Electrical Technology Topic 1 Sem 1 Politeknik
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Transcript of Electrical Technology Topic 1 Sem 1 Politeknik
DET1013 - ELECTRICAL TECHNOLOGY
Chapter 1:
Introduction to Electric Circuit
AUTHORS:AMINAH BINTI OTHMANJA’AFAR BIN SURADIJUNAIDA BINTI SHAARIZULKURNAIN BIN ABDUL HAMID
COURSE LEARNING OUTCOME
1. Apply the concept and principles of DC electrical circuit using different method and approach. (C3, PLO1)
2. Solve DC circuit problems using appropriate DC electrical laws and theorems. (C3, PLO2)
3. Conduct the laboratory activities of DC electrical circuit using appropriate electrical equipment. (P4, PLO5)
4. Demonstrate ability to work in team to complete assigned task during practical work sessions. (A3, PLO11)
TOPIC TITLE (RTA)CHAPTER TITLE RTA
1.0 INTRODUCTION TO ELECTRIC CIRCUIT
08:12
2.0 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
08:10
3.0 CAPACITORS AND CAPACITANCE 05:004.0 INDUCTORS AND INDUCTANCE 05:045.0 MAGNETIC CIRCUIT,
ELECTROMAGNETISM AND ELECTROMAGNETIC INDUCTION
04:04
LEARNING OUTCOME (2 Hours)1.1 Know standard symbols for electrical components.
1.1.1 Identify common symbols in electrical circuit diagrams.
1.2 Understand the general features of cells and batteries.1.2.1 Describe the difference between cells and batteries.1.2.2 Show the effects of different cell connections:
a. seriesb. parallelc. series-parallel
1.2.3 Calculate the total voltage of series sourcesa. with the same polarities.b. with opposite polarities.
Standard Symbol for Electrical Components
G V A
conductor /wire switchground/earth
Cell (dc supply) Battery (dc supply) AC supply
resistor inductor capacitor
galvanometer voltmeter ammeter
Cell
• A single unit of a primary or secondary battery that converts chemical energy into electric energy.
Battery
• A battery is a series of two or more connected cells, which changes chemical energy into electrical energy.
Relationship of Cells & Batteries
• A Battery is a combination of cells• Cell combination could be in
SERIES, PARALLEL & SERIES-PARALLEL
• Practically, a cell is also notified as a battery.
Series Connection Cells
Series Connection Cells
Total e.m.f., ET = E1 + E2 + E3 + E4
= 2.0 + 2.0 + 2.0 + 2.0 = 8V
Example 1.1Calculate total e.m.f. of the circuit below
Parallel Connection Cells
Parallel Connection Cells
Total e.m.f., ET = E1 = E2 = E3 = 2.0V
Example 1.2Calculate total e.m.f. of the circuit below
Series-Parallel Connection Cells
Series-Parallel Connection Cells
Total e.m.f. for series cells, ESeries = E1 + E2 + E3 + E4
= 2.0 + 2.0 + 2.0 + 2.0 = 8V Total e.m.f., ET = ESeries = 8V
Example 1.3Calculate total e.m.f. of the circuit below
Series Connection with same Polarities
Total e.m.f., ET = E1 + E2
= 8 + 6 = 14V
Example 1.4Calculate total e.m.f. of the circuit below
Series Connection with opposite Polarities
Total e.m.f., ET = E1 + E2
= 8 - 6 = 2V
Example 1.5Calculate total e.m.f. of the circuit below
i) 5V 11V 4V
B AAnswer: 20V
44V
ii) 44V B A
44VAnswer: 44V
SELF-EXERCISEQUESTION: Calculate total e.m.f. of each cells connection as follow.
ANSWER
ANSWER
SELF-EXERCISEQUESTION: Calculate total e.m.f. of each cells connection as follow.
iii) 4V 3V
B 2V 5V AAnswer: 7V
6V 1V
iv) 4V 4V 4VB A
Answer: 120V30 cells
ANSWER
ANSWER
SELF-EXERCISEQUESTION: Calculate total e.m.f. of each cells connection as follow.
v) 14V B A 20 cells 14V
Answer: 14V
14V
5V 5V 5V
vi) B 5V 5V 5V A
5V 5V 5V
Answer: 50V10 cells
ANSWER
ANSWER
LEARNING OUTCOME (1 Hour)1.3 Know electric current and quantity of electricity.
1.3.1 State the definition of electric current. 1.3.2 State the unit of charge. 1.3.3 Indicate charge or quantity of electricity Q from Q=It.
1.4 Know the main effects of electric current. 1.4.1 Identify the three main effects of electric current,
giving practical examples of each.
1.5 Understand resistance and resistivity 1.5.1 Explain that electrical resistance depends on four
factors. 1.5.2 Express that resistance R= ρI/A where ρ is the resistivity.
Electric current, I• Current: - motion of charge
- depends on the rate of flow of charge- electric fluid- unit of current is ampere (A)
• Equation: dq = changing of charge
I = dt = changing of timeI = current (ampere)
• For steady state condition: Q = charge (coulomb)I = , thus Q = It t = time (second)
Example 1.6If a current of 5 A flows for 2 minutes, find the charge transferred.
Electric current, I
Q = It = 5 x 2 x 60 = 600 C
Main Effect of Electric Circuit
1. Heat Effect - Example: soldering iron, water heater, fuse, bulb, cookers, electric fires, furnaces, kettles, iron
2. Magnetic Effect - Example: bells, relays, motors, generators, transformers, telephones, lifting magnets, car ignition
3. Chemical Effect - Example: cell and battery, electroplating
Resistance & resistivity
• Resistance – property of a component which restricts the flow of electric current.
• The value of resistance depends upon 4 factors:
1. Length, l2. Cross-sectional area, A3. resistivity, ρ4. temparature
Resistance & resistivity
• Equation:R = [Unit = Ω]
R = resistance [Ω] l = Length [m] A = Cross-sectional area [m2] ρ = resistivity [Ω.m]
• Resistivity is difference for different material
Example 1.7Calculate resistance of a 5m long conductor if it has cross sectional area and resistivity Ω.m
Resistance & resistivity
Resistance, R= =
= 1.5Ω
Resistor (R)
• A device that is manufactured to have specific resistance.
• Used to limit current flow and reduce voltage applied to other components.
• Basic unit is ohm (Ω)
Resistor (R)
• Different examples of resistors
SELF-EXERCISEi) In what time would a current of 1 A transfer a charge of 30 C?
Answer: 30sANSWER
ii) What would be the resistivity of 2m length conductor wire if the resistance value is 500Ω and the cross sectional area 0.5
Answer: 125µΩmANSWER
LEARNING OUTCOME (1 Hour)
1.6 Understand Ohm’s Law. 1.6.1 Explain Ohm’s Law. 1.6.2 Outline the procedure adopted when using Ohm’s Law
1.7 Apply Ohm’s Law in circuit. 1.7.1 Construct circuit to explain Ohm’s Law. 1.7.2 Use Ohm’s Law to find current, voltage and resistance in a circuit.
.
Ohm’s Law
• Ohm’s Law states that the current (I) through a conductor between two points is directly proportional to the potential difference or voltage (V) across the two points, and inversely proportional to the resistance (R) between them.
I =
Ohm’s Law Triangle
V = IR
I =
R =
V
RI
Simple Circuit
E = E.M.F. (Electromotive force)
- Generates from voltage source
- Example: cells / batteries
I =
Current =
From Ohm’s Law:
Simple Circuit
V drop = Voltage drop--------------------------- - appears when current, I flows through resistor,R.- Inverse polarity from E
V drop = IR
From Ohm’s Law:
Voltage = Current x Resistance
Simple Circuit
• A battery possess e.m.f. that produces DC current.
• A complete circuit should consist of at least 1 electricity source (battery) and 1 load (resistor)
Source Load
E R
+
-• Current will only produce when the
source (battery) is connected to the load (resistor) in close loop connection.
I
Simple Circuit
• A battery possess e.m.f. that produces DC current.
• A complete circuit should consist of at least 1 electricity source (battery) and 1 load (resistor)
Source Load
E R
+
- • Current will only produce when the source (battery) is connected to the load (resistor) in close loop connection.
I
• When current flows across resistor, R, voltage drop, Vd will be produced across R
+
-
Vd
Simple Circuit
• A battery possess e.m.f. that produces DC current.
• A complete circuit should consist of at least 1 electricity source (battery) and 1 load (resistor)
Source Load
E R
+
- • Current will only produce when the source (battery) is connected to the load (resistor) in close loop connection.
I
• When current flows across resistor, R, voltage drop, Vd will be produced across R
+
-
Vd
Simple Circuit (Example)
15V 10Ω
+
-
I
+
-
Vdrop
Example 1.8QUESTION: By referring to the circuit below, calculate:i) Current, Iii) Voltage drop across resistor 10Ω, Vdrop
Simple Circuit (Example)
15V 10Ω
+
-
I
+
-
Vdrop
i) Current, I = = = 1.5A
ii) Voltage drop, = IR = 1.5 x 10 = 15V
SELF-EXERCISEA 100 V battery is connected across a resistor and causes a current of 5 mA to flow. Determine the resistance of the resistor. If the voltage is now reduced to 25 V, what will be the new value of the current flowing?
R = 20kΩ
I = 1.25mA
ANSWER
ANSWER
LEARNING OUTCOME (2 Hours)1.8 Understand series, parallel and series-parallel connections.
1.8.1 Identify a series circuit.1.8.2 Explain the flow of current and voltage division in the series circuit.1.8.3 Identify a parallel circuit.1.8.4 Explain the voltage drop and the current division in the parallel circuit.1.8.5 Explain the equivalent resistance in series and parallel
circuits.1.8.6 Identify a combination of series and parallel circuit.1.8.7 Explain the total resistance for the combination of series and parallel circuit.
LEARNING OUTCOME (2 Hours)1.9 Apply series, parallel and series-parallel connections to dc circuit.
1.9.1 Construct a series connection circuit1.9.2 Calculate the flow of current and voltage division in the series circuit.1.9.3 Construct a parallel circuit.1.9.4 Calculate the voltage drop and the current division in the parallel circuit.1.9.5 Construct a series-parallel connection circuit.1.9.6 Calculate the equivalent resistance in series and parallel circuits.1.9.7 Calculate the total resistance for the combination of series and parallel circuit.1.9.8 Use of voltage divider in series circuit and use of current divider in parallel circuit.1.9.9 Solve problems related to series, parallel and combination of series and parallel circuits.
Series Circuit
• Is formed when any number of devices are connected end-to-end so that there is only one path for current to flow.
Series Circuit Characteristics
Series Circuit Characteristics
1. Resistances are additive RT = R1 + R2 + R3
2. The current flows throughout the circuit is same.I = IR1 = IR2 = IR3
3. Different resistors have their individual voltage drop
VR1 ≠ VR2 ≠ VR3
4. Total e.m.f equals to the sum of voltage drops across each resistorE = VR1 + VR2 + VR3
Equivalent resistance in series
RT = R1 + R2 + R3Equ. 1
• Applicable to any means of resistors.• Standard equation of series connection
resistors.
Equivalent resistance in series (resistors with same value)
r = resistance valuen = amount of resistors
Equ. 2 RT = r x n
• Applicable for any means of resistors with same value.
Voltage Divider Rule
VR1 = x E
Series Circuit (Example)
15V
4Ω 6Ω
8Ω
Example 1.9By referring to the circuit above, calculate:i) Total resistance of the circuit, ii) Current, Iiii) Voltage drop across resistor 6Ω,
Series Circuit (Example)
15V
4Ω 6Ω
8Ω
i) Rtotal = 4 + 6 + 8 = 18Ω
ii) I = = = 0.833A
iii) VR2 = IR2 = 0.833 x 6 = 5V or
VR2 = x E = x 15 = 5V
use VDR
Parallel Circuit
• Is formed when two or more devices are arranged in a circuit side by side so that current can flow through more than one path
Parallel Circuit Characteristic
Parallel Circuit Characteristic1. Total resistance can be determined from:
RT =
2. Different resistors have their individual current.IR1 ≠ IR2 ≠ IR3
3. Same voltage acts across all parts of the circuit E = VR1 = VR2 = VR3
4. Supplied current equals to the sum of different current flows through each resistor.
I = IR1 + IR2 + IR3
Equivalent resistance in parallel
RT = Equ. 1
• Applicable to any means of resistors.• Standard equation of parallel connection
resistors.
Equivalent resistance in parallel (2 resistors case)
RT = Equ. 2
• Applicable for 2 resistors connection only.
Equivalent resistance in parallel (same value case)
RT = r = resistance valuen = amount of resistors
Equ. 3
• Applicable for any means of resistors with same value.
Current Divider Rule (CDR)
IR1 = x IEqu. 1
• Applicable to any means of resistors.• Standard equation of current divider rule
Current Divider Rule (2 Resistors case)
IR1 = x IEqu. 2
• Applicable for 2 resistors connection only.
Parallel Circuit (Example)
6Ω
Example 1.10QUESTION: By referring to the circuit above, calculate:i) Total resistance of the circuit, ii) Current, Iiii) Voltage drop across resistor 8Ω, iv) Current through resistor 4Ω,
20V4Ω 8Ω
Parallel Circuit (Example)
20V6Ω4Ω 8Ω
i) Rtotal = = 1.846Ω
ii) I = = = 10.83A
iii) VR3 = E = 20V
iv) IR1 = = = 5A or
IR1 = x I = x 10.83 = 5A
CDR
Series-Parallel Circuit
For this diagram:• R1 is parallel with R2.• Ra is series with equivalent resistance of R1
and R2.
Total Resistance of Series-Parallel Circuit
• RT is the equivalent resistance of Ra, R1 and R2
referencepoint
• Start solving by calculating the total resistance of parts located farthest away from the reference point.
• Exception: if there are any series/parallel connection resistors at any part of circuit which is not farthest from the reference point, solve the total resistance of the series/parallel connection first. Then you can use the tips mentioned above to solve your problem.
Total Resistance of Series-Parallel Circuit (Example )
Example 1.11Calculate equivalent resistance, of the circuit below.
Total Resistance of Series-Parallel Circuit (Example )
Rb =
RT = Ra + Rb
Total Resistance of Series-Parallel Circuit (Example)
A
BRT
10Ω 10Ω 5Ω
6Ω3Ω
4Ω
8Ω
9Ω
Example 1.12Calculate the total resistance, RT of the circuit below.
Total Resistance of Series-Parallel Circuit (Example )
A
BRT
10Ω 10Ω 5Ω
6Ω3Ω
4Ω
8Ω
9Ω
Step 1: Identify any series/parallel connection (in between) and calculate the total resistance.
Ra
Ra = 4 + 8 = 12Ω
Total Resistance of Series-Parallel Circuit (Example )
A
BRT
10Ω 10Ω 5Ω
6Ω3Ω
4Ω
8Ω
9Ω
Step 1: Identify any series connection (in between) and calculate the total resistance.
Ra
Ra = 4 + 8 = 12Ω
12Ω
Total Resistance of Series-Parallel Circuit (Example )
A
BRT
10Ω 10Ω 5Ω
6Ω3Ω
9Ω
Step 2: Identify the farthest part from ref. point and calculate the total resistance.
Ra
Rb = 5 + 6 = 11Ω
12Ω
Rb
Total Resistance of Series-Parallel Circuit (Example )
A
BRT
10Ω 10Ω 5Ω
6Ω3Ω
9Ω
Step 2: Identify the farthest part from ref. point and calculate the total resistance.
Ra
Rb = 5 + 6 = 11Ω
12Ω Rb 11Ω
Total Resistance of Series-Parallel Circuit (Example )
A
BRT
10Ω 10Ω
3Ω
9Ω
Step 3: calculate the total resistance of next portion until reach ref. point.
Ra
Rc = = 2.36Ω
12Ω Rb 11Ω
Rc
Total Resistance of Series-Parallel Circuit (Example )
A
BRT
10Ω 10Ω
3Ω
9Ω
Step 3: calculate the total resistance of next portion until reach ref. point.
Ra
Rc = = 2.36Ω
12Ω Rb 11ΩRc
2.36Ω
Total Resistance of Series-Parallel Circuit (Example )
A
BRT
10Ω 10Ω
9Ω
Step 3: calculate the total resistance of next portion until reach ref. point.
Ra
Rd = 10 + 2.36= 12.36Ω
12ΩRc
2.36Ω
Rd
Total Resistance of Series-Parallel Circuit (Example )
A
BRT
10Ω 10Ω
9Ω
Step 3: calculate the total resistance of next portion until reach ref. point.
Ra
Rd = 10 + 2.36= 12.36Ω
12ΩRc
2.36ΩRd
12.36Ω
Total Resistance of Series-Parallel Circuit (Example )
A
BRT
10Ω
9Ω
Step 3: calculate the total resistance of next portion until reach ref. point.
Ra
Re = = 6.09Ω
12ΩRd
12.36Ω
Re
Total Resistance of Series-Parallel Circuit (Example )
A
BRT
10Ω
9Ω
Step 3: calculate the total resistance of next portion until reach ref. point.
Ra
Re = = 6.09Ω
12ΩRd
12.36Ω
Re6.09Ω
Total Resistance of Series-Parallel Circuit (Example )
A
BRT
10Ω
9Ω
Step 4: Finally, calculate the total resistance, RT of the circuit.
RT = 10 + 6.09 + 9 = 25.09Ω
Re 6.09Ω
Total Resistance of Series-Parallel Circuit (Example )
A B10kΩ 9kΩ3kΩ 3kΩ
6kΩ
6kΩ
Example 1.13Calculate the total resistance across point A - B
Total Resistance of Series-Parallel Circuit (Example )
A B10kΩ 9kΩ3kΩ 3kΩ
6kΩ
6kΩ
RT
Reference point
Ra
Ra = 3k + 3k = 6kΩ
Total Resistance of Series-Parallel Circuit (Example )
A B10kΩ 9kΩ3kΩ 3kΩ
6kΩ
6kΩ
RT
Reference point
Ra
Ra = 3k + 3k = 6kΩ
6kΩ
Total Resistance of Series-Parallel Circuit (Example )
A B10kΩ 9kΩ
6kΩ
6kΩ
RT
Reference point
Rb = = 2kΩ
6kΩ
Rb
Total Resistance of Series-Parallel Circuit (Example )
A B10kΩ 9kΩ
6kΩ
6kΩ
RT
Reference point
Rb = = 2kΩ
6kΩ
Rb
2kΩ
Total Resistance of Series-Parallel Circuit (Example )
A B10kΩ 9kΩ
RT
Reference point
RT = 10k + 2k + 9k = 21kΩ
2kΩ
∴
Total Resistance of Series-Parallel Circuit (Example )
Series-Parallel Circuit (Example)
+- 20V
2kΩ 4kΩ
8kΩ
6kΩ
18kΩ
20kΩ
Is
Example 1.14QUESTION: By referring to the circuit above, calculate:i) Equivalent resistance of the circuit, Rtotalii) Current from supply, Is iii) Current through resistor 18kΩiv)Voltage drop across resistor 8kΩ,
Series-Parallel Circuit (Example)2kΩ 4kΩ
8kΩ
6kΩ
18kΩ
20kΩ
+-
20V
Isi) Rtotal Calculation
• Temporarily, remove voltage source from the circuit.
• The open nodes leaved by your voltage source would be your reference point
Rtotal Ra
Ra = 4k + 8k + 6k = 18kΩ
Series-Parallel Circuit (Example)2kΩ 4kΩ
8kΩ
6kΩ
18kΩ
20kΩ
i) Rtotal Calculation
• Temporarily, remove voltage source from the circuit.
• The open nodes leaved by your voltage source would be your reference point
Rtotal
Ra
Ra = 4k + 8k + 6k = 18kΩ
18kΩ
Series-Parallel Circuit (Example)2kΩ
18kΩ18kΩ
20kΩ
i) Rtotal Calculation
• Temporarily, remove voltage source from the circuit.
• The open nodes leaved by your voltage source would be your reference point
Rtotal
Ra
Rb = = 9kΩ
Rb
Series-Parallel Circuit (Example)2kΩ
18kΩ18kΩ
20kΩ
i) Rtotal Calculation
• Temporarily, remove voltage source from the circuit.
• The open nodes leaved by your voltage source would be your reference point
Rtotal
Ra
Rb = = 9kΩ
Rb 9kΩ
Series-Parallel Circuit (Example)2kΩ
20kΩ
i) Rtotal Calculation
• Temporarily, remove voltage source from the circuit.
• The open nodes leaved by your voltage source would be your reference point
Rtotal
Rtotal = 2k + 9k + 20k= 31kΩ
Rb 9kΩ31kΩ
Series-Parallel Circuit (Example)ii) Is Calculation
• Place voltage source back to the circuit.
• Your current from source is calculated using Ohm’s Law
Rtotal
Is = = = 645.16 μA
31kΩ+-
20V
Is
Series-Parallel Circuit (Example)
+-
20V
2kΩ 4kΩ
8kΩ
6kΩ
18kΩ
20kΩ
645.16μA ii) I18 Calculation
• Use current divider rules (CDR) or any other relevant methods
I18
Series-Parallel Circuit (Example)
+-
20V
2kΩ 4kΩ
8kΩ
6kΩ
18kΩ
20kΩ
645.16μA ii) I18 Calculation
• Use current divider rules or any other methods relevantI18
18kΩRa
If Use CDR: I18 = x 645.16μ = 322.58 μA
Other method: I18 = = 322.58 μA
Series-Parallel Circuit (Example)
+-
20V
2kΩ 4kΩ
8kΩ
6kΩ
18kΩ
20kΩ
645.16μA ii) V8 Calculation
• Calculate the current flows through 8kΩ resistor first
• Use Ohm’s Law to calculate the Voltage drop
• Other method as Voltage Divider Rule (VDR) also could be used here if you understand well the technique
322.58μA 322.58μA
I8 = 645.16μ – 322.58μ= 322.58 μA
Series-Parallel Circuit (Example)
+-
20V
2kΩ 4kΩ
8kΩ
6kΩ
18kΩ
20kΩ
645.16μA ii) V8 Calculation
• Calculate the current flows through 8kΩ resistor first
• Use Ohm’s Law to calculate the Voltage drop
• Other method as Voltage Divider Rule (VDR) also could be used here if you understand well the technique
322.58μA 322.58μA
I8 = 645.16μ – 322.58μ= 322.58 μA
+
V8
-
V8 = IR = 322.58μ x 8k = 2.58V
SELF-EXERCISEFind the value of the total resistance, current from supply and voltage drop across resistor 90Ω in the diagram as below
= 24.5Ω
I = 2.041A
= 45.92V
ANSWER
ANSWER
50V
2Ω 4Ω
90Ω
22Ω
8Ω 8Ω
ANSWER
LEARNING OUTCOME1.10 Understand Delta–Star transformation. 1.10.1 Express formula required to transform from Delta to Star
and Star to Delta1.10.2 Illustrate circuits to show star and delta connections.1.10.3 Explain steps to solve problems involving Star-Delta transformation.
1.11 Apply the concept of Delta–Star transformation.1.11.1 Construct circuits to show star and delta connections.1.11.2 Solve problems involving Star-Delta transformation.
1.12 Understand electrical power and energy.1.12.1 Explain electrical power and energy.1.12.2 Express electrical power formula from Ohm’s Law and the unit.1.12.3 Calculate the electrical power and energy in a circuit.
Delta-Star Transformation
• Standard 3-phase circuits or networks take on two major forms with names that represent the way in which the resistances are connected, a Star connected network which has the symbol of the letter, Υ (wye) and a Delta connected network which has the symbol of a triangle, Δ (delta).
Delta-Star Transformation
R1 R3
R2
a
b
c
Ra
RcRb
Ra = Rb =
Rc =
Star-Delta Transformation
R1 R3
R2
a
b
c
Ra
RcRb
R1 =
R2 =
R3 =
Delta-Star (Example)
x y
4Ω
6Ω
8Ω
12Ω
10Ω
Example 1.15Calculate the total resistance, Rxy of the circuit below.
Rxy
Delta-Star (Example)
x y
4Ω
6Ω
8Ω
12Ω
10Ω
Ra
Rb
Rc
Ra = = 1.78Ω Rb = = 2.67Ω
Rc = = 1.33Ω
Convert --- Y
Delta-Star (Example)
x y
12Ω
10Ω
1.78Ω
2.67Ω
1.33Ω
Rd
Re
Rd = 1.33 + 12= 13.33 Ω
Re = 2.67 + 10= 12.67 Ω
Delta-Star (Example)
x y1.78Ω
10Ω
2.67Ω
12Ω1.33Ω
Rd
Re
Rd = 1.33 + 12= 13.33 Ω
Re = 2.67 + 10= 12.67 Ω
13.33Ω
12.67Ω
Delta-Star (Example)
x y1.78Ω
Rf = = 6.5Ω
13.33Ω
12.67Ω
Rf
Delta-Star (Example)
x y1.78Ω
Rf = = 6.5Ω
13.33Ω
12.67Ω
6.5Ω
Rxy = 1.78 + 6.5 = 8.28Ω
Electrical Power & Energy
• ELECTRICAL POWER is defined as the rate at which electrical energy is transferred by an electric circuit.
• The SI unit of power is Watt.• Equation:
Power, P = VI Equ. 1
V – voltage measured in Volts (V)
I – current measured in Ampere (A)
Electrical Power & Energy• From Ohm’s Law;
I = V/R and V = I*R
Hence Power, P = Equ.2
P = I2R Equ.3
Electrical Power & Energy
• ENERGY can be defined as capacity to do work
• The unit of energy is Joule• Equation :
Energy/Work Done, W = Pt
P – power measured in Watt (W)
t – time measured in seconds (s)
Electrical Power & Energy
20V
15Ω
25Ω
Example 1.16By referring to the circuit below, calculate:i) Power that’s supplied by the batteryii) Power that’s absorbed by 25Ω resistoriii) Energy supplied by the battery after 30siv) Energy absorbed by the 15Ω resistor after 2 hours
Electrical Power & Energy
20V
15Ω
25Ω
i) Power that’s supplied by the battery, PsRT
RT = 15 + 25 = 40ΩIT
IT = = = 0.5A
Use Equ. 1:
Power, Ps = V*I = 20 x 0.5 = 10W
Electrical Power & Energy
20V
15Ω
25Ω
ii) Power that’s absorbed by 25Ω resistor PL
0.5A
Use Equ. 3:
Power, PL = I2*R = 0.52 x 25 = 6.25W
Electrical Power & Energy
20V
15Ω
25Ω
iii) Energy supplied by the battery after 30s
0.5A
Energy, W = P*t = 10 x 30 = 300 J
Electrical Power & Energy
20V
15Ω
25Ω
iv) Energy absorbed by the 15Ω resistor after 2 hours
0.5A
Energy,
W = P*t = I2*R*t = 0.52 x 15 x 2 x 60 x 60 = 27 kJ
SELF-EXERCISEi) Diagrams below show a delta connection circuit
with its equivalent star connection circuit. If R1=20kΩ, R2=40kΩ and R3 =80kΩ, calculate Ra, Rb and Rc
11.43kΩ
5.71kΩ 22.86kΩ
ANSWER
ANSWERANSWER
SELF-EXERCISEii) With refer to the diagram as below, calculate power that supplied by the battery and power dissipation at resistor 40kΩ.
= 1.6mW
Ps = 4.8mWANSWER
ANSWER
RECAP• Cell and battery are sources of DC type of
electricity.• Voltage, current and resistance are recognized
as three basic elements of electrical circuit which contribute in Ohm’s Law.
• Electrical circuit can be constructed in series, parallel and combination of series-parallel connection.
• Star-Delta transformation technique is required to analyze network that involve Star/Delta connection.
• Power and Energy is the product of voltage and current elements of a circuit.
REFERENCES
Main: John Bird (2010). Electrical Circuit Theory & Technology. Fourth
Edition. Newness. (ISBN: 978-0-08-089056-2)
Additional: 1. Allan R. Hambley (2011). Electrical Engineering, Principles
and Applications, Fifth Edition. Prentice Hall. (ISBN-13: 978-0-13-213006-6)
2. B.L. Theraja (2010).Textbook of Electrical Technology .S Chand & Co Ltd. (ISBN: 978-8121924900)
REFERENCES
3. Darren Ashby (2011). Electrical Engineering 101, (3rd Ed ) [Paperback] Elsevier Inc. (ISBN: 978-0123860019)
4. John Bird. (2010). Electrical And Electronic Principles And Technology. Fourth Edition. Newness. (ISBN: 978-1-85617-770-2)
5. Meizhong Wang. (2010). Understandable Electric Circuits First edition © 2005 Higher Education Press, China, English translation ©2010 The Institution of Engineering and Technology. (ISBN 978-0-86341-952-2)
6. V. K. Mehta (2010). Principles of Electrical Engineering and Electrical [Paperback] S Chand & Co Ltd. (ISBN: 978-8121927291)