Ω12

Ω2Ω5

Ω10+

-'V

i3i1

i2

X


Consider the independent source only.

4 A

Ω12

Ω2Ω5

Ω10+

-''V

4 A

Ω2Ω5

Ω10

+- ''V

Ω12


Current divider

V509.2359.23V"V'V

V9.233

10iV"

A2.7692

31012

12A4i

x

x

=−=+=

−=

−=

=

++=

4 A

Ω2Ω310

Ω12

+- ''V

ix

iy


2. Source Transformation:A transformation that allow a voltage source in series

with a resistor to be replaced by a current source in parallel with the same resistor or vice versaHow?

We need to find Is and Vs such that VL and IL is the same in both circuits

VL

R

RL

+

- Vs

IL

+-

a

bKCT 1

+

-VL

ILa

b

IS

KCT 2


sL

L IRR

RI+

=

ss IRV =

In KCT 2,

For IL to be the same , we need

In KCT 1 ,

RRV

IL

sL +

=


R

VS

+-

a

b

RV

I ss =

Where

ss IRV = or

a

b

IS R


Example :Using source transformation, find the power associated with the 6 V source.

1. Consider the 40 V source in series with (5Ω)

+-

40 V+-

Ω4 Ω5

Ω02Ω03

Ω6

6 V

Ω01

+-

Ω4

Ω5Ω02Ω03

Ω6

6 V

Ω01

A8540

=


2. Take (5// 20 Ω)

3. Consider 8A in parallel with (4Ω)

+-

Ω4

Ω402//5

=Ω03

Ω6

6 V

Ω01

A8

+-

Ω4

Ω03

Ω6

6 V

Ω01

A4

+-

(8 A)(4)=32 V


4. Take (4+6+10) in series

5. Consider 32 V in series with (20Ω)

+-

Ω4

Ω036 V

Ω 20

+- 32 V

+-

Ω4

Ω02Ω036 V

A6.12032

=


6. Take (30//20 Ω)

7. Consider 1.6A in parallel with (15 Ω)

+-

Ω4

Ω1220//03

=6 V

A6.1

+-

Ω4

6 V

Ω 12

+- 1.6 A (12)

=19.2 V

i


)(absorbingW4.95P

(0.825)6ivPA0.825124

619.2i

6v

6V

=

==⇒=+

−=

Example :Use source transformation to find V0

+-

Ω52

Ω001

Ω5

250 VA8

+

-Vo Ω15


2. Consider (250 V) in series with (25 Ω)

+-

Ω52

Ω66.16

250 VA8

+

-Vo

1. Take (5//15)//100 = 6.66 Ω

Ω52 Ω66.16A8+

-Vo

A1025

250=


V20Ω)(10A)(2RiV0 ===

3. Find equivalent

A2810i =−=

Ω1025//16.66R

==

+

-Vo


Example:Use source transformation to find V0

+-

60 V

Ω8

Ω5

Ω6.1

Ω02

Ω6120 V

+-

+

-

Vo36 A

Ω8

Ω5

Ω6.1

Ω02 Ω6

+

-

Vo36 AA12

560

=A620

120=


( ) ( ) V48Ω8A6RiV

A6(30)81.62.4

2.4iRRR

Ri

320

321

12

===

=++

=++

=

Ω8

Ω6.1

+

-

VoR1

i = 36 + 6 - 12 = 30 A

Ω2.46//5//20

=

R2

R3

i1 i2


Example : Use source transformation to find V0?

Consider (10V) in series with (1Ω)

+

Vo

--

2 A+

-

Vs10V 1 ohm

1 ohm1 ohm

1 ohm

1 ohm

1 ohm10 A 2 A1 ohm

1 ohm 1 ohm

1 ohm

+

Vo

--


Take (1//1)=0.5

Consider (10A) in parallel with (0.5 Ω)

1 ohm

1 ohm1 ohm

1/2 ohm 2 A10 A

+

Vo

--

+

-5 V

1/2 ohm

2 A

1 ohm 1 ohm

1 ohm

+

Vo

--


Take 0.5Ω in series with 1 Ω

Consider 5V in series with 1.5 Ω

1 ohm

1 ohm

2 A

1.5 ohm

+

-5 V

+

Vo

--

+

Vo

--

1.5 ohm10/3 A 2 A

1 ohm

1 ohm


Add the current sources

7. Take (4/3A) in parallel with (3/2 Ω)

( ) V4/725.111

1V0 =++

=

4/3 A 1.5 ohm

1 ohm

1 ohm

+

Vo

--

+

-2 V

1.5 ohm

1 ohm

1 ohm

+

Vo

--


Thevenin and Norton Theorems

Thevenin Theorem:A portion of the circuit at a pair of nodes can be replaced

by a voltage source Voc in series with a resistor RTH, where Vocis the open circuit voltage and RTH is the Thevenin’s equivalent resistance obtained by considering the open circuit with all independent sources made zero

RLcircuit

a

b

+- RL

a

b

VOC

RTH


Norton Theorem :A portion of the circuit at pair of nodes can be replaced by a current source Isc in a parallel with a resistor RTH. Isc is the short circuit current at the terminals, and RTH is the Thevenin’s equivalent resistance

RLcircuit

a

b

RL

a

b

ISC

RTH

Here we will consider ( 3 ) cases :1. Circuit containing only independent sources.2. Circuit containing only dependent sources.3. Circuit containing both independent and dependent sources.


Case (1): Circuit containing only independent sources:• Procedure of Thevenin’s Theorm:a. Find the open circuit voltage at the terminals , Voc.b. Find the Thevenin’s equivalent resistance, RTH at the

terminals when all independent sources are zero:

Replacing independent voltage sources by short circuitReplacing independent current sources by open circuit

c. Reconnect the load to the Thevenin equivalent circuit

• Procedure of Norton’s Theorm:a. Find the short circuit current at the terminals, Isc.b. Find Thevenin’s equivalent resistance, RTH (as before).c. Reconnect the load to Norton’s equivalent circuit.


Example :Use Thevenin’s and Norton Theorms to find V0

+- RL

VOC

RTH

RL

ISC

RTH

Using Thevenin Theorm:

+ - +-Ωk2

6 V 12 V

Ωk2Ωk4

+

-

Vo


V4k)(4iVAm1k4k2

V6i 14kΩ1 −==⇒=+

=

V8V4V12Voc =−=

First find VOC:

+ - +-

Ωk2

6 V 12 V

Ωk4

+

-

Voc

+

-

i1


Second, find RTH

RTH = 2k//4k = 4/3 k Ω

Thevenin equivalent circuit is

( )

V4.8V

V8k3

10k2

VRk2Ωk2V

0

ocTH

0

=

=

+=

Ωk2 Ωk4RTH

+-

VOC

RTH

+

-Vo Ωk2


+ - +-

Ωk2

6 V 12 V

Ωk4ISC

+ -

Ωk2

6 V

Ωk4 +-

12 V

i1i

i2

+

-V 2 k

-12 V

+

X

Using Norton Theorm

First find Isc


Am3k4V12i1 ==

Am3k26

k2V

i

V6V0V612

2k2

2k2k

−=−

==

−=⇒=+−

KVL around outer loop:

KCL at x :

Am6IAm6i0im3m30iii

sc

21

=⇒=⇒=−+=−−


RTH is the same as before:

Am2.4m)(6k2k3

4k3

4)(I

k2RR

I scTH

TH0 =

+=

+=

V4.8k)(2m)(2.4k)(2IV 00 ===

2 k

ISC

RTH

Io +

-Vo


Example :Use Thevenin and Norton to find V0

Using Thevenin Theorm:

+- Ω4

Ω8Ω5

Ω20

+

-Vo

Ω72

Ω12


KVL around the upper loop :

1. Find Voc :

+-

Ω8Ω5

Ω20

Ω72

Ω12

i1

a

b

(1)0i5i250)i(i5i8i12

21

2111

KK=−=−++


A3iA,0.6i(2)72i25i5

72i20)i(i5

21

21

212

===+−

=+−KK

V64.8V(3)20(0.6)8

i20i8V

oc

21oc

=+=

+=

KCL around lower loop :


2. Find RTH

RTH = (8+4) // 12 = 12 // 12 = 6Ω

Ω8Ω5

Ω20

Ω12

a

b

Ω8( )Ω4

Ω//205=

Ω12

ab


V25.92V

(64.8)64

4

VR44V

o

ocTH

o

=+

=

+=

3. Reconnect the load :

+-

VOC

RTH

+

-Vo Ωk44Ω


Using Norton Theorm:

1. Find ISC :

KVL around upper loop :

(1)0i8i5i250)i(i5)i(i8i12

321

21311

KK=−−=−+−+

+-

Ω8Ω5

Ω20

Ω72

Ω12

i1

i3

i2


KVL around lower loop :

KVL around right loop :

(2)72i20i25i572)i(i20)i(i5

321

3212

KK=−+−=−+−

(3)0i28i20i80)i(i20)i(i8

321

2313

KK=+−−=−+−

A10.8I

A10.8i,A12.72i,A6i

SC

321

=⇒

===


2. Find RTH

From before , RTH = 6 Ω

3. Reconnect the load

( ) V25.9210.846

64V

I4R

RΩ)(4

iΩ)(4V

o

SCTH

TH

2o

=

+

=

+

=

=

a

b

ISC

RTH

i1 i2

Ω4

+

-Vo


Case(2) : Circuits containing only dependent sourcesHere there is NO energy source in the circuit.

VOC is always zero and ISC is always zeroSo we can only find RTH

Procedure for finding RTH

1. Connect an independent voltage ( or current) source at the terminals ,Vx (or Ix)

2. Find the corresponding current ( or voltage) at the terminal , Io ( or Vo)

3. Find RTH = Vx /Io or RTH = Vo/Ix a

b

RTH


Example:Find the Thevenin equivalent circuit

1. Apply voltage source at the terminals (Vx=1V)

Ωk3

2000 Ix

Ωk2

Ωk4

Ix

a

b

Ωk3

2000 Ix

Ωk2

Ωk4

Ix

+-

Vx = 1Vi1 i2

V1


0IIk3

4k3

1I2I

0I3000

I400012000

I4000I2000Ik)(4Vwhere

XXXX

XXXX

X1

=−−+−

=−−

+−

=

KCL at node V1 :

0Ik3V1

k2VI2000

0Iii

X11X

X21

=−−

+−

=−+

Am0.1I3000

1342I

X

X

=

=

+


( ) ( ) ( )k3

Am0.1k41k3

Ik4Vk3VV

i

XX

1X2

−=

−=

−=

Ωk5Am0.2

V1iV

R

Am0.2i

2

XTH

2

===

=

a

b

RTH


Case (3) : Circuits containing both independent and dependent sources

Procedure of Thevenin or Norton Theorms:a. Find the open circuit voltage and the terminals ,VOCb. Find the short circuit current at the terminals, ISC .c. Compute RTH = VOC/ISC

Note :RTH can not be found as in the case of only independent sources

d. Construct the Thevenin or Norton circuits a

b

ISC

RTH

Norton circuitThevenin circuit

a

b

RTH

+-Voc


Example :Find the Thevenin equivalent circuit with respect to the terminals a, b

1. Find VOC :

+

-

Voc

Ω20

Ω80Ω60

4 A Ω40

160 ix

ix

a

b

+

-

Voc

Ω20

Ω80

Ω60

Ω40

160 ix

ix+-

i240 V i1

Z


KVL around the lift loop :

(1)240i200i800i40i160i80240

x

xx

KK=+=+++−

(2)0ii2i40i80

x1

x1

KK=−=

KVL around right loop :

KCL at Z:(3)0iii x1 KK=−−

A0.75iA0.375i

A1.125i

x

1

==

=


V30VΩ)(40(0.75A)

Ω)(40iV

OC

xOC

===∴

2. Find ISC :

Since we have short circuit , 80 // 40 // 0 = 0

Ω20

Ω80Ω60

4 A Ω40

160 ix

ix

V


Current divider

( ) A342060

60ISC =+

=

Ω10A3V30

IV

RSC

OCTH ===

Ω20

Ω604 A ISC

3. Find RTH

ix=0160ix source is zero

a

b

RTH

+-Voc

4.

V


Example :Use Thevenin theorem to find the Thevenin equivalent circuit with respect to a, b

1. Find VOC

KCL at node z :

+- Ω1

Ω5V40

ix

2 ix

i1 a

b

Z

8 A

(1)8ii30i8ii2

1x

1xx

KK−=−=−++


(2)40ii50i1i540

1x

1x

KK=+=++−

KVL around outer loop

V20i1VA20i,A4i

1OC

1x

==⇒==⇒

Find ISC :

+- Ω1

Ω5V40

ix

2 ix

a

b

8 AISC


KVL around outer loop :A8i0i540 xx =⇒=+−

( ) ( ) A32883II8i3

I8ii2

SC

SCx

SCxx

=+=⇒=+

=++

Ω0.6253220

IV

RSC

OCTH ===

KCL at z :

3. Find RTH :+

-

Ω5V40

ix

2 ix

8 AISC

Z

a

b

RTH

+-Voc

Thevenin equivalent circuit is


4. Maximum Power Transfer• A technique in which the load is selected to maximize the power transfer.• This technique is based on the Thevenin equivalent circuit.

L

2

LTH

OC

L2

LL

RRR

V

RiivP

+

=

==

RL

RTH

+

-Voc

i +VL--


We wish to select RL to maximize PL:

( ) ( )( )

[ ]

THL

LTH

LLTH

4LTH

LLTHLTH2oc

4LTH

LTH2

OCL2

OC2

LTH

L

L

L

L

RR0RR

0R2-RR

0)R(R

R2-)R(R)R(RV

0RR

)R2(RVRV)R(RdRdP

0dRdPTake

==−⇒

=+⇒

=+

++

=+

+−+=

=

If RL = RTH , what is the maximum Power Transfer?


( ) ( )

TH

2OC

maxL

TH

2OC

2TH

TH2

OC

TH

2

TH

OC

L2

maxL

R4VP

R4V

R4RV

RR2

V

RiP

=

==

=

=RL

RTH

+

-Voc i


Example:

•Find RL for maximum Power Transfer ?

•Find the maximum Power transfer to RL ?

Ωk4Am3

+-

10 V+ -

Ωk01

Ωk02

Ωk5.2Ωk8

RL

10 V


Ωk4Am3

+-

10 V+ -

Ωk01

Ωk02

Ωk5.2Ωk8

10 V

i1 i4

i3

i2 +

-

Voc

V1 V2

Let’s find Thevenin equivalent circuit .

KCL at node V1 :

0Ωk8VV

Ωk4VAm3

0iiAm3

211

21

=−

−−

=−−


(1)m3Vm0.125Vm0.375

m3Vk8

1k8

1k4

1V

21

21

KK=−

=

−

+

KCL at node V2:

0k12.5

Vk2010V

k8VV

0iii

2221

432

=−−

−−

=−−

(2)m0.5Vm0.255Vm0.125

m0.5Vk12.5

1k20

1k8

1k8

1V

21

21

KK−=−

−=

++−

V7.03VV10.34V

2

1

==


( )

V4.375V

7.0312.51010

k12.5V2k1010

ik1010V

OC

4OC

−=

+−=

+−=

+−=

To find RTH :Ωk8 Ωk5.2

Ωk4 Ωk02 Ωk01

RTH


( )[ ] ( )[ ]( )

Ωk5Rk10//k10

k10//k2.5k7.5k10//k2.5k20//k12

k10//k2.5k20//k4k8R

TH

TH

==

+=+=

++=

( )

Wm0.957Pk)(54

4.375R4

VP

maxL

2

TH

2OC

maxL

=

−==

RTH

+-Voc RL = RTH


Example :

1. Find RL for maximum Power Transfer?

2. Find max. power transfer to RL ?

First , find Thevenin equivalent:

RL3k ohm

1k ohm

4m A

+-

2000 Ix

2k ohm

4k ohm

Ix


+

Voc

--

Using source transformation

+

Voc

--

4k ohm

2k ohm

+-

2000 Ix'

4m A

1k ohm

3k ohm Ix’

+

-16V

4k ohm

+-

2000 Ix'

2k ohm

4k ohm

Ix’


Now, find Isc:

KVL around the loop:-16 + 4k Ix’ - 2k Ix’ + 2k Ix’=0Ix’= 4mA.

Voc= (2kΩ) Ix’= 8V.

4k ohm

2k ohm+-

2000 Ix''4k ohm

+

-16V Ix’’ Isc

I1

V1


KCL at V1:I1 - Ix’’ – Isc = 0

04

12

1416

=−−kV

kV

k

04

12

14

)''21(16=−−

−−kV

kV

kkIxV

And

Where V1=2k Ix’’

Hence,Or V1=5.333V

mAkVIsc 333.14

1==


( )( )

Wm38P

k2464

k648

R4V

P

Ωk6Am1.333

V8IV

R

L(max)

2

TH

2OC

L(max)

SC

OCTH

=

===

===


Capacitors and Inductors Chapter (5)

Capacitors :A circuit element that is composed of two conducting plates or surfaces separated by a dielectric (non conducting) materials

A+

-

+-C

circuit symbol


Let A : surface area of each plated : distance between the two plates

Capacitance

Capacitance

As Area

As distance

dAC ∝∴

dAε

C 0=

≡0ε

It is found that

Where :

Permittivity of free space


• If a voltage source (v) is connected to the capacitor , +ve charge will be transferred to one plate while –ve charge will be transferred to the other plate.

mF10*8.85ε 12

0−≡

Let the charge stored at the capacitor q≡

qv ∝∴If v q,

vcq =It has been found that


vqc =⇒

C is the capacitance

Current in capacitor :

We know that

( )

td(t)dvC(t)i

)(vctd

d(t)i

tdq(t)di(t)

cc

cc

=⇒

=∴

=

t


(t)dvCdt(t)itd(t)dv

C(t)i

cc

cc

=∴

=Q

Voltage of capacitors

(t)dtiC1(t)vd cc =⇒

dτ)(τiC1)(tv(t)v

tτ

tτc0cc

0

∫=

=

+=

Where t0 : initial time


The capacitor is a passive element and follows the passive sign convention

Capacitors only store and releaseELECTROSTATIC energy. They do not “create”

Linear capacitor circuit representation

)()( tdtdvCti =

Note:


Power of the capacitors :

dt(t)dv(t)vC

td(t)dvC(t)v(t)P

(t)i(t)v(t)P

cc

ccc

ccc

==

=

( )

+= ∫

=

=

tτ

tτc0ccc

0

dττiC1)(tv(t)i(t)P

or


Energy of capacitor

( )

( ) ( )dτ

τdτdv

τvC

dττP(t)w

tτ

τ

cc

tτ

τcc

∫

∫=

−∞=

=

−∞=

=

=

( ) ( )

( )

( ) ( )∞−−=

=

=

∞

−∞∫

2c

2c

(t)v

)(-v

2c

c

(t)v

)(vc

vC21tvC

21

τvC21

τdvτvC

| c

c

c

c


Assuming

( ) 0vc =∞−

(t)vC21(t)w 2

cc =∴

( ) (t)q2C1tw

C(t)qC

21(t)w

C(t)q(t)v

2c

2

2

c

c

=

=∴

=


Example:The following voltage is imposed across the terminals of a 0.5 µF capacitor.

( )

∞<≤

≤≤≤

=−− t1V,e4

1t0,Vt40t,V0

(t)v1t

c

1

4

2 3

v (t)

t


Find the following:1. ic(t)2. Pc(t)3. wc(t)

( ) ( )

( )[ ] ( )( )

∞⟨⟨−=−=

⟨⟨==

⟨

=∴

−−−−−− t1Aµe2e1C4e4dtdC

1t0Aµ2C4t4dtdC

0t0

ti

1)(t1)(t1t

c

dt(t)dv

C(t)i cc =(1)


(2)

( )( ) ( )

( )( )( )( ) ( )( )( ) ( ) ( )

∞<<−=−

<≤=

≤=

=

=

−−−−−− t1µWe8e4e4Fµ0.51t0µWt84t4Fµ0.5

0tW0dt

tdv0Fµ0.5

dt(t)dv(t)vC(t)P

1t21t1t

c

ccc

Aµ2

Aµ2-

i C (t)

t


Aµ8

Aµ8-

P C

t

supplied power

absorbed power

1

( )( )

( ) ( )[ ] ( )

∞<≤=

≤≤=

≤

=

=

−−−− t1µJ4e4eFµ0.521

1t0µJt4t4Fµ0.521

0t0

(t)w

(t)vC21(t)w

1t221t

22c

2cc

(3)

Charging Power

Discharging Power


)1(24 −− te

wC (t)

t

4 t 2

1

Jµ4

Example :

The voltage at the terminals of a 0.5 µF capacitor is

≥

≤=

− 0tVt)sin(40000e1000t0

(t)v20000tc


Find:1. i(0)2. Power delivered to the capacitors at t = Π/80 m S.3. Energy stored in the capacitor at t = Π/80 m S

dt(t)dv

C(t)i c=


( )[ ][ ])e10*2(t)sin(40,000t)(40,000)(40,000cose100C

t40,000sine100dtdC

20,000t620,000t

t20,000

−−

−

−+=

=

[ ]A2(0)i

0(40,000)(1)(1)10010*0.5(0)i

c

6c

=+= −

m

80ΠPC(2) Find ?

( ) ( )( )[ ])e10*2(t)(40,000sin(40,000)t)(40,000cose100*

t40,000sine100Fµ0.5

0t,dt

(t)dv(t)vC(t)P

20,000t620,000t

20,000t

ccc

−−

−

−+

=

≥=


ng)(dischargiW20.79m)80Π(PC −=

[ ]

Jµ519.2m)80Π(W

t)(40,000sin100eC21(t)vC

21(t)W

C

220,000t2CC

=

== −

m)80Π(WC(3) Find ?


Inductors :Inductors are circuit elements that consist of a conducting wire in the shape of a coil

Circuit representation for an inductor


• If a current is flowing in the inductor, it produce a magnetic field ,Φ.

•The direction of (Φ) depends on the right-hand rule.•As the current increases or decreases, the magnetic field spreads or collapse •The change in magnetic field induces a voltage across the inductor.

dt(t)diL(t)V

dt(t)d(t)V

LL

L

=∴

Φ=

Li(t))( =Φ t

Where L is the inductance and measured in Henry [H]


Current in inductors :

dt(t)vL1(t)di

dt(t)diL(t)v

LL

LL

=

=

( )dττvL1)(ti(t)i

tτ

tτ0LL

0

∫=

=

+=

Integrate both sides as before


=

=

dt(t)di

L(t)i

(t)i(t)v(t)P

LL

LLL

Power in inductor :

( )

+=

=

∫=

=

tτ

tτ0LLL

LLL

0

dττvL1)(ti(t)v(t)P

dt(t)di(t)iL(t)P


Energy in inductors:

( )

( ) ( )

(t)iL21(t)w

beforeAs

dτdττdi

τiL

dττP(t)w

2LL

Ltτ

τL

tτ

τLL

=

=

=

∫

∫=

−∞=

=

−∞=


Example :The current flow through an 100 m H inductor

≥

≤=

− 0tAet100tA0

(t)i t5L

( ) ( ) ( )[ ]sec0.2t

010t50e010e5et10

0dt

(t)dilet

max

t5

t5t5

L

==+−∴

=+−

=

−

−−

Find :(1) Maximum value of current. (2) vL(t) , (3) PL(t) , (4) wL(t)

First, find t max

t

iL(max)

iL(t)

t max


( ) ( ) ( )

A0.736e2

e0.2100.2ii1

0.25LLmax

==

==−

−

( ) [ ]( ) ( )

( )t51e10t50e0.1

et10dtdH0.1

dt(t)diL(t)v(2)

t5

t5

t5

LL

−=

+−=

=

=

−

−

−

>−

<=

− 0t,t)5(1e0t,0

(t)v t5L


( ) ( ) ( )[ ] 0t,10t50eet100.1(t)P

0t,0(t)Pdt

(t)di(t)iL(t)P

?(t)P(3)

t5t5L

L

LLL

L

≥+−=

≤=

=

−−

( )

≥−

≤=

−=

−

−

0t,t)5(1et100t,0

(t)P

t5010et(t)P

10tL

10tL


( ) ( )

≥=

≤=

=

−− 0t,et5et100.121

0t,0

(t)iL21(t)w

?(t)w(4)

t1022t5

2LL

L


w (t)

P (t)

i (t)

v (t)

Inductor Capacitor

( )∫=

=

+tτ

tτc0C

0

d ττic1)(tv

dt(t)diL L

dt(t)vd

c C ( ) dττvL1)(ti

tτ

tτL0L

0

∫=

=

+

dt(t)dv

(t)vc CC dt

(t)di(t)iL LL

(t)vc21 2

C (t)iL21 2

L

Summary of results :


Notes on capacitor :

1. If vC(t) = constant , iC (t) = 0

Capacitor will be open circuit

2. vC(t) cannot change instantaneously ( no sudden change)

( ) 0dττibecausetτ

tτC

0

=∫=

=

3. Capacitors can store energy

(t)vc21(t)w 2

CC =


Notes on inductors :

1. If iL(t) = constant , vL(t) = 0

Inductor will be short circuit

2. iL(t) cannot change instantaneously ( no sudden change)

( ) 0dττvbecausetτ

tτL

0

=∫=

=

3. Inductors can store energy

(t)iL21(t)w 2

LL =


C1

+ -V1 (t)

CNC3C2

VN (t)V3 (t)V2 (t)

V (t)

Capacitors and Inductors combinations :

1. Series capacitors :

+ -

CS

V (t)


For each capacitor ,

(t)v(t)v(t)v(t)v(t)v N321 ++++= L

( )

N21

tτ

t0τk0kk

vvvv(t)

dττic1)(tv(t)v

+++=

+= ∫=

=

L

( ) dττic1) (tvv(t)

tτ

tτ

N

1k k

N

1k0k

0

∫∑∑=

===

+

=

The equivalent capacitance CS is

N21

N

1k kS C1

C1

C1

C1

C1

+++== ∑=

L


dt(t)dvC

dt(t)dvC

dt(t)dvC

(t)i(t)i(t)i(t)i

N21

N21

+++=

+++=

L

L

dt(t)dvCC

dt(t)dv(t)i P

N

1kk =

= ∑

=

Parallel capacitors :

N321P CCCCC ++++= L

The equivalent capacitance ,CP

C1

+

-

CNC2V (t)i1(t) i2(t) iN(t)

i(t)

CP


Series Inductors :

(t)v(t)v(t)v(t)v N21 +++= L

dt(t)diL

dt(t)diL

dt(t)diL N21 +++= L

dt(t)diL

dt(t)diL(t)v S

N

1kk =

= ∑

=

N321S LLLLL ++++= L

L1

+

-

V1 (t)

LNL2

VN (t)V2 (t)

V (t)


Parallel Inductors :

(t)i(t)i(t)i(t)i N21 +++= L

( ) dττvL1)(ti(t)iwhere

tτ

tτk0kk

0

∫=

=

+=

L1

+

-LNL2

V (t)

i2(t) iN(t)

i(t)

i1(t)LP


N21P L1

L1

L1

L1

+++= L

The equivalent inductance , LP

( )

( ) dττvL1)i(t(t)i

dττvL1)(ti(t)i

tτ

tτP0

tτ

tτ

N

1k k

N

1k0k

0

0

∫

∫∑∑=

=

=

===

+=

+

=


a

b

L1 = 2 H

L7 = 8 H

L6 = 30 H

L5 = 10.4 H

L4 = 5 H

L3= 6 H

L2 = 15 HL8 = 16 H

Example:

Find the equivalent inductance with respect to the terminals a ,b?

• L89=L8 in parallel with L9

H9.6L9L8

L9L8L89 =+

=∴

L9=24H


• L589=L5 in series with L89


H209.610.4LLL 895589 =+=+=

H12LL

LLL

5896

58966589 =

+=



H20128LLL 6589776589 =+=+=

H4LL

LLL

765894

765894476589 =

+=


H1046LLL 34658933476589 =+=+=

H6LL

LLL

34765892

3476589223476589 =

+=



• Lab=L1 in series with L23476589

H862LLL 23465891ab =+=+=


a

b

C1 = 5 µF

C3 = 10 µF

C2 = 4 µF C6 = 48 µF

C5 = 3 µF

C4 = 30 µF

C7 = 16 µF

Example :Find the equivalent capacitance at the terminals a and b ?

• C67 = C6 in series with C7

Fµ12CC

CCC

76

7667 =

+=


• C567 = C5 in parallel with C67

• C4567 = C4 in series with C567

Fµ10CC

CCC

5674

56744567 =

+=

Fµ15Fµ12)(3CCC 675567 =+=+=

• C34567 = C3 in parallel with C4567

• Cab = C1 in series with C34567 in series with C2

Fµ20F10)µ(10CCC 4567334567 =+=+=

Fµ2C4µ1

µ201

µ51

C1

C1

C1

C1

ab

2345671ab

=⇒

++=++=


L1

L2

L6

L5

L4

L3

a

b

in all inductorsare 4 mH

Example :Find the equivalent inductance at a,b

• L36 = L3 in parallel with L6

Hm2LL

LLL

63

6336 =

+=



Hm2LL

LLL42

4224 =

+=

a

b

L5

L4

L1

L2

L36

X y

y

a

b

L5

L4

L1

L36

X yL24



Hm2.4LL

LLL

5124

51241245 =

+=

• L124536 = L1245 in series with L36

Hm4.4LH m2)(2.4LLL

124536

361245124536

=+=+=

• L124 = L1 in series with L24

Hm6Hm2)(4LLL 241124 =+=+=


Example :Find the equivalent inductance at a , b if all L are 6 m H

a

b

L5L4

L2

L3

L1

L6L12 = L1 in parallel with L2

Hm3LL

LLL

21

2112 =

+=


L123 = L12 in parallel with L3

( ) ( )( ) ( ) Hm2Hm

6363

LLLLL

312

312123 =

+=

+=

a

b

L5L4

L12

L3

L6

a

a

X

y


L1234 = L123 in series with L4

Hm862LLL 41231234 =+=+=

b

L5

L4

L12

L6

aa

X

y

L123


L12345 = L1234 in parallel with L5

Lab = L6 in series with L12345

( ) ( )( ) ( ) Hm42.3Hm

6868

L5LLLL

1234

5123412345 =

+=

+=

Hm9.429Hm3.4296LLL 123456ab

=+=+=


Example:Find the equivalent capacitance w.r.t. a, b if all C’s are 4 µ F

C45 = C4 in parallel with C5

= C4 + C5 = 8 µ F

a

b

C5

aa

b b

bXC4

C3C2

C1


a

b

C45

X b

b

a a

C3C1

C2


= C2 + C3 = 8 µ F


C123 = C1 in series with C23

( ) ( ) Fµ38Fµ

1232Fµ

8484

CCCCC

231

231123 ==

+=

+=

a

b

C45

C1

C2

X

C23


Cab = C123 in parallel with C45

Fµ3

32

Fµ3248Fµ8Fµ

38CCC 45123ab

=

+=+=+=


a

b

C1 = 3 µF

C3 = 4 µFC2 = 2 µF

C4 = 12 µF

C6 = 2 µF

C5 = 3 µF

X X

Z

Example :Find the equivalent capacitance w.r.t a, b ?


= C2 + C3 = 6 µ F



= C235 + C6 = (2+2) µ F= 4 µ F

Cab = C1 in series with C2356 in series with C4

Fµ1.5C

µ121

µ41

µ31

C1

C1

C1

C1

ab

423561ab

=

++=++=

C235 = C23 in series with C5

( ) ( ) Fµ2Fµ3636

CCCCC

523

523235 =

+=

+=


Chapter (6)First Order Transient Circuits

• This chapter deals with the transient response of first order circuit .

• first order circuit are those circuits whose response can be expressed by a first order differential equations.• Example of such circuits include the RL and RC circuitsRL circuit : circuit that contains an indicator and a resistor.RC circuit : circuit that contains a capacitor and resistor

For example: RC circuitConsider the following circuit . +

- iVS

C

R


S

S

Vi(t)R(t)v0i(t)R(t)vV

c

C

=+∴=++−

Sc

c

c

Vdt

(t)dvCR(t)v

dt(t)dv

Ci(t)since

=+⇒

=

Scc V

CR1(t)v

CR1

dt(t)dv

=+∴

b(t)vadt

(t)dv

VCR

1b,CR

1alet

cc

S

=+⇒

==

This is the general form of a linearfirst order differential equation


This is a first order linear ordinary non-homogenous differential equation describing the response of the capacitor voltage.

•It is first order because the highest degree of derivative is one.

• it is called linear because the differential equation is linearfunction of (dvc /dt) and vc(t)

Example of non-linear differential equation :

b(t)vdt

(t)dv(t)v 2

cc

c =+

• it is called ordinary because it deals only with ordinary derivatives ( not partial derivative )


• It is called non-homogenous because b ≠ 0.

•Example of first order linear ordinary homogenous differential equation is

0(t)vadt

(t)dvc

c =+

Example :R L circuit

+- iL(t)

R

VSL

KVL around the loop


LV

(t)iLR

dt(t)di

0dt

(t)diL(t)iRV

SL

L

LLS

=+

=++−

b(t)iadt

(t)diLV

b,LRalet

LL

S

=+

==

This is a first order linear ordinary non-homogenous differential equation

To find the response of the Vc(t) in RC circuit or iL(t) in the RL circuit we need to solve these differentials


Solution of the first order differential equation :

Consider the first order linear ordinary non-homogenous differential equation :

We want to find X(t) that satisfies (*)

(*)bx(t)adt

dx(t)LL=+

Theorem : ( in differential equation)If x (t) = xP (t) is any solution of equation (*) and x (t) = xc (t) is any solution of the homogenous differential equation

(**)0(t)xadt

(t)dxc

c LL=+


then(t)x(t)xx(t) cP +=

Where :

xP(t) = particular solution ( forced solution)

xC(t) = complementary solution ( natural solution)

• Hence we need to solve 2 differential equations

(*)b(t)xadt

(t)dxP

P LL=+

What is the function xP(t) that if its differential is summed to a*xP(t) will give a constant (b)• The solution xp(t) must be constant

xP(t) = k1


Use xP(t) in the non-homogenous differential equation

abk(t)x

abk

bka)(kdtd

1P

1

11

==⇒

=

=+

Particular (forced) response

• Consider the homogenous differential equation:

adt

(t)dx(t)x

1

(*)0(t)xadt

(t)dx

c

c

cc

−=⋅∴

=+ LL


( )[ ]

( )[ ] a(t)xlndtd

dt(t)dx

(t)x1(t)xln

dtdsince

c

c

cc

−=⇒

⋅=

[ ][ ]

Catc

c

c

e(t)x

Cta(t)xln

dta(t)xln

+−=

+−=

−= ∫Take the integral of both sides

at2c

c2

Catc

ek(t)x

eklet

ee(t)x

−

−

=∴

=

=


τt/21 ekkx(t) −+=

Hence : let τ = 1/a ≡ time constant

Time constant : a parameter that determines the rate of decrease ofx(t)

Let’s find the solution of RC & RL circuits :

at21

cP

ekkx(t)

(t)x(t)xx(t)−+=

+=⇒

Hence, a general form of the solution is:


Example :

+- ic(t) RVS

C

S1

CRt

21at

21c

Sc

c

VCR

1CR

Vs

abk

ekkekk(t)v

CRV

(t)vCR

1dt

(t)dv

===

+=+=⇒

=+

−−

Assume vc(0) = v0

To find k2 , we need the initial condition of vc(t)

For example , if we know vc(0) = V0


S02

2S0

CRt

2Sc

VVk(1)kVV

ekV(0)v

−=+=

+=⇒−

CRt

S0S

CRt

21c

e)V(VV

ekk(t)v−

−

−+=

+=

As a special case , let’s consider the natural response

Natural response :Circuit response when no source is affecting the response


Case 1 : Natural RC response

ic(t) R

C

0(t)vCR

1dt

(t)dv

0Rdt

(t)dvC(t)v

0R(t)i(t)v

cc

cc

cc

=+

=+

=+Assume vc(0)=V0

First order linear ordinary homogenous differential equation :CR

t

2c ek(t)v−

=

We can find k2 from initial conditions


CRt

0c

02

2CR

t

20c

0c

eV(t)v

VkkekV(0)v

thenV(0)vAssume

−

−

=

=∴===

=vc(t)

t

V0

RCt

21 ekkx(t)−

+=

Note : the general solution of the forced response is

RCt

0c

0S02

S1S

eV(t)V

V)V(Vkand0Vk0Vsince

−=⇒

=−===⇒=


Case 2 : Natural response of RL circuit

iL(t)

R

L

0(t)iLR

dt(t)di

0dt

(t)diL(t)iR

LL

LL

=+

=+

Assume : iL(0) = i0

KVL :

This is a first order linear ordinary homogenous differential equation

tLR

2L ek(t)i−

=


To find k2, we need initial condition iL(0)

tLR

0L

02

2

(0)LR

20L

ei(t)i

ikkeki(0)i

−

−

=

==== iL(t)

t

i0

Note :• In the forced response , we have

RV

k

ekk(t)i

S1

tLR

21L

=

+=−


Since VS here is 0 then k1 = 0

tLR

0L

S0S

02

ei(t)i

0)V(sinceiRV

ik

−

=⇒

==

−=

Analysis Techniques :

1. The differential equation approach

•Here , a differential equation that describe the behavior of thecircuit is used.•This first order differential equation is expressed in tems of the voltage across the capacitor or current through the inductor.•Then the solution of this differential equation is obtained


Example :

Find iL(t) , t ≥ 0 ?

• First, find the initial condition iL(0- )

At t = 0- , the inductor behaves as a short circuit .

2 H20 A Ω1.0Ω40

iL (t)

Ω2

Ω10


•Hence , we assume that the current through the inductor doesn’t change instantaneously

iL(0 - ) = iL(0) = iL(0+) = 20 A

20 A Ω1.0Ω40

iL (0-)= 20 A

Ω2

Ω10

iL(0 - ) = 20 A


At t = 0 the switch is open

Where Req = (40 // 10) + 2 = 400/5 + 2 =10 Ω

L= 2 H iL (t) Ω10

R eq

=+

-

2 HΩ40

iL (0-)= 20 A

Ω10

Ω2

OR


KVL around the loop :

0(t)i5dt

(t)di

0(t)i10dt

(t)di2

0(t)iRdt

(t)diL

0v(t)v

LL

LL

LeqL

eqL

=+

=+

=+

=+


We know that

5t2L

τt

2at

2L

ek(t)i51

a1τ5awhere

ekek(t)i

−

−−

=∴

==⇒=

==

We can find k2 from the initial condition

iL(0 ) = 20 = k2 e0 = k2

iL(t) = 20 e-5t A

iL(t)

t

20


Example :

Find vc(t) , t ≥ 0 ?

•The switch has been closed for long time .•The capacitor behave as open circuit .

7.5 m AΩk80 Ωk50

Ωk20

Fµ0.4

t = 0

+

-

vc (t)

7.5 m AΩk80 Ωk50

Ωk20

vc (0-)


At t = 0 , the switch is open


0(t)v50dt

(t)dv

Fµ0.4C,0dt

(t)dvCk50(t)v

0(t)ik50(t)v

cc

cc

cc

=+

==

+

=+

( )

( ) V200k150k80Am7.5Ωk50

k70k80k80Am7.5Ωk50)(0v(0)v)(0v ccc

=

=

+

=== +−

Ωk50+

-

vc(t)Fµ0.4

Vc(t) = 200 e-50t V.


Example :

Find vc(t) , t ≥ 0 ?

For t < 0 , the capacitor behave as open circuit .

V40)(0v(0)v)(0v ccc === +−

At t = 0 , the switch is moved

Ωk60

Ωk601

Ωk20

Fµ0.25+

-

vc (t)

+- +

-

40 V75 V

Ωk8 Ωk40t = 0


Ωk601

+

-

vc (t) +-

75 V

Ωk40Ωk8

Ωk601+

-vc (t)

Ωk40Ωk8

1.875 m A

Source transformation


+

-vc (t) +

-60 VFµ0.25

32 k

8 k

+

-vc (t)

Ωk23

Ωk8

1.875 m A



6000-b,100a

6000(t)v100dt

(t)dv

0(t)vdt

(t)dvCk4060

0(t)v(t)ik4060

cc

cc

cc

==

−=+

=++

=++

601006000

abk

ekk(t)v

1

at21c

−=−

==

+= −


Ve10060(t)v

100k40k6040kk(0)v

t100c

22

21c

−+−=

=⇒=+−=+=

vc(t)

t

30

- 60

To find k2 , we use initial condition:

40


Case 2 : Step by step approach

1. Assume the solution is x(t) = k1 + k2 e-t/τ

2. Assume that the circuit is in steady state before the switch moves replace a capacitor by open circuit

replace a inductor by short circuitThen find vc(0

-) or iL(0-)3. The switch is now in the new location :

Replace the capacitor by a voltage source = vc(0-)

Replace the inductor by a current source = iL(0-)And solve for x(0)

4. Assume t = ∞ , find x (t = ∞)replace capacitor by open circuit and inductor by short circuit


5. Find the time constant :

HOW ??- Find the Thevenin equivalent resistance w.r.t the terminals of

the capacitor or inductor.− τ = RTH C or τ = L/RTH

6. Find the constants :k1 = x(∞)k1+k2 = x(0) k2 = x(0)-x(∞)

x(t) = x(∞)+[x(0) - x(∞)] e-t/ τ

x(t) = final value + [ initial value – final value ] e –t/τ


Example :

[ ] τt

000

τt

210

e)(V(0)V)(V

ekk(t)v(1)−

−

∞−+∞=

+=

Find V0 (t) ?

Fµ2

+

-

v0 (t)+-

12 V Ω2

+-

8 V

Ω2

Ω2

Ω1


(1)20i2-i508)i-(i2i312-

21

211

LL==−++

(2) Assume steady state , replace capacitor by open circuit .Mesh

V812412(1)i)(0v

A0i,A4i(2)8i4i2

0)i(i28i2

1c

21

21

122

=+−=+−=

===+−

=−++

−

LL

v0 (0-)12 V

Ω2

+-

8 V

Ω2

Ω2

Ω1

+-


3. The switch is moved now t = 0 ,replace the capacitor by a voltage source = vc(0

-) and solve for V0(0)

V4218

222)(0v(0)V c0 =

=

+

= −

vc (0-)

= 10 V

12 VΩ2

Ω2Ω1

+- +

-

v0 (t)

+

-=8V


4. At t = ∞ replace capacitor by open circuit

V524

5212

122212)(V0 =

=

++

=∞

12 VΩ2

Ω2Ω1

+- v0 (t)

+

-)(vc ∞


5. Find the time constant find the Thevenin equivalent resistance w.r.t x ,y

RTH = 1 // (2+2) = 1 // 4 = 4/5 Ω

12 VΩ2

Ω2Ω1

+- x

y

( )58F2Ω

54CRτ TH =

==


[ ]

t85

0

t85

τt

0000

e54

524(t)v

e5244

524

e)(v(0)v)(v(t)V6.

−

−

−

−=

−+=

∞−+∞=


Example :

[ ] τt

000

τt

210

e)(v)(0v)(v

ekk(t)v:Step(1)−+

−

∞−+∞=

+=

+

-

v0 (t)

24 V

VxΩ4

Ω4

Fµ2+

-

2 Vx

t = 03 A

+-

Step (2) : assume steady-state , replace capacitor by open circuit

Find v0(0-) ?


V60)(0vV60243624(12)3

24v3v24v2)(0v

V12(3)4v

0

xxx0

x

=

=+=+=+=++=

==

−

−

+

-v0 (0

-)

24 V

VxΩ4

Ω4

+

-

2 Vx

3 A

+-


Step (3) : now switch is moved .Find v0(0)

(t)vx(t)henceV60)(0v(0)v)(0v

c

000

==== −+

24 V

VxΩ4

Ω4

+

-

2 Vx+-

( )∞0v

Step (4) : assume t = ∞Find V0(∞) ?

V24)(V24

v24v2)(V0Vx

0

xx0

=∞=

++=∞=


Step (5) : Find the time constantfind the RTH w.r.t the terminals of capacitor

Voc = 24 V

24 V

VxΩ4

+

-

2 Vx+-

+

-

vocΩ4


A428

2V

I

V8V24V3

0V24V2

xsc

x

x

xx

===

−=−=

=++

24 V

VxΩ4

+

-

2 Vx+-

Ω4 I SC

24 V

VxΩ2

+

-

2 Vx+-

I SC

Now, find Isc

sec. 12F)(2Ω)(6CRτ

Ω64

24IVR

TH

SC

OCTH

µµ ===

===


[ ]

[ ]

Ve3624(t)V

e246024(t)V

e)(V(0)V)(V(t)V:Step(6)

12t

0

12t

0

τt

0000

µ

µ

−

−

−

+=

−+=

∞−+∞=

τt

210 ekk(t)i−

+=

Example : Find i0(t) , t > 0 ?

Step (1) :

12 V

Ωk4

Ωk2

Fµ200

t = 0+-

i0(t)

Ωk2

Ωk2

Ωk2


Step 2 : assume steady state ( for t < 0 ) replace capacitor by open circuit .

V4)(0v(0)v)(0v

V4k6

2k)12()(0v

ccc

c

===

==

+−

−

12 V

Ωk2

+-

i0(t)

Ωk2

Ωk2

Ωk2

vC(0-) -+


Step 3 : now switch is moved , replace capacitor by voltage source = vc(0) ,Now find i0(0)

12 V

Ωk2

+-i0(0+)

Ωk2

Ωk2

Ωk2

+ -4 V

Ωk4

i 2


Am2.66k3k1Am8(0)i0 =

=

i0

Ωk2Ωk1

8 m A

i0(0+)

Ωk2Ωk2

6 m A2 m A

Ωk2


Step 4 : assume t = ∞ , find i0(∞) . Steady state Replace capacitor by open circuit

12 V

Ωk2

+-

Ωk2

Ωk2

Ωk2

)(i0 ∞

Ωk4

12 V

+-

Ωk2 Ωk2

)(i0 ∞


Ωk2Ωk2

Ωk2

Ωk2Ωk4

RTH

Am3k4

12)(i0 ==∞

Step 5 : find time constant .First find RTH at terminals of the capacitor

( )

( ) ( ) sec0.6Fµ200Ωk3CRτΩk3

k2k2//k4//k4R

TH

TH

====

+=


Step 6 : find the solution i0(t)

[ ]

Ame0.333(t)i

Ame3)(2.663

e)(i(0)i)(i(t)i

0.6t

0

0.6t

τt

0000

−

−

−

−=

−+=

∞−+∞=


Example:

Ωk21

Fµ50

Ωk4 Ωk8Ωk3

Ωk21

V12 Ωk4+-t = 0

i0 (t)

Find io(t) , using Step by step approach.


Step 1 : assume i0(t) = k1+ k2 e –t/ τ

Step 2 : assume t < 0 ( steady state) Replace capacitor by open circuit and find voc(0

-)As we have done before , vc(0

-) = vc(0)= vc(0+) = - 4 V.

Step 3 : now the switch is moved replace the capacitor by voltage source of value -4 and find i0(0)

Ωk4 Ωk8Ωk21

Ωk4

i0 (0+)

iΩk3

4-

+-+-

Ωk21


( )

Am21

1612

32

k4k12k12

32(0)i

Am32

k64

k12//k4k34i(0)

0−

=

−

=

+

−=

−=

−=

+−

=

Step 4 : assume t = ∞ , steady state replace capacitor by open circuit and find i0(∞)

i0(∞) = 0Ωk12Ωk3 Ωk4

)(i0 ∞


Step 5 : find τ [ ]

( ) ( ) sec0.3µ50k6CRτk6k3k4//k12R

TH

TH

====+=

[ ]

0tAme21(t)i

Ame210

e)(i(0)i)(i(t)i

0.3t

0

0.3t

τt

0000

≥−=

−=

∞−+∞=

−

−

−

Step 6 : Ωk12Ωk3 Ωk4

RTH


Example :

Step 1 : assume i0(t) = k1 + k2 e -t/τ

Step 2 : assume t < 0 ( steady state )Replace inductor by short circuit and find

iL(0-) = i0(0-)

Ωk4 Ωk2

t = 0

i0 (t)

Am10Ωk4

Ωk2Ωk5

Hm10

find i0(t) using step by step approach ?


Am5k2k2

k2Am10)(0i0 =

+

=−

Ωk4 Ωk2i0 (0)

Am10Ωk4

Ωk2

Ωk5

Am10Ωk2

Ωk2Ωk5


Step 3 : now the switch is moved t = 0 replace inductor by a current source of value = 5 mA.Since it is inductor i0(0

-) = i0(0)

Step 4 : assume t = ∞ ( steady state )Replace inductor by short circuit and find i0(∞)

Ωk4 Ωk2i0 (t)

Am10Ωk4

Ωk2Ωk5


Am5k2k2

k2Am10)(i0 =

+

=∞

Step 5 : find τ where τ = L / RTHLet’s find RTH at the terminals of the inductor RTH = (4 k // 4 k) + 2 k

= 4 k Ωτ = L / RTH = 10m/4k = 2.5 µ sec

i0 (t)

Am10Ωk4

Ωk2Ωk5

Ωk4

Ωk2Ωk5

Ωk4 RTH


[ ]

[ ]Am5(t)i

Aem5m5m5

e)(i(0)i)(i(t)i

0

sµ2.5t

τt

0000

=−+=

∞−+∞=−

−

Step 6 : find i0(t)


Example :

Find i0(t) using step by step approach ?

Step 1 : assume i0(t) = k1 + k2 e –t / τ

Step 2 : assume t < 0 ( steady state )Replace inductor by short circuit and find iL (0-)

Ω4

Ω8Ω6

Ω6

Ω21

Ω4

+-

t = 0

24 V

1 H

i L(t)

i 0(t)


iL(0-) = 24 / 6 = 4 Ai0(0

-) = 0 A

Step 3 : switch is moved Replace inductor by current source of value (4 A) and find i0(0)

Ω4

Ω8Ω6

Ω6

Ω21

Ω4

+-

24 Vi L(0-)

i 0(0-)

Ω4

Ω4

Ω6

+-

24 Vi L(0-)

i 0(0-)


Ω4

Ω8Ω6

Ω4

i L(0-)

i 0(0+)

Ω21 4 A

Ω4

Ω8Ω6

Ω4

i 0(0+)

Ω21

+-

48 V

Ω4

Ω16

i 0(0+)

48 V +-i0(0) = - 48 / 20

i0(0) = -2.4 A


Step 4 : assume t = ∞ ( steady state)Replace inductor by short circuit and find i0(∞)

i0(∞) = 0 A

Step 5 : find τ , τ = L / RTH So find RTH across the terminal of the inductor

Ω4

Ω8Ω6

Ω4

Ω21RTH


Step 6 : find i0(t)

[ ]( )

0tfor,0(t)iAe2.4(t)i

0t,Ae02.40

Ae)(i(0)i)(i(t)i

0

t4.80

t4.8

τt

0000

⟨=−=

⟩−−+=

∞−+∞=

−

−

−

i0(t)

t

- 2.4

( )[ ]

sec4.81

RLτ

4.812//46//84R

TH

TH

==

=++=


Example :Solve previous example using differential equation approach

For t < 0 , steady state Replace the inductor by short circuit and find iL(0-) and i0(0

-)

As before iL(0-) = 4 A = iL(0+)i0(0

-) = 0 A

For t > 0 ,Ω4

Ω8Ω6

Ω4

i 0(t)

i L(t)

L = 1 HΩ21


(t)i53(t)i

(t)i2012

k8k12k12(t)i(t)i

L0

LL0

−=

−=

+

−=

Ω4

Ω4

i 0(t)

i L(t)Ω21

So we need to find iL(t) first


KVL :

0(t)i4.8dt

(t)di

0(t)i4.8dt

(t)diL

0(4.8)(t)i(t)V

LL

LL

LL

=+

=+

=+

Ω21Ω8

i 0(t)

i L(t)Ω4.8

i L(t)L


( )

0t,0(t)i0t,e2.4(t)i

e512(t)i

e453(t)i

53(t)i

0

t4.80

t4.80

t4.8L0

⟨=⟩−=

−=

−=

−=⇒

−

−

−

i0(t)

t

- 2.4

0t,e4(t)i

k4(0)i0t,ek(t)i

t4.8L

L

t4.8L

⟩=⇒

==⟩=

−

−


Chapter (7)Second Order Transient Circuits

•Here , we consider circuits with both capacitor , inductor , andresistors (RLC)•We expect the circuit to be descried by a second order differential equation.•Consider the parallel RLC circuit

iS (t)

iR (t) iL (t) iC (t)

L CR

+

-

VC (t)


Take first derivative

( ) (t)idt

(t)dvCdττv

L1)(ti

R(t)v

sc

t

tc0L

c

0

=+++∴ ∫

dt(t)di

dt(t)vd

C(t)vL1

dt(t)dv

R1 s

2c

2

cc =++

Let’s assume that the voltage across the capacitor is VC(t)

(t)i(t)i(t)i(t)i CLRS ++=KCL :


dt(t)di

C1(t)v

LC1

dt(t)dv

CR1

dt(t)vd s

cc

2c

2

=++

0dt

(t)diS =

0(t)vLC1

dt(t)dv

CR1

dt(t)vd

cc

2c

2

=++

If IS(t) = constant (DC)

Consider the series RLC :

VS(t)

+-

R L

Ci


KVL :

( ) (t)VdττiC1)(tV

dtdi(t)Li(t)R

0V(t)V(t)V(t)V

S

t

t0C

CLRS

0

=+++

=+++−

∫

dt(t)dV

(t)iC1

dti(t)dL

dtdi(t)R S

2

2

=++

0dt

(t)dVS =

Take first derivative :

If VS = constant

0(t)iCL

1dt

di(t)LR

dti(t)d

2

2

=++ Series RLC


Hence, let’s assume that the differential equation we wish to solve is

A(t)xadt

dx(t)adt

(t)d212

X2

=++

It is known that the solution x (t) can be expressed as

(t)x(t)xx(t) CP +=

xP(t) : particular ( forced solution )xC(t) : complementary ( natural solution)

To find xP(t)

The only solution is the constant xP(t) = k0


( ) ( )

2P

20

020102

2

aA(t)x

aAk

Akakdtdak

dtd

=⇒

=

=++

0(t)xadt

(t)dxadt

(t)dc2

c12

cX2

=++

To find xC(t) : ( natural solution )

For simplicity let’s define :

202

01

a

ξ2a

ω

ω

=

=


Where :ξ : damping ratio.w0 : underdamped natural frequency

0(t)xdt

(t)dxξ2dt

(t)dc

20

c02

cX2

=++ ωω

Assume xc(t) = k e s t

[ ] [ ] [ ]

[ ] 0sξ2sek

0ekeskξ2esk

0ekekdtdξ2ek

dtd

200

2ts

ts20

ts0

ts2

ts20

ts0

ts2

2

=++

=++

=++

ωω

ωω

ωω


k : non-zeroest : non-zero

0sξ2s 200

2 =++ ωω

1ξξs2

4ξ4ξ2s

200

20

20

20

−±−=

−±−=

ωω

ωωω

Characteristic polynomial

The roots :

1ξξs

1ξξs2

002

2001

−−−=

−+−=

ωω

ωω Complex frequencies[ rad / sec]


ts2

ts1c

21 ekek(t)x +=⇒

ts2

ts10

CP

21 ekekkx(t)

(t)x(t)xx(t)

++=

+=Q

210 kkkx(0) ++=

k1 , k2 can be found from initial conditions

From x(0):

From dx(0) / dt:

2211

ts22

ts11

skskdt

dx(0)

eskeskdt

dx(t)21

+=

+=


Note :There are 3 types of responses based on the values of (S)

1. Overdamped response ( ξ > 1)

S1 and S2 are real and distinct .

20

ts2

ts10

aAk

ekekkx(t) 21

=

++=

k1 and k2 can be found from initial conditions

2211

210

skskdt

dx(0)kkkx(0)

+=

++=


2. Underdamped response ( ξ < 1 )

S1 & S2 are complex conjugate

2001,2

2001,2

2001,2

ξ-1ωjωξS

1ξ-1ωωξS

1ξωωξS

m

m

m

−=

−−=⇒

−−=⇒

−≡

≡2

0d

0

ξ1ωω

ωξσLet Damped radian frequency

[ rad / sec ]

d1,2 ωjσS m−=⇒


( ) ( )

( )( ) ( )ΘsinjΘcose

ekekek

ekekk

ekekkx(t)

Θj

tωj2

tωj1

tσ0

tωjσ2

tωjσ10

ts2

ts10

dd

dd

21

±=

++=

++=

++=

±

−−

+−−−

Q

( ) ( )( ) ( ) ( )( )[ ]( ) ( ) ( )[ ]tωsinkjkjtωcos)k(kek

ωsinωcoskωsinωcoskekx(t)

d21d21tσ

0

dd2dd1tσ

0

−+++=

−+++=−

− tjttjt

( )212

211

kkjAkkALet−=

+=


( ) ( )

20

dtσ

2dtσ

10

aAk

tωsinetωcoseAkx(t)

=

++= −− A

A1 & A2 can be found from initial conditions :

( ) ( )

( ) ( )

2d1

tσ-2dd

tσd2

tσ1

tσddd

tσ1

10

AωAσdt

dx(0)σeAtωsinωetωcosA

eσAetωcosωtωsineAdt

dx(0)Akx(0)

+−=

−+

−=

+=

−

−−


+−=

+=

=

⇒

2d1

10

20

AωAσdt

dx(0)Akx(0)

aAk

021

2001,2

ωξSS1ξωωξSSince

−==⇒

−−= m

3. Critical damped response ( ξ = 1 )

*0(t)xαdt

(t)dxα2dt

(t)xd

ωξαLet

0(t)xωdt

(t)dxωξ2dt

(t)xd

c2

0c

2c

20

c20

c02

c2

=++⇒

≡

=++⇒


Which means thattα

2tα

1c ekek(t)x −− +=

tα2

tα1c ekek(t)x −− += t

It is impossible for this solution to satisfy two initial conditions :

It can be shown that the following solution also satisfies the differential equation:

tα2

tα10

cP

etkekkx(t)

(t)x(t)xx(t)−− ++=

+=⇒


21

tα2

tα2

tα1

10

20

kαkdt

dx(0)

ekeαtkeαkdt

dx(t)kkx(0)

aAk

+−=

+−−=

+=

=

−−−

+−=

+=

=

⇒

21

10

20

kαkdt

dx(0)kkx(0)

aAk


V (0)

+

-

RLC

Example :

A3(0)i,V12v(0)AssumeF1C

H0.5L

Ω31R

L ====

=

Find v (t) ?


0CL

1sCR

1s

0(t)vLC1

dt(t)dv

CR1

dt(t)vd

2

cc

2c

2

=

++

=++

Characteristic polynomial is :

11.0622

3ω23ξ

3ωξ22ω2ω

0CL

1SCR

1S

0

0

02

0

2

>===

==⇒=

=++

Overdamped

From before we know


A363

112(0)i12v(0)Since

(1)kk12v(0)ekekv(t)

ekekv(t)

2s,1s

R

21

t2-2

t-1

tS2

tS1

21

21

==⇒=

+==+=

+=

−=−=

LL

39dt

dv(0)

0dt

dv(0)139

0dt

dv(0)C336

0(0)i(0)i(0)i CLR

−=

=+

=++

=++


t2t2

1

21

21

t22

t1

e27e15-v(t)

27k15-k

(2)39k2k

39k2kdt

dv(0)

ek2ekdt

dv(t)

−−

−−

+=

==

=+

−=−−=

−−=

KK


24 V

+-

RL

Ct = 0

F4.0CH0.1LΩ802R

µ===

Find vc (t) , t > 0 ?

Example :

If vc(0) = 0iL(0) = 0


KVL :

( )dττiC1(0)Vi(t)R

dtdi(t)L24

V(t)V(t)V24t

tC

CRL

0

∫+++=

++=

0(t)i10*25dt

di(t)2800dt

i(t)d

0(t)iCL

1dt

di(t)LR

dti(t)d

(t)iC1

dtdi(t)R

dti(t)dL0

62

2

2

2

2

2

=++

=++

++=


Char. Poly. is

128.0100002800

ω228008002ωξ2

rad/sec0005ω10*52ω

010*52s8002s

00

062

0

62

<===⇒=

=⇒=

=++

ξ

Underdamped

4800j1400ωjσ

ξ-1wjwξs

d

20021,

m

m

m

−=−=

−=


( ) ( )0k

tωsineAtωcoseAki(t)

0

dtσ

2dtσ

10

=++= −−

( ) ( )t4800sinet4800coseAi(t) t14002

t14001

−− += A

A1 & A2 can be found from initial conditions

420R(0)i(0)V42(0)V(0)V(0)V

L

CRL

=++=++

i(0) = A1 = 0

Since homogenous


t)sin(4800e0.05i(t)

0.05AA4800Aω402

4800ω

AωAσ402dt

di(0)

42dt

di(0)L(0)v

t1400-2

22d

d

2d1

L

=

===⇒

=

+−==

==

dtdi(t)0.1i(t)28042(t)v

(t)Vi(t)Rdt

di(t)L24

V(t)V(t)V24

c

C

CRL

−−=

++=

++=From KVL


( )[ ]( )

( ) ( )

+−

−=

−

−

−

t1400

t1400

t1400c

e1400-0.05*t4800sin4800t4800cose0.05

0.1

t4800sine0.0528042(t)v

( )( ) ( )

( ) ( )t4800cose42t4800sine742(t)vt4800sine7t4800cose42

t4800sine1442(t)v

t1400t1400c

t1400t1400

t1400c

−−

−−

−

−−=

+−

−=


+

-

R

L Ci L(0) vc(0)

Example :

F91C

H1LΩ6R

=

==

Find vc (t)?

vc(0) = 1 V , iL(0) = 0

0(t)i9dt

di(t)6dt

i(t)d

0(t)iCL

1dt

di(t)LR

dti(t)d

2

2

2

2

=++

=++


α−==⇒−==⇒=+

=++

21

212

2

ss3ss03)(s

09s6s

0ki(0)etkeki(t)

1

t32

t31

==+= −−

1dt

di(0)

010dt

di(0)

01R(0)idt

di(0)L

0(0)V(0)V(0)V CRL

−=

=++

=++

=++From KVL :


t32

21

eti(t)

1k

1kkdt

di(0)

−−=

−=

−=+−= α

dtdi(t)L(t)iR(t)V

0(t)V(t)V(t)V

C

CRL

−−=

=++From KVL :

Now, we need to find Vc(t):


[ ] ( ) ( )[ ]

t3t3c

t3t3t3

t3t3t3c

eet3(t)veet3et6

1e3et-et-6(t)v

−−

−−−

−−−

+=

+−=

−+−−−=


Chapter 8AC Steady–State Analysis

-So far , we have discussed the response of circuits due to DC source.-Here , we discuss circuits with sinusoidal sources.

* Sinusoidal

- Here we study sinusoidal functions :Consider the sinusoidal function :

( ) ( )tωsinXtωx M=Where :XM = amplitudeω = radian ( angular ) frequency


X M

- X M

2π π

2π3

π2

tω

t)X(ω

Tω

Note , X (t) repeats itself every (2 π) radius.

DEF : Period , TTime it takes the signal to repeat itself

f1T =


Where: f ≡ frequency in Hertz ( Hz )

• since ω T = 2 π

ω = 2 π / T = 2 π f

Consider the two signals x 1 (t) & x 2 (t)

Φ)t(ωsinX(t)XΘ)t(ωsinX(t)X

M22

M11

+=+=

Subtract Φ from both signals

t)(ωsinX(t)XΦ)-Θt(ωsinX(t)X

M22

M11

=+=∴


If Θ = Φ X1(t) & X2 (t) are in phase

If Θ ≠ Φ X1(t) leads X2 (t) by Θ – ΦOr X2(t) lags X1 (t) by Θ – Φ

tω

t)(ωsinX(t)X M22 =Φ)-Θt(ωsinX(t)X M11 +=

ΦΘ −

Θ)t(ωsinXX(t) M +=Q


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )βsinαsinβcosαcosβαcos

αcosβsinβcosαsinβαsin±=

=m

mm

( ) ( ) ( ) ( )[ ]( ) ( ) ( ) ( )

( ) ( )tωcosBtωsinAX(t)tωcosΘsinXΘcostωsinXX(t)

tωcosΘsinΘcostωsinXX(t)

MM

M

+=⇒+=+=

From trigonometry , we have

Where :A = XM cos (Θ)B = XM sin (Θ)

Also : tan (Θ) = sin (Θ) / cos (Θ) = B / A

Θ = tan -1 ( B / A )

B

A

X M

Θ


( )( )

( )

++=⇒

+=⇒

+=⇒

−

ABtantωsinBAX(t)

ΘtωsinXX(t)BAX

122

M

22M


Example :

30)t(50cos20(t)X60)t(50sin10(t)X

2

1

+=+=

120)t(50sin20(t)X90)30t(50sin2030)t(50cos20(t)X

2

2

+=++=+=Q

Find the frequency , phase angle between X1 and X2

Subtract 120 from both sides

t)(50sin20(t)X'60)t(50sin10(t)X'

2

1

=−=

Let

Note :cos (x) = sin ( x + 90 o)sin (x) = cos ( x – 90 o)

Frequency is50 rad/sec.


X 1 (t) leads X 2 (t) by -60 ( X2 leads by 60 )

X 2 (t) lags X 1 (t) by -60 ( X1 lags by 60 )

* Relationship between sinusoidal and complex numbers

We know that if the complex number z = x + j y = r e j Θ

Where :

=

+=

−

xytanΘ

yxr

1

22


Example : Consider the circuit : V(t) = VM cos (ω t)

Find i (t) ?

V (t)

+-

R

Li (t)

( ) (1)dt

di(t)L(t)iRtωcosV

dtdi(t)L(t)iRV(t)

M LLL+=

+=Using KVL :

Assume that the solution is a sinusoidal function

( )( ) ( )

)sin(A - A2 ),cos(A A1 where(2)tωsinAtωcosAi(t)

tωcosAi(t)

21

Θ=Θ=+=⇒

Θ+=LL


Now find A1 and A2

Plug (2) in (1)

( ) ( ) ( )( ) ( )[ ]

[ ] ( ) [ ] ( )tωsinωALARtωcosωALARtωcosωAtωsinωA-L

tωsinARtωcosARtωcosV

1221

21

21M

−++=++

+=

ωALAR0ωALARV

12

21M

−=+=⇒

Solving for A1 and A2 , we find

222M

2222M

1 LωRVLω

A,LωR

VRA

+=

+=


( )

22

21

1

M

M

1

2

AAA

RLωtanΘ

RLω

VRVLω

AAΦtan

)sin(A -A2 and )cos(A A1 since

+=

−

=

−=

−=

−=

Θ=Θ=

−

( ) ( )( )

( ) 222

2M

2222

2M

222

2222

2M

22

2222

2M

22

22

1

LωRV

LωR

VLωR

LωR

VLω

LωR

VRAA

+=

+

+=

++

+=+


( )

−

++

=

Θ+=

+=+=

−

RLωtantωcos

LωRVi(t)

tωcosAi(t)LωR

VAAA

1

222M

222M2

22

1

•It is clear that it is very complicated to find the solution using sinusoidal function


[ ][ ][ ]

ΘVV whereeVv(t)

eΘVRe

eeVRe

eVReΘ)t(ωcosVv(t)

Θ)t(ωsinVjΘ)t(ωcosVeV x)sin( j (x) cos e

formula sEuler’ Using) t w( cos V (t) v

that assumed Earlier we

M

tωj

tωjM

ΘjtωjM

Θ)t(ωjMM

MMΘ)t(ωj

M

jx

M

∠==⇒

∠=

=

=+=⇒

+++=⇒

+=

Θ+=

+

+

Dropping e jωt since itexists in all terms

Phasor form


Phasor representation :

( )( ) )90(AΘtωsinA

AΘtωcosAformPhsor domainTime

o−Θ±∠↔±

Θ±∠↔±↔


Example :

V (t)

+-

R

Lo0VVis formphasor The

t)(ωcosVv(t)

M

M

∠=

=

Assume :

tωjM

tωjM

eIi(t)eIi(t)

Φ)t(ωcosIi(t)

oΦ∠=

=

+=

We know that

tjoM

tj

eVtvVetv

ω

ω

0)()(

∠=

=


KVL :

[ ] [ ] [ ]

ωjILIRVeωjILeIReV

eIdtdLeIReV

dtdi(t)L(t)iRv(t)

tωjtωjtωj

tωjtωjtωj

+=⇒+=

+=

+=

[ ]

+−

=

+=⇒+=

222 ωLRωjLRVI

ωjLRVIωjLRIV


( )

−

∠

+

+=

+−

∠=

−

RLωtanωLR

ωLRV

ωLRωjLR0VI

1222222

M

222Mo

Example :Convert from time domain to phasor form

( )( )

Aooo

o

o

o

3012)90(12012IV)(-4524V

120t337sin12i(t)45t337cos24v(t)

∠=−∠=

∠=

+=

−=


Example :Convert from phasor form to time domain

zHK1fif)(-7510IV12016V

o

o

=∠=

∠=

( ) π20001000π2fπ2ω ===

( )( )o

o

75tπ2000cos10i(t)120tπ2000cos16v(t)

−=

+=


* Phasor relationships for circuit elements

1. Resistor :

V (t)

+

-

R

i (t)

Assume :

iv

tωjiM

tωjvM

)Θtω(jM

)Θtω(jM

)Θtω(jM

)Θtω(jM

ΘΘ&IRV

eIReV

eIReV

i(t)Rv(t)eVv(t)&

eIi(t)

iv

v

i

==

Θ∠=Θ∠

=

==

=

++

+

+

In phasor form


tω

i (t)

v (t)

)Θtω(jM

)Θtω(jM

i

v

eIi(t)

eVv(t)+

+

=

=

2. Inductor :

V (t)

+

-

L

i (t)Assume :


[ ])Θt(ωjM

ieIdtdL

dtdi(t)Lv(t) +==

oiv

tωjiM

tωjvM

)Θtω(jM

)Θtω(jM

90ΘΘ&ILωjV

eIωLjeV

eωILjeV iv

+=

=Θ∠=Θ∠

= ++

tω

i(t)v(t)

o90

The voltage leads theCurrent by 90o.


3. Capacitor i (t)

V (t)

+

-)Θtω(j

M

)Θtω(jM

i

v

eIi(t)

eVv(t)+

+

=

=

[ ])Θt(ωjM

veVdtdC

dtdv(t)Ci(t) +==

Assume :

oiv

tωjvM

tωjiM

)Θtω(jM

)Θtω(jM

90ΘΘ

ICωj

1Vor

VCωjIeVCωjeI

eVCωjeI vi

−=

=

=Θ∠=Θ∠

= ++

tω

i(t) v(t)

o90

The current leadsThe voltage by 90o.


Example : If the voltage across the 20 mH inductor isv (t) = 12 cos (377 t + 20o ) find i (t) ?

( ) ( )

( ) ( )( )o

oo

o

o

70t377cos1.59i(t)

)07(1.5990m20377

2012I

Im20377j2012ILωjV

−=

−∠=∠

∠=

=∠

=

L


Example : If the voltage across the capacitor isv (t) = 100 cos (314 t + 15o ) find i (t) ??C = 100 µ F

( ) ( ) ( )( ) ( )( )

( )o

o6-4

6-

105t314cos3.14i(t)1053.14)9015(1010314I

1510*100100314jVCωjI

+=

∠=+∠=

∠=

=

oo

o


In summary

1. Resistor : RR IRV =

LL ILωjV =

CC ICωj

1V =

2. Inductor :

3. Capacitor

* Impedance and admittance :Definition : impedance : ( Z )The ratio of the voltage over the current

)ΘΘ(IV

ΘIΘV

IVZ

VI

Z1y&

IVZ

iVm

m

im

Vm −∠=∠∠

==

===


ivzm

m ΘΘΘ&IV

Zwhere −==

In complex form :

( )( )

=

+=

=

=

+=

−

RXtanΘ

XRZ

ΘsinZX

ΘcosZRwhereXjRZ

1z

22

z

z


In summary :

=−=

+==

+=∠=

−

RXtanΘΘΘΘ

XRZIVZ

XjRZΘzZZ

formcomplexformphasor

1zivz

22

m

m


Impedance of R,L,C

Cω1j

Cωj10Z)90(

Cω1ZC

Lωj0Z90LωZL

0jRZ0RZR

form)(complexZform)(phasorZElement

Co

C

Lo

L

Ro

R

−=+=−∠=

+=∠=

+=∠=


* Parallel and series connections of ( Z )

• Series : Z 1 Z 2 Z 3 Z n

Z eq

Z 1 Z 2 Z n

• Parallel

Z eq

n21eq Z1

Z1

Z1

Z1

+++= L

n21eq ZZZZ +++= L


Definition : admittance ( y )It is the reciprocal of Z

BjGy

xRxj

xRRy

xRxjR

xjR1

Z1y

VI

Z1y

2222

22

+=⇒

+

−+

+

=

+−

=+

==

== [S] siemens

Conductance Susceptance


* In terms of R ,L ,and C

oCC

o

LLL

RRR

90CωCωjyCωj

1Z-

)90(Lω

1Lωj

1Z1yLωjZ-

GR1

Z1yRZ-

∠==⇒=

−∠===⇒=

===⇒=


* Parallel and series connections of ( y )

• Series : y 1 y 2 y 3 y n

y S

y 1 y 2 y n

• Parallel

y P

n21S y11

y1

y1

+++= Ly

n21P yyyy +++= L


Example :Find the equivalent impedance

sec/rad2ωF2C,F1C

H2L,Ω2RH1L,Ω1R

21

22

11

===

====

L 2

L 1 C 1

C 2

R 1

R 2

Z eq


( ) ( )

Ω41j

Cω1jZ,Ω

21j

Cω1jZ

Ω4jLωjZ,Ω2j12jLωjZ2RZ,1RZ

2C2

1C1

2L21L1

2R21R1

−=−=−=−=

=========

( )( ) ( )

Ω22.33.243037j30

30845j3Z

23j315

4j23j3

415j1

221j2j1

41j4j

41j4j

ZZZZZ//ZZ

oeq

R2C1L1R1C2L2eq

∠=+=−

+=

++−=++=

+−++−

−=

++++=


v (t)

40 m H

Ω20

+- Fµ50

i (t)

Example : Find the current i (t) in the circuit

( )( )( )

sec/rad377ωV)-30(120V

30t377cos120v(t)9060t377cos120v(t)

60t377sin120v(t)

o

o

oo

o

=∠=

−=

−+=

+=


V

Z L

R

+-

CZ

I( ) ( )

( )( )53.05jZ

µ503771j

Cω1jZ

Ω15.08jZ,m 40377jLωjZΩ20RZ

C

C

LL

R

−=

−=−=

=====

Z eq

+-V

( )( ) ( )

( )o

oo

o

o

CLReq

eq

39.23t377cos3.87i(t)

A)39.23(3.87Ω9.2330.96Ω)30(120I

Ω9.2330.9653.05j//15.08j20

Z//ZZZ

ZVI

−=

−∠=∠−∠

=

∠=−+=

+=

=


Example :Find the equivalent impedance :

Ω4Z eq

Ω2

Ω2

Ω2j

Ω6j

4j-

( ) ( )[ ]( ) ( )[ ]( ) ( )

( )Ω17.653.551.076j3.38Z

24j6

2j42j222j4//2j2

24j6j4//2j2Z

oeq

eq

∠=+=

++

++=

+++=

+−++=


Phasor diagram :

A diagram that shows the magnitude and phase of various voltage and currents in the circuit

For example :

o1

o2

o1

1307I

605V

4510V

∠=

∠=

∠=

V 1 V 1I 1

60o

45o

130o105

7


Example :

Draw the phasor diagram of all currents and voltage

I 1 I 2

j42 −=ZΩ= 21Z

o454I =

+

-

V

A18.433.574j2

4j454ZZ

ZII

454I

oo

21

21

o

∠=

−

−∠=

+

=

∠=


( ) V18.437.1418.433.572ZIV

A108.431.784j2

2454ZZ

ZII

oo11

oo

21

12

∠=∠==

∠=

−

∠=

+

=

| I | = 4|V | = 7.14

| I2 | = 1.79

45o108.4o

| I1 | = 3.57

18.43o


Circuit Analysis Technique Note :All analysis techniques discussed before can be used in AC steady-state analysis Example :Find Io using nodal analysis ?

I 1I 2

Ω1j

V2+-

A901 o

I o

V1Ω1j−

Ω1j−

o01

Ω1


Here , we have a supernode :Applying KCL at supernode

( ) ( ) (1)901V450.707V450.707

1j11

1j1V

1j1V901

1j1V

1jV

1j1V901

III901

o2

o1

o

21o

221o

2o1o

LL∠=−∠+∠

−

++−

=

−++

−=

++=

)2(01VV

01VVo

21

o12

LL∠−=−

∠+=


A)18.44(1.581j

VI

V71.561.58V

V108.41.58V

o2o

o2

o1

−∠==

∠=

∠=


Example : use mesh analysis to find Vo ?

I 2

+

-

Ω2

+-

Ω2jΩ2j−

Ω2

A902 o

o024V o

I 1

SupermeshKVL around supermesh

( )( ) (1)024I2j2I2

0I2I2jI2024o

21

221o

LL∠=−+

=+−+∠−


V36.0210.88I2V

A36.025.44I

A15.254.56I

(2)902II

902II

o20

o2

o1

o21

o12

∠==

∠=

∠=

∠=+−

∠=−

LL


Example : find Vo ?

I 2

+

-

Ω2 Ω1j−

V012 o

A06 o

V o+-

V 1 V 2

Ω2j Ω1

Ω2

I 3I 1

03

V2j

V1jVV

0III

2221

321

=−−−

−=−−

V012V o1 ∠=

KCL at node V2


( )

V33.76.65V31

211VV

V33.719.969V

j21

31

9012

2j1

31

9012V

90122j21

31V

9012j012

31

j21

j1V

03

V2j

V1j

V012

o22o

o2

oo

2

o2

oo

2

222o

∠==

+

=

∠=

+∠

=

−

∠=

∠=

+

−+

∠=−∠

=

++

−

=−−−

−∠


Example :Use Thevenin Theorm to find Vo ?

+-

V012 o

Ω1 Ω2 Ω1

Ω1Ωj− Ω1jVo

+

-

First find VOC

+-

V012 o

Ω1 Ω2 Ω1

Ωj− Ω1j Voc

+

-

I1I3

I2

Vx

KCL around V x


0III 321 =−−

V33.693.32j2

j1VV

V)-29.7(7.44V3.69j6.460.8j1.4

12V

511j

521V012

14j2j1V012

j21j1V012

0j2

Vj

V1

V012

oxoc

oxx

xo

xo

xo

xxxo

∠=

+

=

∠=⇒−=+

=

−+

+=∠

+−

++=∠

+

++=∠

=+

−−

−−∠


Ω1 Vo

+

-+

-

ZTH

Voc

V12.541.3V

33.71.661133.693.32

Z11VV

oo

oo

THOCo

∠=

∠+

∠=

+

=

Find ZTH:

Ω 33.7 1.6611||]2)||1[( oTH ∠=++−= jJZ

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