Program of instruction

Teaching calendar

• Instruct circuit’s basic concepts ,principals and analysis method

• Some important principals from steady state (steady state) （ direct current’s steady state ）

• Analyze direct current’s transient state (transient state) （ the transition process of direct current ）

• Analysis and compute alternating current (alternating current) （ phasor method ， analysis current’s steady state ）

• Operation method （ the transient state of analysis current 、 second-order ）

• exchange easily• transport cconomically be convenient to distribute• swift and precisely to contral measure• transmit by electromagnetic wave in space

To prepare necessary knowledge for ulterior major course

Listen in heart,read book ,write homework,take notes

The purpose to Study the course:

The Requirement of learing the course ：

§1.1 Circuit and Model

一、 Electric-energy’s Transportation and Transformation) 。Electric current’s Routecircuit

二、 Transmit and dispose signal

E R

Requirement ： Reduce Loss) 、 Improve Efficiency

For example ： power system)--Dynamoelectric 、 substation 、 conduct electricity 、 contribute electricity 、 consume electricity to constitute integrity 。 Question ：Why to conduct electricity in high voltage?

Requirement ： no distortion 。 For Example ： Megaphone——microphone) 、 amplifier) 、 loudspeaker)

The function of circuit

If circuit dimension is far shorter thanelectromagnetic wave,this sort of circuit is called

Lumped Circuitm

fc 6

8

10650103

If electrical wire is very long ,this sort of circuit is called Distributed Circuit

Important assumption in electric circuit) ：

Fact circuit: It is composed of electric component to change energy,transmit and dispose electric energy 。

bulb

Circuit Model is a circuit graph which is used to analyse and compute and a model which is composed of ideal circuit component.,also as a component which neglects inferior factor and emphasize main factor 。

loadpower supply

E R

Circuit Model ： For short circuit

Battery

bulb

一、 voltage and electric current’s fact direction

R UE

a

b

Put forward a question ： In complicated circuit,it is difficult to judge physical quantity’s fact direction of components .How to solve?

E1

A B

R

E2

IR

Uab=Va- Vb

§1.2 voltage and current’s reference direction

Current’s directionAB ？

Current’s

directionBA ？

current’s direction

positive charge motion’s direction

Voltage’s direction Potential step down’s direction

I

Electromotive force’s direction Potential step up’s direction

E

I

R Uab

+_

+

_

Non-associated reference direction

People rule a direction,And only having the reference direction voltage and current have the deference between Positive and Negative

Assume at will

Voltage and current’s reference Identical

Voltage and current’s reference contrary

二、 Reference Direction ：

Associated

reference

directionU U

component

component

component

component

I

UI

I

IU

一、 Energy ：In t0 ~ t ， electric field force do work that transfer Positive Charge form A to B

tq

tqudqW

0

dtdqi

Junits 0

、：JoulediuWt

t

二、 Power ： If the voltage between any two point is U ,current flowing in the part circuit is I ， so the power consuming by the part circuit :

W Wattunits 、：titudt

dWtp

§1.3 Power and Energy

For Direct Current ： P=UI

1. u 、 i have associated reference direction

P =ui> 0 a load ， absorb electric powerP =ui ＜ 0 a electric source ， release electric power

2. u 、 I have non-associated reference direction

P=ui > 0 a electric source,release electric powerP =ui ＜ 0 a load,absorb electric power

I +

–UE

+–

I

Conclusion: when u 、 i have identical fact direction,it is a load. When u, I have contrary fact direction,it is a electric source

Electric power has release and absorption to judge a component which is a voltage or a load.

一、 Linear resistance Ri

uResistance units ： ohm 、 Ωdefinition ： conductance G=1/R

G units ： 西门子、 S

Gi

u

Ohm’s law ：u 、 i associate refer

ence direction u=Ri

u 、 i Non-associated reference direction u=－ Ri

u

i

V-A characteristicOpen circuit:R ∞ ，

Short circuit:R 0

§1.4 Circuit Component

i

u

C

Capacitance units : Farad 、F u

qC eCapacitanc q

C-V characteristic

1. Current and voltage’s relation

charge electric passed ofquantity thesecond one(In dtdqi

dtduCi

dtduCi associated

reference direction

Non-associated reference direction

The expression in time field

Direct current ： capacitance is considered as open circuit

二、 Linear Capacitor ：

di

Cdu

t

t

tu

tu 00

1

Suppose that capacitance is working from t0 to t ,what is the voltage ?

diC

tutut

t0

10

initial value

Conclusion ： the voltage of a capacitance is decided by current and initial value

2. the capacitance’s Active Power and Energy ：

dttduCtutitutpcousins tantan

d

dduCupd

t

t

t

t 00

CWenergy field electric

022

20

tutuCduuCt

t

Capacitance is a store component

Assignment 1-1 、 1-4

interview of last class ：some problems in assignment:Power system ?

Non_associate

Reference direction ？Load ？

Electric source ？

resistance ：capacitance ：

associate

1 、 Make sure draw designs and designate the reference direction

2 、 Distinguish a load or a electric source ？1-3 lesson

Some problems in assignment:

Li

uInductance units ： Henry 、H

i

Ψ

W-A characteristic

iL

inductance

1. Current and voltage’s relation ：Electromagnetic Induction law

dt

du

dtdiLu

dtdiLu

direct current ： inductance is viewed as short circuit

associated reference direction

Non-associated reference direction

The expression in time field

三、 linear inductance component:

Suppose inductance is working from t0 to t ,what is voltage?

du

Ldi

t

t

ti

ti 00

1 duL

titit

t0

10

initial value

Conclusion ： inductance’s current is decided by voltage and initial value.

dt

tdiLtititutpneouspowerins tantan

d

ddiLipdelectric

t

t

t

t 00

Cenergy W field

022

20

titiLdiiLt

t

Inductance is a store

component

2. Inductance’s Active Power and Energy ：

1. Independent voltage source:

+

–US

I +

–

Us

i

Peculiarity ：（ 1 ） Whenever voltage source has constant voltage containing value and direction（ 2 ） voltage’s current can change with outer electric circuit altering.its value is decided by electric source’s voltage and outer circuit

Ideal voltage source

+

–US

I

R

+–

uS

i

R

Fact voltage source

i

u

USi

u

US

§1.8 Voltage Source and Current Source(ideal)

+

–U

IS

ideal current source

Peculiarity ：（ 1 ） Whenever electric source’s current keeps constant including value and direction

（ 2 ） electric source’s voltage can change with the change of outer circuit and its value is decided by electric source’s current and outer circuit

i

u

Ideal current source V-A characteristic

IS

i

u

Fact current source V-A characteristic

IS

+

–u

i

R

Fact current source

iS

2. Independent Current Source ：

+

–ri1

Called “dependent electric source”its value and direction is controlled by another voltage or current.

+

–u1

+

–μu1

VCVS(Voltage-controlled voltage source )

i1

CCVS(Current-controlled voltage source)μ 、 Β the number without units 、 r units is Ω 、 g units is S

+

–u1 gu1

VCCS(Voltage-controlled current source)

i1 βi1

CCCS(Current-controlled current source)

3. Controlled Source ：

§1.10 Kirchhoff’s Law

E1

A B

R

E2

C D

① ②

0

Reference node

noun ：Branch ： The same current pass a branch in circuit.

Node ： it connects more than three branch.Loop ： Any Closed Route in electric circuit 。

in Lumped Circuit ， at any time ， for any node ， the algebra sum of Branch current is zero ：

0 iGiven ： i flowing out as positive ＋ flowing in negative －

0 4321 iiiiSo

4321 namely iiii

In lumped circuit at any time ,the sum of current of flowing in any node is equal to the sum of current of flowing out the node 。

i1

i2

i3 i4

important

一、 Kirchhoff’s current law ：（ KCL ）

In lumped current at any time , if circle along a loop, Algebraic Sum of loop voltage is zero ：

0 u1. Giving a loop direction

2. Stipulation ： when voltages is identical to the loop direction,its direction is positive,else negative 。

0621 uuu0662211 iRiRiR

i1

i2

i3

i4

i5

i6

+

–u1 +

–u6+

–u2

R1

R2

R6

二、 Kirchhoff’s voltage law ：（ KVL ）

Solve I 、 US 、 R

I

+

-us

12A

12

3

5A

R

1

15A

6A

IR

uR

I1

solve ： I=6-5 =1A

IR=15-1=14AI1=12+6=18A

I3

I3=18-15=3A US - 3I1 - 12I3 =0

US=90V

US - 3I1 – 15 *1- UR =0

UR=21V R=1.5 Assignment ： 1-4 、11 、 14 、 20 、 21

examole

Interview of Last Class) ： 1 、 R 、 L 、 C component’s voltage and electric current relationship ？

2 、 power source

Voltage source

electric source

independence

controlled

controlled

independence

3 、 Kirchhoff’s Law ：Key point