BASICS OF ELECTRICITY 1 WHAT IS ELECTRICITY? 1) Generating electricity + - Hydrogen atom Helium Lithium Atomic nucleus Electron Atomic.

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1 BASICS OF ELECTRICITY 1 WHAT IS ELECTRICITY? 1) Generating electricity + - Hydrogen atom + + - - Helium + + - - - Lithium Atomic nucleus Electron Atomic nucleus Proton Neutron

2 6. BASICS OF ELECTRICITY 2) Frictional electricity (1) When two different objects are rubbed together, electricity is produced. (2) Frictional electricity is positive or negative. (3) Similarly charged bodies repel each other. (4) Oppositely charged bodies attract each other.

3 6. BASICS OF ELECTRICITY 1 CURRENT, VOLTAGE, RESISTANCE 1) Current (1) What is Current? Electric cable Free electron Electron move. Neutralized Supplied from B ABAB

4 6. BASICS OF ELECTRICITY (2) Current unit 1 Second1 Ampere1 Coulomb= 1 ampere = 1000 milliamperes 1 milliampere = 1000 microamperes Ampere(A) Milliampere(mA) Microampere(µA)

5 6. BASICS OF ELECTRICITY 2) Voltage Water pressure Voltage Electron flow Current No water pressure Voltage is zero

6 6. BASICS OF ELECTRICITY 3) Resistance Generator Current Load resistance Cable internal resistance

7 6. BASICS OF ELECTRICITY

8 4) Ohm's Law 132

9 6. BASICS OF ELECTRICITY : Current : Voltage : Resistance IERIER

10 6. BASICS OF ELECTRICITY 3 ELECTRIC CIRCUIT (APPLYING OHM'S LAW) 1) Basic circuit Switch Fuse Lamp +- 12 V I = ?

11 (1) When voltage and current are known and resistance is determined 6. BASICS OF ELECTRICITY When a resistance was applied to a voltage of 24V, a current of 8A was produced. What was the resistance value? 24 V I = 8 A R = ? From Ohm's Law I = 8 (A)E = 24 (V)

12 (2) When resistance and current are known and voltage is determined 6. BASICS OF ELECTRICITY How many volts is required to conduct a current of 2A through a resistance of 50 100 V I = 2 A R = 50 From Ohm's Law R = 50I = 2 (A) E = I x R E = 2 x 50 = 100 (V)

13 6. BASICS OF ELECTRICITY 2) Series circuit

14 6. BASICS OF ELECTRICITY R1R1 R2R2 R3R3 E E1E1 E2E2 E3E3 Total voltage E = E 1 + E 2 + E 3 = IR 1 + IR 2 + IR 3 = I (R 1 + R 2 + R 3 ) Combined resistance E = E 1 + E 2 + E 3

15 6. BASICS OF ELECTRICITY Let's obtain the values of the combined resistance and the current. 12 V R = R 1 + R 2 + R 3 6 + 4 + 2 = 12

16 6. BASICS OF ELECTRICITY 3) Parallel circuit

17 R1R1 R2R2 E I2I2 I1I1 I I = I 1 + I 2 = Total current 6. BASICS OF ELECTRICITY Combined resistance Ohms’ Low

18 6. BASICS OF ELECTRICITY What is the combined resistance value in the figure below?

19 6. BASICS OF ELECTRICITY 2) Battery structure and function Terminal Lid Cell plug Connector Separator Glass mat Positive plate Negative plate Battery jar Glass mat Positive plate Pole Negative plate Separator Pole

20 6. BASICS OF ELECTRICITY Pb + 2H 2 SO 4 +PbO 2 PbSO 4 + 2H 2 O + PbSO 4 Positive pole platesNegative pole plates + + + + - - - - + + + + Hydrogen ion + Lead ion + Negative polePositive pole Current Electron

21 6. BASICS OF ELECTRICITY +- Larger terminal is positive

22 6. BASICS OF ELECTRICITY 3) In-service maintenance Diluted sulfuric acid (Specific gravity : 1.400) Purified water Replaceable quantity (CC) (per 1 liter of electrolyte) Battery electricity specific gravity (20 degrees) (1) Adjusting electrolyte level and specific gravity

23 6. BASICS OF ELECTRICITY (2) Measuring electrolyte specific gravity A B Rubber ball Glass tube Floating scale Suction tube

24 6. BASICS OF ELECTRICITY 4) Storage cautions

25 6. BASICS OF ELECTRICITY 5) Charging (1) Charging type Supplementary chargingInitial charging

26 6. BASICS OF ELECTRICITY (2) Charging methods Constant-current charging method Fast charging method 1 2

27 6. BASICS OF ELECTRICITY (3) Voltage and specific gravity changes during charging 02468101214161820 1.1 1.2 1.3 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 1.9 Charging time (H) Terminal voltage (V) Specific gravity Voltage Specific gravity

28 6. BASICS OF ELECTRICITY (4) Charging-related cautions + - - + + + - -

29 6. BASICS OF ELECTRICITY (5) Battery capacity UnitAmpere-hour (Ah) Ampere-hour capacity (Ah) = discharge current (A) x discharge time (h) Watt-hour capacity (Wh) = discharge current (A) x discharge time (h) x mean discharge voltage (V) (Example) Discharge:5 hours Current:6A 6 (A) x 5 (h) = 30 (Ah) Voltage:12V Battery:40Ah 40 (Ah) x 12 (V) = 480 (Ah) Ampere-hour capacity (Ah)

30 6. BASICS OF ELECTRICITY 100Ah 20 hours at 5 A 5 A 100Ah 10 hours at 10 A Heat loss Chemical reactions 10 A