lecture 1 fundamental of electricity

By SY CheungEL / IVE (HW)

LECTURE 1

Fundamental of Electricity

EEE3404 – Electrical Engineering Principles 1

2© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2

1. Atom Model Nucleus

Electron



2. Atom

• Protons

• Neutrons

• Electrons


“Each atom has the same number of protons and electrons.”


3. Proton and Electron

Protons carries positive charge

– it is relatively large mass

– does not play active part in electrical current flow

Electrons carries negative charge

– light mass

– play an important role in electrical current flow



4. Unit of Charge

• Unit of Charge is called Coulomb (C)

• An electron and a proton have exactly same amount of charge

• One coulomb of charge is equal to approximately 628 x 1016 electron charge



5. Free Electrons

Free Electrons

Applying Heat or Light



6. Electrical Materials

All material may be classified into three major classes:


• Conductors (copper, aluminum, siliver,

platinum, bronz, gold)

• Semiconductors (germanium, silicon)

• Insulators (glass, rubber, plastic, air,

varnish, paper, wood, mica, ceramic,

certain oils)


• Conductors have many free electrons which will be drifting in a random manner within the material

6. Electrical Materials


• Insulators have very few free electrons

• Semiconductors falls somewhere between

these two extremes


7. Electric current

• Electric current is the movement, or flow of electrons through a conductive material

• It is measured as the rate at which the charge is moved around a circuit, its unit is ampere (A)

I=Q/t or Q=It



8. Electromotive Force

• In order to cause the 'free' electrons to drift in a given direction an electromotive force must be applied

• The emf is the 'driving' force in an electrical circuit

• The symbol for emf is E and the unit of measurement is the volt (V)



8. Electromotive Force

Typical sources of emf are cells, batteries and generators

The amount of current that will flow through a circuit is related to the size of the emfapplied to it



9. Potential Difference (p.d.)

• Whenever current flows through a circuit element in a circuit such as resistor, there will be a potential difference(p.d.) developed across it

• The unit of p.d. is volts(V) and is measured as the difference in voltage levels between two points in a circuit




• Emf (being the driving force) causes current to flow

• Potential difference is the result of current flowing through a circuit element

• Thus emf is a cause and p.d. is an effect




LOAD

P.D.=1.4 V

E.m.f.=1.5V, Rint.

Equivalent circuit of a battery



10. Resistance

• Resistance is the 'opposition' to the current flow measured in ohms (Ω)

• Conductors have a low value of resistance

• Insulators have a very high resistance

• Load in DC/AC circuits



Circuit convention

+

-

+

-

I

E VR

11. Ohm’s Law



11. Ohm’s Law

It states that current in a resistive circuit is directly proportional to its applied voltage and inversely proportional to its resistance provided that all other factors (e.g. temperature) remain constant.

IRVorI

VRor

R

VI.e.i ===

V

IR

Slope = 1/R

V

I

V α I



12. Power

Power (P) is defined as the rate of doing work (W).

havewet

QIand

Q

WVSince

Wwattst

WPei

==

= )(..

Additional relationships are obtained by substituting

V=IR and I=V/R, we have:

P = VI

R

VPandRIP

22

==



13. Resistors in Series

E

I

V1

V2

V3

R1

R2

R3



By Ohm's law

• V1 = IR1 volts;

• V2 = IR2 volts; and

• V3 = IR3 volts

• E = V1 + V2 + V3

• E = I (R1 + R2 + R3)




• E = IReq and

• Req = R1 + R2 + R3 ohm

• where Req is the total circuit resistance

when resistors are connected in series the total resistance is found simply by adding together the resistor values




14. Potential Divider

The voltage drops shared by the resistors

are given as follows:

• V1=E x R1/ (R1+ R2+ R3)

• V2=E x R2 / (R1+ R2+ R3)

• V3=E x R3 / (R1+ R2+ R3)



15. Resistors in Parallel

R1

R2

R3

I1

I2

I3

I

E



By Ohm’s Law

I1=E/ R1

I2=E/ R2

I3=E/ R3




The total current of the circuit I is the sum of I1, I2 and I3 , thus

I = I1 + I2 +I3




The total resistance or the equivalent resistance(Req) of the circuit is defined to be

Req= E/I




By substituting the above expression for the currents, we have

E/Req=I=E(1/R1+1/R2+ 1/R3)

Thus we found

1/Req=(1/R1+1/R2+ 1/R3)




Quiz

Calculate the total resistance of the following resistors connected in parallel:

a)30 Ω, 60 Ω

b)180 Ω, 90 Ω, 60 Ω, 60 Ω



16. Current Divider

The current in each branch of the circuit is

given as

• I1= I x Req/ R1

• I2= I x Req/ R2

• I3= I x Req/ R3



17. Power and Energy in a resistive

circuit

Power is equal to the current multiplied by the voltage and the unit of power is watt (W)

P = I x E (W)

Energy is equal to the power multiplied by the time for the circuit being energized. The unit is Joule (J).

Energy = P x t (J)



18. Power in a resistive circuit

By Ohm's law E=IR, the above equation can be modify to be

P = I2R

Power is equal to the current squared, multiplied by the resistance.



Use Ohm's law again, where I =E/R, we have

P=E2/R

Power is equal to the voltage squared, divided by the resistance.

18. Power in a resistive circuit



Example 1

RBC

E=10V

I

VAB

VBC

VCD

2Ω 5Ω 3Ω

A B C D

RAB

RCD



•VAB + VBC + VCD is exactly equal to the emf=10V

•The total resistance of the circuit is 2+5+3=10Ω

•By Ohm's law V=IR, the current I should be equal to 1A

Example 1 - Answer



• VAB = IRAB =1 x 2 = 2V.

• VBC = I RBC =1 x 5 = 5V.

• VCD = IRCD =1 x 3 = 3V.

• Power dissipation in RAB = I2RAB = 12 x 2 = 2W.

• Power dissipation in RBC = I2RBC = 12 x 5 = 5W.

• Power dissipation in RCD = I2RCD = 12 x 3 = 3W.

• Total power dissipated = 2+5+3 =10W

Example 1 - Answer



Example 2

R1

R2

R3

I1

I2

I3

I

E=6V

3Ω

6Ω

2Ω



–The potential difference across each of the three resistors is equal to the battery emf 6V

–Apply Ohm's Law

• E=I1 R1 ; I1=E/R1 = 6/2 = 3A

• E=I2 R2 ; I2=E/R2 = 6/3 = 2A

• E=I3 R3 ; I3=E/R3 = 6/6 = 1A

Example 2 - Answer



• The total current I is equal to the sum of currents

• I1+I2+I3 = 3+2+1 =6A.

• Power dissipation in R1 = I12R1 = 32 x 2 = 18W.

• Power dissipation in R2 = I22R2 = 22 x 3 = 12W.

• Power dissipation in R3 = I32R3 = 12 x 6= 6W.

• Total power dissipated =18+12+6 =36W

Example 2 - Answer


lecture 1 fundamental of electricity

Education

vocational training

electrical circuit

flow of electrons

free electrons free

unit of charge unit

circuit element

current flows

equivalent circuit