01BASCI

Service Training MALAGA

EPG -ITYPICAL ELECTRICAL CIRCUIT

TYPICAL HYDRAULIC CIRCUIT

I

Q LOAD

LOAD POS

NEG

SOURCE

SOURC E

0I- BASIC ELECTRICITY - COMPARISON - OHM FORM - SERIES CIRCUITS - PARALLEL CIRCUITS

Claude Dec. 1997

EPG -ICHAPTER : COMPARISON CIRCUIT

TYPICAL ELECTRICAL CIRCUIT

I

TYPICAL HYDRAULIC CIRCUIT

Q LOAD

LOAD POS

NEG

SOURCE

SOURCE

Circuit

Circuit

• Voltage Source (battery) • Conductors (wire) • Load (resistor, lamp, etc.)

• Pressure Source (pump) • Tubing, hoses • Load (hydraulic cylinder)

Current Flow • Measured in ampere • Symbol is "I" (intensity) • The movement of electrons

Oil Flow • Measured in gpm, cfm, L/min • Symbol is "Q" (quantity) • Movement of fluid molecules

Positive Source

Positive Source

• Battery (generates DC voltage) • Produces certain voltage regardless of the load

• Pump (when pump shaft is rotated, fluid is positively expelled from the output port, no matter how restrictive the load)

P

2

EPG -ICHAPTER : COMPARISON OPPOSITION

Q

I +

High

High

∆P

Vd Low

Low

RESISTOR

ORIFICE

-

• • • •

P

Opposition to Current Flow

Opposition to Oil Flow

Resistance

Resistance

Opposes current flow Measured in Ohm Symbol is R or Ω Dependent on length, diameter, material and temperature

3

• • • •

Opposes oil flow Measured in psi, kPa, L/min Symbol is P Usually measured as a pressure drop (∆P) in a hydraulic circuit

EPG -ICHAPTER : COMPARISON STORING

Plate

Gas

Piston

Oil P

Dielectric Plate Fluid Input

CAPACITOR

Storage Devices • • • • •

P

Stores electrical charge Measured in Capacitance Symbol is C Unit is Microfarad (µF) Pair of conductors separated by a dielectric material

4

ACCUMULATOR

Storage Devices • Develops and stores pressure • Measured in pressure • Stores pressure as a result of forcing a volume of oil into a accumulator chamber

EPG -ICHAPTER : COMPARISON DIRECTIONAL

+

Q

I

Anode

Cathode

-

DIODE

Directional Controls • Current flows in one direction • Current flows when anode is more positive than the cathode • Use multimeter on "diode check" function and measure voltage drop

P

5

CHECK VALVE

Directional Controls • Fluid flows in one direction • Symbol indicates direction of oil flow

EPG -ICHAPTER : BASIC ELECTRICITY OHM FORM

Ohm's Law I = E/R

E=IxR

R = E/I

The above formulas will be used when describing electric and electronic circuits. Three types of electrical circuits will be discussed.

P

•

Series Circuits - Current can flow in only one path.

•

Parallel Circuits - Current can flow in more than one path.

•

Series-Parallel Circuits - Has both series and parallel paths for current to flow.

6

EPG -ICHAPTER : BASIC ELECTRICITY UNKNOWN CALCULATION

E I

R

E= I x R

E I

E R

E = I R

I

R

R= E I

Given two known in any electrical or electronic circuit, the unknown can be calculated. This slide shows an example of solving an unknown by placing a finger over the unknown and then performing the mathematical equation as shown. The Ohm's Law circle is a memory aid to help solve the equation for either voltage, current or resistance.

P

7

EPG -ICHAPTER : BASIC ELECTRICITY CALCULATE

I = 200 mA A ? E = _____

E I

R

E= I x R

Filament Resistance = 60 Ω

P

8

EPG -ICHAPTER : BASIC ELECTRICITY CALCULATE

I = 600 mA A

E R

I

E = 12V

R= E I

? Ω Filament Resistance = _____

P

9

EPG -ICHAPTER : BASIC ELECTRICITY CALCULATE

? mA I = ____ A

E R

I

E = 24 V

I= E R

Filament Resistance = 240 Ω

P

10

EPG -ICHAPTER : BASIC ELECTRICITY SERIES CIRCUITS

R1

R2

R3

• Sum of all voltage drops equal source voltage • Current flow through each load is the same • Total resistance is equal to sum of all the resistors

POS NEG

POS NEG

12V

12V

Series Circuits Series circuits may have several resistors (loads) connected to a voltage source. The important point to demonstrate when explaining series circuits is current flow which has only one path, and as such, the current flowing in the circuit passes through all resistances equally. Series circuits have the following features: •

The current through each resistor is the same.

• The voltage drop across each resistor will be different if the resistance values are different. •

P

The sum of all the voltage drops equal the source voltage.

11

EPG -ICHAPTER : BASIC ELECTRICITY CALCULATE

V1

V2

V3

8Ω

24Ω

16Ω

R1

R2

R3

A

POS NEG

POS NEG

12V

12V

Solution: • • •

P

Total circuit resistance = the sum of all the resistors or 8 + 24 + 16 = 48Ω Source voltage = the sum of the two batteries (connected in series) or 12 + 12 = 24V Current flow = source voltage divided by total resistance or 24 ÷ 48 = 500 mA or 0.5 amps

12

EPG -ICHAPTER : BASIC ELECTRICITY CALCULATE

6Ω

? ___Ω

4Ω

R1

R2

R3

2A

POS NEG

POS NEG

12V

12V

P

13

EPG -ICHAPTER : BASIC ELECTRICITY CALCULATE

6Ω

2Ω

4Ω

R1

R2

R3

?

POS NEG

POS NEG

12V

12V

P

14

short

EPG -ICHAPTER : BASIC ELECTRICITY CALCULATE

? 6Ω

2Ω

4Ω

R1

R2

R3

A

unwanted path POS NEG

POS NEG

12V

12V

P

15

EPG -ICHAPTER : BASIC ELECTRICITY PARALLEL CIRCUIT

R2

R1

POS NEG

12V

RULES

POS NEG

12V

• Voltage drop across each resistor is the same • Current flow through each resistor is different if the resistor values are different • The sum of the separate currents equals the total current flow in the circuit

Parallel Circuits In parallel circuits, the voltage drop across each resistor is equal to the potential of the current source since there is more than one path for current to flow through each resistor. Parallel circuits have the following features: •

The voltage drop across each resistor (load) is the same.

•

The current through each resistor will be different if the resistance are different.

• The sum of the separate currents equals the total current in the circuit.

P

16

R3

EPG -ICHAPTER : BASIC ELECTRICITY CALCULATE

R2

R1

6Ω

3Ω A

V2

V1 A1

POS NEG

12V

A2

POS NEG

12V

Voltage drop each resistor is the same as source voltage (24V).

• Total current flow is the sum of the separate currents (each path), not enough information to solve individual current flows without using Ohm's Law to solve other elements of the circuit. •

Solve the current flow through each load by using Ohm's Law. V1 = source voltage (24V) R1 = 3 Ω A1 = E1 ÷ R1 = 24 ÷ 3 = 8 (A1 = 8 amps) V2 = source voltage (24V) R2 = 6 Ω A2 = E2 ÷ R2 = 24 ÷ 6 = 4 (A2 = 4 amps)

P

17

2Ω V3

Solution: •

R3

A3

EPG -ICHAPTER : BASIC ELECTRICITY PAGE

R2

R1

3Ω ?

A

V1

V2 A1

POS NEG

12V

P

6Ω

18

A2

EPG -ICHAPTER : BASIC ELECTRICITY PAGE

R2

R1 ? ___

8A

POS NEG

24V

P

19

6Ω

EPG -ICHAPTER : BASIC ELECTRICITY CALCULATE

R2

R1

3Ω ?

A

V1

V2 A1

POS NEG

12V

P

6Ω

20

A2

EPG -ICHAPTER : BASIC ELECTRICITY CALCULATE

R2

R1 ? ___

8A

POS NEG

24V

P

21

6Ω

EPG -ICHAPTER : BASIC ELECTRICITY SERIES - PARALLEL

R1 R3

12V POS

NEG

R2

DRAWING EQUIVALENT CIRCUITS IS IMPORTANT IN SOLVING SERIESPARALLELS CIRCUITS

P

22

EPG -ICHAPTER : BASIC ELECTRICITY EQUIVALENT CIRCUIT STEPS

R1 R3 R2

12V POS

NEG

R1 R3 Re R2

12V

Step 1

POS

NEG

Re =

R1 x R2 R1 + R2

Re + R3 = Rt Re 12V

Step 2

POS

R3

Equivalent Circuit

NEG

Rt

12V

Step 3 P

23

POS

NEG

Equivalent Circuit

EPG -ICHAPTER : BASIC ELECTRICITY CALCULATE Rt

R1 = 6Ω

2+2=4 R3 = 2Ω Re = 2Ω

Re

POS

12V

R2 = 3Ω

12V NEG

Re =

POS

6 x 3 18 = =2 9 6+3 Rt = 4Ω

12V POS

NEG

TOTAL RESISTANCE IS 4 OHM THEN I = E / R = 12 / 4 = 3 amps

P

24

Equivalent Circuit

NEG

R3 = 2Ω

Equivalent Circuit

EPG -ICHAPTER : BASIC ELECTRICITY RECONSTRUCTING

Re = 2Ω

Et = 12 volts I t = 3 amps Rt = 4 ohms

E3 = I x R = 3 x 2 = 6V I3 = 3 A R3 = 2Ω

12V POS

NEG

6 volt drop

NOW IT IS NECESSARY TO RECONSTRUCT THE ORIGINAL CIRCUIT

P

25

EPG -ICHAPTER : BASIC ELECTRICITY FINAL CALCULATION

6V 1A 6Ω R1 R2

12V POS

P

NEG

26

6V 2A 3Ω

6V 3A 2Ω R3