BACHELOR OF TECHNOLOGY (BTech) IN MANUFACTURING
TECHNOLOGY
DEPARTMENT: Mechanical Engineering
OPTION: Manufacturing Technology
MODULE TITLE AND CODE: Electrical and Electronics Concepts in
Manufacturing (GENEE801)
NUMBER OF CREDITS: 10
RQF LEVEL: 8
ACADEMIC YEAR: 2023 - 2024
i
TABLE OF CONTENT
LEARNING OUTCOME 1. APPLY ELECTRICITY .......................................................................- 1 1.1. INTRODUCTION ...................................................................................................................- 1 1.1.1. Electrical circuit ................................................................................................................- 1 1.1.2. Basic electrical parameters................................................................................................- 3 1.1.3. The instruments used to measure electrical quantities ......................................................- 6 1.1.4. Some of the terminology related to electricity and electrical circuit ................................- 6 1.1.5. Electrical effects................................................................................................................- 8 1.1.6. Difference between DC and AC electricity ......................................................................- 9 1.2. DC ELECTRICAL CIRCUIT................................................................................................- 11 1.2.1. Ohm’s Law......................................................................................................................- 11 1.2.2. Calculation of electrical power and electrical energy .....................................................- 12 1.2.3. Connection of resistors....................................................................................................- 13 1.2.4. Current divider rule.........................................................................................................- 16 1.2.5. Voltage divider rule ........................................................................................................- 17 1.2.6. Superposition Theorem ...................................................................................................- 20 1.3. AC ELECTRICAL CIRCUIT................................................................................................- 24 1.3.1. Generation of single phase emf.......................................................................................- 24 1.3.2. Factors affecting AC voltage ..........................................................................................- 25 1.3.3. Equation of alternating voltage and current ....................................................................- 25 1.3.4. Important A.C. terminology............................................................................................- 28 1.3.5. Values of alternating voltage and current .......................................................................- 29 1.3.6. Important relations ..........................................................................................................- 37 1.3.7. Phase difference ..............................................................................................................- 38 1.3.8. Types of electrical loads .................................................................................................- 40 1.3.9. Single phase AC voltage to series circuits ......................................................................- 46 1.3.10. Power in AC circuits .....................................................................................................- 55 1.3.11. Energy in AC circuits....................................................................................................- 56 1.3.12. Power factor (𝑪𝒐𝒔 𝜃) ....................................................................................................- 56 1.3.13. Active and reactive components of current...................................................................- 57 1.3.14. Single phase AC voltage to parallel circuits .................................................................- 62 -
ii
LEARNING OUTCOME 1. APPLY ELECTRICITY
1.1. INTRODUCTION
1.1.1. Electrical circuit
Definition
Electrical circuit is a system consisting of conductors connected to components which use
electron flow for their operation.
It may also be defined as the path taken by electric current to flow through a consuming
device.
The main parts of an electrical circuit
The electrical circuit comprises the following main parts:
(i)
Electrical source / supply: It supplies the flow of electric charges (electricity).
The source may be dc source or ac source.

DC source: It is the source that produces direct voltage continuously and
has the ability to deliver direct current.
Examples: Battery, dc generator, etc.

AC source: It is the source that produces alternating voltage continuously
and has the ability to deliver alternating current.
Example: ac generator (alternator)
(ii)
Conducting devices: Provide an electrical path taken by electric current.
Examples: Wires or cables.
(iii)
Controlling devices: Allow the user to control by turning the circuit on or off.
Example: Switches
(iv)
Protective or Safety devices: Prevent damage to the circuit in the event of short
circuit, etc.
Examples: Fuses, circuit breakers, etc.
(v)
Consuming devices (Electrical loads): Electrical Load is a device or an electrical
component that consumes electrical energy and convert it into another form of
energy. The load changes electrical energy into useful form.
Examples: Electric lamps, electric heaters, iron, electric kettle, electric stove, fans,
motors, etc.
-1-
The wire or conductor through which the current starts from the dynamo to the bulb is called
positive wire, main wire or live wire. In a.c. is called the phase wire.
The wire through which the current returns back to the source of supply after passing the bulb
is called negative wire or return wire. In a.c. it is called neutral wire.
The two conductors through which the current is taken from the supply source and returns
back are known as supply mains.
-2-
1.1.2. Basic electrical parameters
(i) Electromotive force (e.m.f) and voltage (potential difference)
Electromotive force (e.m.f) may be defined in one of the following ways:

The force that causes an electric current to flow in an electric circuit

The force in volts provided by the source (battery or generator) to move the electrons
around the circuit.

The electric potential difference developed by any source of electrical energy such as
battery or generator.
The emf is produced by chemical action in a battery/cell or conversion of mechanical energy
into electrical energy in a generator.
In the case of battery, it is the potential difference between two electrodes.
A device that maintains potential difference between two points is said to develop
electromotive force (e.m.f.). A simple example is that of a cell.
Voltage, also called Potential difference, electric pressure or electric tension, is defined as
the difference of electric potentials of the two points in an electric circuit.
In other words, it is defined as the amount of energy required to move an electric charge from
one point to another in an electric circuit.
Mathematically, voltage is defined as the ratio of the work done to the electric charge.
Voltage, V =
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 (𝑾)
𝑪𝒉𝒂𝒓𝒈𝒆 (𝑸)
According to Ohm’s law, the voltage is given by the product of resistance of the conductor
and the current flowing through it, i.e., Voltage, V = IR
In a circuit, when the current flows through a resistor, the potential difference across the
resistor is known as voltage drop across it, i.e. V = IR
-3-
Types of voltage
Depending on the variation in polarity of the voltage with respect to time, it is classified into
two basic type, namely,

Direct voltage

Alternating voltage
When the polarity of voltage in an electric circuit does not change with time, it is called
direct voltage. It is named so because it acts in only one direction.
The direct voltage may be a constant voltage or a pulsating voltage.
In case of constant voltage, the magnitude of voltage does not change with time, while in case
of the pulsating voltage, the magnitude changes with respect to time.
Pulsating voltage is a periodic voltage which changes in value but never changes direction.
It is caused by voltage rectified from AC that has not been filtered.
On the other hand, when magnitude of the voltage changes continuously and direction
changes periodically, then voltage is known as alternating voltage.
The alternating voltage can have different types of waveforms like sinusoidal waveform,
square waveform, triangular waveform and many more.
In actual practice, the alternating voltage mostly has sinusoidal waveform due to several
technical and economic reasons.
-4-
Sources of voltage
The energy sources that we use in electrical circuits and systems are voltage sources.
The most common sources of voltage are cell, battery, dc generator, ac generator (alternator),
etc.
In electric circuits, the voltage sources are represented by the following symbols:
-5-
(ii) Electric current (I)
It may be defined in one of the following ways:
(i)
The movement of electric charge in an electric circuit.
(ii)
The rate of flow of charge flow past a given point in an electric circuit.
Current (I) = 𝑪𝒉𝒂𝒓𝒈𝒆 (𝑸)
𝑻𝒊𝒎𝒆 (𝒕)
Thus, Q = It
(iii) Resistance (R)
It is the opposition offered by a substance to the flow of electric current.
The unit of resistance is ohm (Ω).
(iv) Electric Power (P)
It is the rate at which work is done in an electric circuit.
(v) Electrical Energy (W)
It is the total work done in pushing charges round an electric circuit.
1.1.3. The instruments used to measure electrical quantities
Electrical Quantity
Instrument
Voltage
Voltmeter
Current
Ammeter
Resistance
Ohmmeter
Power
Wattmeter
Energy
Energy meter
1.1.4. Some of the terminology related to electricity and electrical circuit

Closed circuit: The complete path for the flow of electric current through the load.

Open circuit: The circuit is called open circuit if one of the supply wires is
disconnected or the fuse burns out, then the current will not flow through the bulb.
Open circuit is a gap or break or interruption in a circuit path.
When there is a break in any part of a circuit, that part is said to be open-circuited.
No current can flow through an open.
Since no current can flow through an open, according to Ohm’s Law, an open has
-6-
infinite resistance (𝑅 = 𝐸 = 𝐸 = ∞).
𝐼
0
An open circuit may be as a result of component failure or disintegration of a
conducting path such as the breaking of a wire.

Short circuit: The circuit is called a short circuit if the supply mains are connected
directly by a piece of wire without any load.
Since in this circuit the value of the current is much greater than the closed circuit, the
fuse gets blown off.
A short circuit or short is a path of low resistance.
When a short circuit occurs, the resistance of the circuit becomes low.
As a result, current greater than the normal flows which can cause damage to circuit
components.
The short circuit may be due to insulation failure, components get shorted, etc.

Electrical overload: Electrical overload refers to a situation where an electrical
circuit is carrying more current than it is designed to handle.
This can occur when too many electrical devices are connected to a single circuit.
Electrical overloads can lead to overheating or excessive generation of heat, and the
risk of fire or damage to equipment.

Resistor: The term resistor refers to a device that acts as a two-terminal passive
electrical component that is used to limit or regulate the flow of electrical current in
electrical circuits.
It allows us to introduce a controlled amount of resistance into electrical circuit.

Power rating: Electrical components are often given power rating.
The power rating, in watts, indicates the rate at which the device converts electrical
energy into another form of energy, such as light, heat, or motion.

Efficiency: Efficiency of an electrical device is the ratio of power converted to useful
energy divided by the power consumed by the device.

Equivalent circuit: In the study of electricity, it is often necessary to reduce a
complex circuit into a simpler form.
Any complex circuit consisting of resistances can be redrawn to a basic equivalent
circuit containing the voltage source and a single resistance representing total
resistance. This process is called reduction to an EQUIVALENT CIRCUIT.
-7-
1.1.5. Electrical effects
Electricity is the most widely used form of energy we cannot see, but we can see its effects.
Electricity can produce different types of effects.
According to the law of conservation of energy, energy can be transformed from one form to
another. Basically the effect of electricity is conversion of electrical energy into other form of
energy. When electric current flows in a circuit, it produces various effects such as:
(i)
Lighting effect: When electric current is passed through a conductor, it becomes
very hot due to which it emits light.
Appliances which use lighting effect of electric current are incandescent lamps,
fluorescent lamps, etc.
(ii)
Heating effect: When current flows through a conductor, heat energy is generated
in the conductor.
The heating effect of an electric current depends on the following factors:
 The resistance 𝑅 of the conductor. A higher resistance produces more heat.
 The amount of current 𝐼.
The higher the current the larger the amount of heat generated.
 The time 𝑡 for which current flows.
The longer the time the larger the amount of heat produced.
Hence the heating effect produced by an electric current 𝐼, through a conductor of
resistance 𝑅, for a time 𝑡, is given by: 𝐻 = 𝐼2𝑅𝑡.
This equation is called the Joule’s equation of electrical heating.
Some appliances which use the heating effect of electric current are: Electric
kettle, electric cooker, electric oven, toaster, electric heater, electric iron, etc.
(iii)
Chemical effect: When current flows through a liquid known as electrolyte,
chemical changes occur in the liquid and in metals immersed in it and connected
to the circuit. This is known as the chemical effect as in battery charging.
(iv)
Magnetic effect: The current flowing in a conductor always produces a magnetic
field around the conductor. This is known as the magnetic effect of electric
current. Electromagnets are used in the following situations: electric bells, circuit
breakers, electromagnetic relays, electromagnetic door locks, etc.
(v)
Mechanical effect: Working of an electric motor is based on the mechanical
effect of electric current.
-8-
In the motor, when a current is passed through a rectangular coil of wire placed in
a magnetic field, the coil rotates continuously.
1.1.6. Difference between DC and AC electricity
Definition
DC electricity
AC electricity
DC is defined as
Alternating voltage or current is one whose
unidirectional flow of
magnitude changes continuously with time
current; current only flows in
between zero and a maximum value and whose
one direction.
direction reverses periodically.
It may also be defined as the voltage or current
which changes its polarity as well as magnitude
with respect to time.
Direction of
Current flows only in one
Current reverses direction in equal intervals of
voltage and
direction.
time.
current
There is change of the polarities every halfcycle.
Magnitude
Magnitude of voltage and
Magnitude of voltage and current changes
of voltage
current remains constant.
continuously with time between zero and
and current
maximum value.
Voltage and current oscillate in a sinusoidal
waveform
-9-
Type
Pure and pulsating
Sinusoidal
Frequency
It has no frequency
It has frequency
(Frequency is zero)
Generation
It is produced by:
It is produced by alternators (AC generator).

Batteries
Generator operates on the principle of

Solar cells
electromagnetic induction.

DC generators
Electromagnetic induction is the phenomenon

Conversions from AC
in which a changing magnetic field induces an
to DC by rectifiers
electromotive force (emf).
etc.
The change in magnetic field can be achieved by

Batteries provide DC, which
either rotating the coil within a stationary
is generated from a chemical
magnetic field or rotating a magnetic field
reaction inside of the battery.
around a stationary coil.
Generator converts mechanical energy (rotation
of the armature) into electrical energy
(alternating current) through the creation of a
changing magnetic field.
Conversion
It can be converted into AC
It can be converted into DC by using rectifier.
by using inverter.
Load type
DC load is generally resistive
AC load is resistive, inductive or capacitive
in nature
Types of
In dc circuits, there is only
In ac circuits, there are three types of power:
power
active power (P).
Active power (P), reactive power (Q) and
- 10 -
DC does not generate
apparent power (S)
reactive power.
Power
Power factor is 1
Power factor varies between 0 and 1
It can be stored in batteries
It cannot be stored
factor
Storage
Applications It is commonly used in
electronic devices and low-
It is widely used in homes, businesses and
industries
voltage applications
1.2. DC ELECTRICAL CIRCUIT
DC circuit is a closed path in which the direct current flows.
DC circuits are ones powered by a DC voltage source or voltage source that pushes current in
one direction only.
The current flows in only one direction and it is mostly used in low voltage applications.
The resistor is the main component of the DC circuit.
Examples of dc voltage sources: Batteries, dc generators, solar cells, etc.
Examples of dc loads: Lap top, cell phone, radio, television, etc.
A simple DC circuit is shown in the figure below which contains a DC source (battery), a
load lamp, a switch, connecting leads, and measuring instruments like ammeter and
voltmeter.
1.2.1. Ohm’s Law
The ratio of voltage (𝑽) between the ends of a conductor to the current (𝑰) flowing between
them is constant, provided the physical conditions (e.g., temperature etc.) do not change.
i.e. 𝑽𝟏 = 𝑽𝟐 = 𝑽𝟑 = … = 𝑽𝒏 = 𝑹 = Constant
𝑰𝟏
𝑰𝟐
𝑰𝟑
𝑰𝒏
The current through any two points of the conductor is directly proportional to the potential
- 11 -
difference applied across the conductor, and indirectly proportional to the resistance of the
conductor, provided the physical conditions (e.g., temperature etc.) do not change.
Mathematical equation that describes this relationship: 𝑰 =
𝑽
𝑽
or 𝑹 = or 𝑽 = 𝑹𝑰
𝑹
𝑰
where:
𝑹 is the resistance of the conductor between the two points considered.
1.2.2. Calculation of electrical power and electrical energy
Electrical Power (P)
𝑷=𝑾
𝒕
Where 𝑾: is electrical energy in joules or watt-seconds
t: is the time in seconds
P: the electric power in watt (W).
Electric power can also be calculated as the product of voltage and current. i.e., P = VI
By Ohm’s Law, 𝑽 = 𝑹 or 𝑽 = 𝑹𝑰
𝑰
The electrical power consumed by an electrical appliance is given by:
P = VI, by Ohm’s Law, V = RI, thus: P = (RI) I = R𝐼2
𝑉
𝑉
𝑉2
𝑅
𝑅
𝑅
By Ohm’s Law, I = , thus: P = V ( ) =
P=
𝑽𝟐
P = R𝑰𝟐 or I =√
𝑷
𝑹
or V = √𝑷𝑹
𝑹
Electrical Energy (W)
Electrical energy is calculated by the product of power and time. 𝑾 = 𝑷 x 𝒕
It is measured in joules or watt-seconds, but when dealing with large amount of energy, the
unit used is kilowatt-hour (𝑲𝒘𝒉).
Where: 1Kwh = 1Kw x 1h or Energy in Kwh = Power in Kw x time in hours.
1Kwh = 1000W x 3600 Seconds or 1 Kwh = 3.6 x 𝟏𝟎𝟔 watt-seconds or joules
𝑄
𝑄
𝑡
𝑡
It is known that: I = 𝑄 and P = 𝑊, W = Pt and P = VI, therefore, P = 𝑉 and W = 𝑉 𝑡 = VQ
𝑡
𝑡
W = VQ or V = 𝑊
𝑄
- 12 -
EXERCISES
1. An electric kettle has a resistance of 30Ω.
(i)
What current will flow when it is connected to a 240V supply?
(ii)
Find the power rating of the kettle
(iii)
Calculate the energy consumed by the kettle in one hour.
2. An electric lamp consumes 100W at a supply voltage of 220V. Calculate:
(i)
Current flowing through the lamp
(ii)
Its resistance
(iii)
Energy consumed by the lamp in 50 minutes.
(Express the answer in both joules and Kwh).
3. An electrical installation consists of the following loads:
(i)
Six lamps of 40W each working 4 hours per day.
(ii)
Two fluorescent tubes of 125W each working 2 hours per day.
(iii)
One 1000W electric heater working 3hours per day.
Calculate the total energy consumed (in Kwh) for a month of June.
4. A voltage of 100V is applied across a resistor of 500Ω.
(i)
Calculate the current through the resistor.
(ii)
If the voltage is changed to 110V, determine the new value of resistance
required to keep the same current.
(iii)
If the resistance is changed to 3300Ω, calculate the voltage required to pass
10mA through the resistance.
5. The current in a circuit of a certain voltage and resistance is 10A.
What will be the current when the voltage is tripled and the resistance is doubled?
1.2.3. Connection of resistors
Series connection
Series circuits are types of electrical circuits in which all components are connected in a
chain so that the same current flows through all of them.
A series circuit is defined as a circuit that has only one path for the current flow.
- 13 -
𝑉𝑆 = 𝑉1 + 𝑉2 + 𝑉3 + ... + 𝑉𝑛
𝑉𝑆 = 𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3 + ... + 𝐼𝑅𝑛
𝐼𝑅𝑒𝑞 = 𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3 + ... + 𝐼𝑅𝑛
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 + ... + 𝑅𝑛
Main characteristics of series connection of resistors
1. The same current flows through each part of the circuit.
𝐼𝑡 = 𝐼1 = 𝐼2 = 𝐼3 = ... = 𝐼𝑛
2. The equivalent resistance is equal to the sum of the individual resistances.
𝑅𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 𝑅1 + 𝑅2 + 𝑅3 + ... + 𝑅𝑛
3. The equivalent resistance is greater than the greatest of the resistances.
4. Supply voltage is equal to the sum of individual voltage drops.
𝑉𝑆 = 𝑉1 + 𝑉2 + 𝑉3 + ... + 𝑉𝑛
5. Different resistors have their individual voltage drop.
6. If n resistors, each of resistance R Ω, are connected in series, then, 𝑅𝑇 = nR
7. A series resistor circuit can be considered to be a voltage divider circuit.
8. The total power is equal to the sum of the individual powers used by each circuit
component. 𝑃𝑡 = 𝑃1 + 𝑃2 + 𝑃3 + ... + 𝑃𝑛
9. If one component in a series circuit fails, it interrupts the entire circuit, and no current
can flow.
10. In a series connection of identical light bulbs, the brightness of each bulb decreases as
more bulbs are added, as the total voltage across the circuit is shared among them.
Parallel connection
It is an electrical circuit in which multiple components are connected in such a way that, one
end of each component is attached to a common point, and the other end is attached to a
different common point.
- 14 -
Parallel circuits are type of electrical circuits in which the same voltage occurs in all
components, with current divided among the components based on their resistances.
Resistors in parallel connection are connected to the same nodes from both ends.
A parallel circuit is a circuit having more than one current path connected to a common
voltage source.
𝐼𝑡 = 𝐼1 + 𝐼2 + 𝐼3 + ... + 𝐼𝑛
𝐼𝑡 =
𝑽
𝑹𝒆𝒒
𝟏
𝑹𝒆𝒒
𝑽
𝑹𝟏
=
=
+
𝑽
𝑹𝟏
𝟏
𝑹𝟏
𝑽
𝑹𝟐
+
+
+
𝑽
𝑹𝟐
𝟏
𝑹𝟐
𝑽
𝑹𝟑
+
+
+…+
𝑽
𝑹𝟑
𝟏
𝑹𝟑
𝑽
𝑹𝒏
+…+
+…+
𝑽
𝑹𝒏
𝟏
𝑹𝒏
Main characteristics of parallel connection of resistors
1. The same voltage acts across each part of the circuits.
𝑉𝑠 = 𝑉1 = 𝑉2 = 𝑉3 = … = 𝑉𝑛
2. The reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the
individual resistances.
𝟏
𝑹𝒆𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕
=
𝟏
𝑹𝟏
+
𝟏
𝑹𝟐
+
𝟏
𝑹𝟑
+…+
𝟏
𝑹𝒏
3. The equivalent resistance is always less than the smallest of the resistances.
4. The total current is equal to the sum of branch currents.
𝐼𝑡 = 𝐼1 + 𝐼2 + 𝐼3 + ... + 𝐼𝑛
5. Different resistors have their individual currents.
6. If n resistors, each of resistance R Ω, are connected in parallel, then, 𝑅𝑇 =
𝑅
𝑛
7. A parallel resistor circuit can be considered to be a current divider circuit.
8. The total power is equal to the sum of the individual powers used by each circuit
component. 𝑃𝑡 = 𝑃1 + 𝑃2 + 𝑃3 + ... + 𝑃𝑛
- 15 -
9. Parallel connections offer flexibility in terms of adding or removing components
without affecting the operation of the other components.
This makes parallel connections suitable for the applications that requires modularity
and expandability.
10. In a parallel connection of multiple light bulbs, the combined illumination of all the
bulbs will be brighter than a single bulb.
1.2.4. Current divider rule
Let consider the parallel circuit with two resistors connected in parallel to a voltage source 𝑉.
1
𝑅𝑡
=
𝐼 =
1
𝐼 =
2
1
𝑅1
𝑉
+
=
𝑅1
𝑉
=
1
𝑅2
=
𝑅1 + 𝑅2
𝑅1 𝑅2
𝑅1 𝑅2
𝐼𝑡 (𝑅 + 𝑅 )
1
2
𝑅1
𝐼𝑡 ( 𝑅1 𝑅2 )
𝑅2
𝑅1 + 𝑅2
𝑅2
𝑅𝑡 =
,
=𝐼 (
𝑡
𝑅1 𝑅2
=𝐼 (
𝑡
𝑅1 𝑅2
)
𝑅1 + 𝑅2
𝑅1 + 𝑅2
)
1
𝑅1
1
𝑅2
𝑉 = 𝐼 𝑅 = 𝐼 ( 𝑅1 𝑅2 )
𝑡 𝑡
𝑡
𝑅1 𝑅 2
𝑅1 + 𝑅 2
𝑅1+ 𝑅2
=𝐼 (
𝑡
𝑅2
=𝐼 (
𝑡
𝑅1
)
𝑰 =𝑰 (
𝑹𝟐
)
𝑰 =𝑰 (
𝑹𝟏
𝑅1+ 𝑅2
𝑅1+ 𝑅2
𝟏
𝟐
𝒕 𝑹𝟏 + 𝑹𝟐
𝒕 𝑹𝟏 + 𝑹𝟐
)
)
Current distribution in parallel circuit:
Current in one of the two parallel resistors = Total current x 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟
𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟𝑠
Example
For the following circuit, find the current through each resistor.
- 16 -
Solution
𝑅𝑡 =
𝐼𝑡 =
𝑅1 𝑅2
𝑅1 + 𝑅 2
𝑉
48
𝑅𝑡
=
20 + 30
= 12Ω
= 4A
𝑅2
𝑅1
𝑡 𝑅 +𝑅
1
2
) = 2.4 𝐴
30
)=4(
𝑡 𝑅 +𝑅
1
2
𝐼 =𝐼 (
2
20 𝑋 30
12
𝐼 =𝐼 (
1
=
20 + 30
)=4(
20
) = 1.6 𝐴
20 + 30
1.2.5. Voltage divider rule
Consider a circuit that has three resistors connected in series.
𝑅𝑡 = 𝑅1 + 𝑅2 + 𝑅3
𝐼=
𝑉𝑆
𝑅𝑡
=
𝑉𝑆
𝑅1 + 𝑅2 + 𝑅 3
𝑉 = 𝐼𝑅 = (𝑉𝑆) 𝑅 ,
1
1
𝑅𝑡
Or: 𝑽𝟏 = (𝑹𝟏) 𝑽𝑺
𝑹𝒕
1
𝑉 = 𝐼𝑅 = (𝑉𝑆) 𝑅 ,
2
2
𝑅𝑡
2
𝑽𝟐 = (𝑹𝟐) 𝑽𝑺 ,
𝑹𝒕
𝑉 = 𝐼𝑅 = (𝑉𝑆) 𝑅
3
3
𝑅𝑡
3
𝑽𝟑 = (𝑹𝟑) 𝑽𝑺
𝑹𝒕
Exercises
1. Two resistors connected in series have a resistance of 18 Ω and when connected in
parallel have a resistance of 4 Ω. Find the value of resistances.
- 17 -
2. When a resistor is placed across a 240V supply, the current is 8A.
What is the value of resistor that must be placed in parallel to increase the supply
current to 20A?
3. Two resistors of 4Ω and 6Ω are connected in parallel.
If the total current is 30A, find the current through each resistor.
4. Three resistors 4Ω, 12Ω and 6Ω are connected in parallel.
If the total current taken is 12A, find the current through each resistor.
5. A 50Ω resistor is in parallel with a 100Ω resistor. The current in 50Ω resistor is 8A.
What is the value of third resistance to be added in parallel to make the total current
28A?
Exercises
For the following circuits, find:
(i)
Total resistance and current
(ii)
Current through each resistor
(iii)
Voltage across each resistor
(iv)
Power consumed by each resistor
(v)
Total power consumed.
1.
2.
3.
- 18 -
4.
5.
6.
7.
8.
9.
- 19 -
10.
1.2.6. Superposition Theorem
The essence of superposition theorem is that an emf acting in a linear network produces the
same effect whether it acts alone or in conjunction with other emfs.
The theorem may be stated as under:
“In a linear network containing more than one source of emf, the resultant current in any
branch is the algebraic sum of the currents that would be produced by each emf acting alone,
all other sources of emf, being replaced meanwhile by their respective internal resistances”.
Example:
In the network shown in the figure, find the magnitude and direction of current in each
resistor by superposition theorem:
Solution:
Since there are two sources of emf, two circuits are required for analysis by superposition
theorem.
1.
- 20 -
Total resistance (𝑅 ) = 3 + 4 𝑋 2 = 4.33Ω
𝑡
4+2
35
𝐼1 = 4.33 = 8.08A
𝐼 = 8.08 (
2
𝐼 = 8.08 (
3
4
) = 5.39A
2+4
2
) = 2.69A
2+4
2.
Total resistance (𝑅′) = 4 + 3 𝑋 2 = 5.2Ω
𝑡
𝐼′
3
=
40
= 7.69 A
5.2
𝐼′ = 7.69 (
2
𝐼′ = 7.69 (
1
3+2
3
) = 4.61A
2+3
2
) = 3.08A
2+3
The currents in 𝑅1, 𝑅2 and 𝑅3:
Current in 𝑅1 = 𝐼1 - 𝐼′1= 8.08 – 3.08 = 5A
Current in 𝑅2 = 𝐼2 + 𝐼2′ = 5.39 + 4.61 = 10A
Current in 𝑅3 = 𝐼3′ - 𝐼3 = 7.69 – 2.69 = 5A
- 21 -
Exercises:
In the networks shown in the figures, find the magnitude and direction of current in each
resistor by superposition theorem:
1.
2.
3.
4.
- 22 -
5.
6.
- 23 -
1.3. AC ELECTRICAL CIRCUIT
1.3.1. Generation of single phase emf
Electromotive force (emf) is produced by alternators (AC generator).
Generator operates on the principle of electromagnetic induction.
Electromagnetic induction is the induction of electromotive force (e.m.f.) in a conductor by
varying the magnetic flux linking with the conductor.
It is the phenomenon in which a changing magnetic field induces an electromotive force.
Faraday’s Laws of Electromagnetic Induction
First Law
It is stated as follows:
(i) When the magnetic flux linking a conductor or coil changes, an electromotive force
(e.m.f.) is induced in the conductor or coil.
(ii) When a conductor or coil moves through a magnetic field, an electromotive force
(e.m.f) is induced across the conductor or coil.
This change in magnetic flux linkages can be brought about in the following two ways:
(i)
The conductor is moved in a stationary magnetic field in such a way that the
magnetic flux linking it changes in magnitude.
The emf induced in this way is called dynamically induced emf (as in an a d.c.
generator).
It is so called because emf is induced in the conductor which is in motion.
(ii)
The conductor is stationary and the magnetic field is moving or changing.
The emf induced in this way is called statically induced e.m.f. (as in alternator and
- 24 -
a transformer).
It is so called because the emf is induced in a conductor which is stationary.
Second Law
The magnitude of induced emf in a coil is equal to the rate of change of magnetic flux
associated with that coil.
Mathematically, e = N ∆ɸ = N ∆ (ɸ2 −
∆𝑡
ɸ1)
𝑡2− 𝑡1
In differential form, we have, e = - N 𝑑ɸ
𝑑𝑡
1.3.2. Factors affecting AC voltage
The magnitude of generated voltage will depend upon:
(i)
The number of turns / conductors
(ii)
The strength of magnetic field
(iii)
The speed of rotation (𝑤).
1.3.3. Equation of alternating voltage and current
Consider a rectangular coil, having N turns and rotating in anticlockwise direction with an
angular velocity of 𝑤 radian/second in a uniform magnetic field as shown in the figure:
- 25 -
- 26 -
An alternating voltage or current changes continuously in magnitude and alternates in
direction at regular intervals of time.
It raises from zero to maximum positive value, falls to zero, increases to a maximum in the
reverse direction and fall back to zero again.
From this point, the voltage or current repeats the procedure.
The e.m.f. induced in the coil will be sinusoidal. This can be readily established.
Let the time be measured from the instant the plane of the coil coincides with OX-axis.
In this position of the coil, the flux linking the coil has its maximum value ɸ𝑚𝑎𝑥.
Let the coil turn through an angle Ɵ (= 𝑤𝑡) in anticlockwise direction in 𝑡 seconds.
In this position, flux linkages of the coil at the considered instant (i.e. Ɵ°)
= Number of turns x Flux linking
= N ɸ𝒎 𝑪𝒐𝒔 𝒘𝒕
- 27 -
According to Second Faraday’s Law of Electromagnetic Induction, the e.m.f. induced in a
coil is equal to the rate of change of flux linkages of the coil. (i.e., e = - N 𝒅𝝓)
𝒅𝒕
Hence the e.m.f. e at the considered instant is given by:
e = -N
𝑑
𝑑𝑡
(ɸ𝑚 𝐶𝑜𝑠 𝑤𝑡)
e = - N ɸ𝑚 𝑤 (−𝑆𝑖𝑛 𝑤𝑡)
e = N ɸ𝑚 𝑤 𝑆𝑖𝑛 𝑤𝑡
The value of e will be maximum (call it 𝐸𝑚) when 𝑆𝑖𝑛 𝑤𝑡 = 1
i.e. when the coil has turned through 90° in anticlockwise direction from the reference axis
(OX-axis).
𝐸𝑚= N ɸ𝑚𝑤 or e = 𝑬𝒎 𝑺𝒊𝒏 𝒘𝒕 or e = 𝑬𝒎 𝑺𝒊𝒏 𝜃
It is clear that e.m.f. induced in the coil is sinusoidal i.e. instantaneous values of e.m.f. varies
as the sine function of time angle (Ɵ or 𝑤𝑡).
If this alternating voltage (𝐸𝑚 𝑆𝑖𝑛 𝑤𝑡) is applied across a load, alternating current flows
through the circuit which would also vary sinusoidally i.e. following a sine law.
The equation of the alternating current is given by:
𝒊 = 𝑰𝒎 𝑺𝒊𝒏 𝒘𝒕 provided the load is resistive.
1.3.4. Important A.C. terminology
(i) Cycle
It is one complete set of positive and negative values of an alternating quantity.
A cycle can also be defined in terms of angular measure.
One cycle corresponds to 360° electrical or 2π radians.
- 28 -
(ii) Alternation
It is one-half cycle of an alternating quantity.
An alternation spans 180° electrical.
Thus, the positive or negative half of an alternating voltage is the alternation.
(iii) Time period
It is the time taken in seconds to complete one cycle of an alternating quantity.
It is generally represented by T.
(iv) Frequency
It is the number of cycles that occur in one second.
It is measured in hertz (Hz = Cycles/second). 1 Hz = 1 cycle/second.
The 50 Hz frequency is the most popular because it gives the best results when used for
operating both lights and machinery.
It means that alternating voltage or current completes 50 cycles in one second.
1.3.5. Values of alternating voltage and current
(i) Instantaneous value
It is the value of an alternating quantity at particular instant in time.
It is different from instant to instant.
The instantaneous values of alternating voltage and current are represented by e and 𝒊
respectively.
- 29 -
(ii) Maximum value or Amplitude value or Peak value
It is the maximum value (positive or negative) attained by an alternating quantity during one
cycle.
It is the greatest value reached during alternation, usually occurring once in each half-cycle.
The amplitude of an alternating voltage or current is designated by 𝐸𝑚 (or 𝑉𝑚) for voltage or
𝐼𝑚 for current.
- 30 -
(iii) Average value or Mean value
It is the arithmetical average of all the instantaneous values of an alternating quantity over
one cycle.
Average value =
(i)
𝑨𝒓𝒆𝒂 𝒖𝒏𝒅𝒆𝒓 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆
𝑩𝒂𝒔𝒆
In case of symmetrical waves (e.g. sinusoidal voltage or current), the average the
average value over one cycle is zero.
It is because positive half is exactly equal to the negative half-cycle so that net
area is zero. However, the average value of positive or negative half-cycle is not
zero. Hence, in case of symmetrical waves, average value means the average
value of half-cycle or one alternation.
Average value of a symmetrical wave
=
=
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑜𝑛
𝐵𝑎𝑠𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑛𝑒 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑜𝑛
𝑆𝑢𝑚 𝑜𝑓 𝑚𝑖𝑑−𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑣𝑒𝑟 𝑜𝑛𝑒 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑜𝑛
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑖𝑑−𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠
A symmetrical wave is one which has positive half-cycle exactly equal to the
negative half-cycle.
Suppose positive half-cycle is divided into 𝑛 equal parts.
The mid value of each part is the mid-ordinate.
The average value will be the sum of mid-ordinates divided by the number of mid-
- 31 -
ordinates (i.e. 𝑛 in this case)
=
Example:
𝐼𝑎𝑣
(ii)
𝑖1 + 𝑖 2 + 𝑖3 + … + 𝑖 𝑛
𝑛
In case of unsymmetrical waves (e.g. half-wave rectified voltage, etc), the average
value is taken over the full-cycle.
Average value of an unsymmetrical wave =
𝐴𝑟𝑒𝑎 𝑜𝑣𝑒𝑟 𝑜𝑛𝑒 𝑐𝑦𝑐𝑙𝑒
𝐵𝑎𝑠𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑛𝑒 𝑐𝑦𝑐𝑙𝑒
Average value of sinusoidal current
The equation of an alternating current varying sinusoidally is given by: 𝑖 = 𝐼𝑚 sin 𝜃
Consider an elementary strip of width 𝑑𝜃 in the first half-cycle of current wave as shown in
the figure:
Let 𝑖 be the mid-ordinate of the strip.
Area of strip = 𝑖𝑑𝜃
𝜋
𝜋
Area of half-cycle = ∫ 𝑖𝑑𝜃 = ∫ 𝐼 𝑆𝑖𝑛 𝜃 𝑑𝜃 = 𝐼
0
0
𝑚
∫𝜋
𝑚 0
𝑆𝑖𝑛𝜃 𝑑𝜃 = 𝐼 [−𝐶𝑜𝑠]𝜋 = 2𝐼
𝑚
0
𝑚
Area of half-cycle = 2𝐼𝑚
𝑰𝒂𝒗 = 𝟐𝑰𝒎 = 0.637 𝑰𝒎
𝝅
(iv) R.M.S. value or Effective value
It is the value of d.c. current which when flowing through a given resistance for a given time
produces the same amount of heat as produced by the alternating current when flowing
through the same resistance for the same time.
- 32 -
For example, when we say that the r.m.s. or effective value of an alternating current is 5 A, it
means that the alternating current will do work (or produce heat) at the same rate as 5 A
direct current under similar conditions.
The rms or effective value of an alternating current (or voltage) can be determined as follows:
Consider the half-cycle of a sinusoidal alternating current 𝑖 flowing though a resistance 𝑅 Ω
for 𝑡 seconds.
Divide the time 𝑡 in 𝑛 equal intervals of time, each of duration 𝑡 seconds.
𝑛
Let the mid-ordinates be 𝑖1, 𝑖2, 𝑖3, . . . 𝑖𝑛.
Each current 𝑖1, 𝑖2, 𝑖3, . . . 𝑖𝑛 will produce heating effect when passed through the resistance
𝑅. Suppose the heating effect produced by current 𝑖 in 𝑅 is the same as produced by some
direct current 𝐼 flowing through the resistance 𝑅 for the same time 𝑡 seconds.
Then direct current 𝐼 is the rms or effective value of alternating current 𝐼.
The heating effect of various components of alternating current will be
𝑡
𝑡
𝑡
𝑡
𝑖2 𝑅 , 𝑖2 𝑅 , 𝑖2 𝑅 ,.… 𝑖2 𝑅 joules.
1
𝑛
2
3
𝑛
𝑛
𝑛
𝑛
Since the alternating current is varying, the heating effect will also vary.
Total heating produced by alternating current 𝑖 is
= (𝑖2 𝑅, 𝑖2 𝑅, 𝑖2 𝑅,.… 𝑖2 𝑅) 𝑡
1
=(
𝑖2
1
2
𝑅,
𝑖2
2
𝑅,
3
𝑖2
3
𝑛
𝑅,.… 𝑖2
𝑛
𝑛
𝑅
𝑛
) 𝑡 joules
Heat produced by equivalent direct current 𝐼 = 𝐼2 𝑅𝑡 joules.
Since the heat produced in both cases is the same,
𝑖2 𝑅 + 𝑖2 𝑅 + 𝑖2 𝑅 + .… + 𝑖2 𝑅
𝐼2𝑅𝑡 = ( 1
2
3
𝑛
𝑛
𝑖2 + 𝑖2 + 𝑖2 + … + 𝑖2
) 𝑡 or 𝐼 = √ 1
2
3
𝑛
= √𝑚𝑒𝑎𝑛 𝑜𝑓 𝑖2
- 33 -
𝑛
= Square root of the mean of the squares of the current
= root-mean-square (r.m.s.) value
The rms or effective value of alternating voltage can similarly be expressed as:
𝐸
=√
1
𝑒2 + 𝑒2 + 𝑒2+ … + 𝑒2
2
3
𝑛
𝑛
(i)
For symmetrical waves, the rms or effective value can be found by considering
half-cycles or full-cycles.
However, for unsymmetrical waves, full-cycle should be considered.
(ii)
The rms value of a wave can also be expressed as:
R.M.S. value = √
𝐴𝑟𝑒𝑎 𝑜𝑓 ℎ𝑎𝑙𝑓−𝑐𝑦𝑐𝑙𝑒 𝑤𝑎𝑣𝑒 𝑠𝑞𝑢𝑎𝑟𝑒𝑑
𝐻𝑎𝑙𝑓−𝑐𝑦𝑐𝑙𝑒 𝑏𝑎𝑠𝑒
R.M.S. value of sinusoidal current
The equation of the alternating current varying sinusoidally is given by: 𝑖 = 𝐼𝑚 sin 𝜃
Consider an elementary strip of width 𝑑𝜃 in the first half-cycle of the squared current wave
as shown in the figure:
𝑰𝒓.𝒎.𝒔.
√
=
𝐴𝑟𝑒𝑎 𝑜𝑓 ℎ𝑎𝑙𝑓−𝑐𝑦𝑐𝑙𝑒 𝑠𝑞𝑢𝑎𝑟𝑒𝑑 𝑤𝑎𝑣𝑒
𝜋
=√
𝜋 𝐼2
𝑚
2
𝜋
𝐼𝑚
= √2 = 0.707 𝑰𝒎
Root Mean Square is the actual value of an alternating quantity which tells us an energy
transfer capability of an AC source.
The ammeter records the rms value of an alternating current and voltmeter records the rms
value of an alternating voltage.
It is designated by 𝑰 𝒓.𝒎.𝒔. and is equal to 𝐼𝑚 or 𝑰𝒓.𝒎.𝒔. = 𝑰𝒎 = 0.707 𝑰𝒎
√𝟐
√2
Similarly, for alternating voltage varying sinusoidally, 𝑬 𝒓.𝒎.𝒔. = 𝑬𝒎 = 0.707 𝑬𝒎
√𝟐
In a.c. circuits, voltages and currents are normally given in r.m.s. values unless stated
otherwise.
- 34 -
Example of rms voltage and its equivalent dc voltage level:
(v) Peak-to-peak value
It is the difference between the highest and the lowest values in a waveform.
It is the sum of peak values of positive and negative half cycle.
- 35 -
Example:
- 36 -
1.3.6. Important relations
(i) Time period (𝑻) and frequency (𝒇)
Consider an alternating quantity having a frequency of 𝑓 cycles/second and time period of 𝑇
seconds.
Time taken to complete 𝑓 cycles = 1 second.
Time taken to complete 1 cycle = 1 second.
𝑓
But the time taken to complete one cycle is the time period 𝑇 (by definition).
Therefore, 𝑻 = 𝟏 or 𝒇 = 𝟏
𝒇
𝑻
(ii) Angular velocity (𝒘) and frequency (𝒇)
Referring to the following figure, the coil is rotating with an angular velocity of 𝑤
radians/second.
In one revolution of the coil, the angle turned is 2π radians and the voltage wave complete 1
cycle. The time taken to complete one cycle is the time period 𝑇 of the alternating voltage.
Angular velocity, 𝑤 = 𝐴𝑛𝑔𝑙𝑒 𝑡𝑢𝑟𝑛𝑒𝑑 = 2𝜋 or 𝒘 = 𝟐𝝅 = 2 𝝅𝒇
𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝑇
𝑻
EXERCISES
1. An alternating current is given by: 𝑖 = 141.4 sin 314 𝑡. Find:
i.
The maximum and r.m.s. values of current.
ii.
Angular velocity
iii.
Frequency
iv.
Time period
v.
The instantaneous value of current when t is 3 m second.
- 37 -
2. An alternating voltage of frequency 60 Hz has a maximum value of 120 V.
i.
Write the equation for the instantaneous value of voltage.
ii.
The instantaneous value of voltage when t is 2 m second.
3. A 60Hz voltage of 115V (r.m.s.) is applied on a 100 Ω resistance.
Write the equations for voltage and current.
4. In an ac circuit, the instantaneous values of voltage and current are given by:
𝑣 = 250 Sin 314t and 𝑖 = 10 Sin 314t. Determine:
(i)
Resistance
(ii)
r.m.s. values of voltage and current
1.3.7. Phase difference
The phase difference between two waveforms is a measure of the angular difference between
their respective phase positions.
It indicates how much one waveform is shifted relative to the other waveform in terms of
time or angle. The phase difference is typically expressed in degrees or radians.
Consider three similar coils displaced from each other by angles α and β in a uniform
magnetic field with the same angular velocity.
These three e.m.f.s induced in these three coils do not reach their maximum or zero values at
the same time but one after another.
Here, the phase difference between A and B is β and between B and C is α but between A and
C is α + β.
Phase difference or phase angle between two alternating quantities tells us how much a
quantity is in front or behind another.
- 38 -
Phase difference of 0° or 0 radians
A phase difference of 0° or 0 radians means that the two waveforms are in phase and reach
their corresponding maximum and zero values at the same time.
Lagging and Leading alternating quantities
A leading alternating quantity is one which reaches its maximum value earlier than the
other quantity.
A lagging alternating quantity is one which reaches its maximum value later than the other
quantity.
Examples:
1.
The current 𝑖 lags the voltage 𝑣 by 30°.
- 39 -
2.
3.
4.
Phase angle
A phase angle describes the phase difference that exists between supply voltage and current.
1.3.8. Types of electrical loads
An electrical load is a device or an electrical component that consumes electrical energy and
convert it into another form of energy.
Examples: Electric lamps, electric heaters, motors, etc.
- 40 -
Types of electrical load according to their nature
(i) Resistive load
A resistive load is a type of load that primarily dissipates electrical energy in the form of heat.
Loads consisting of any heating element are classified as resistive loads.
Their impedance remains constant regardless of the frequency of the applied voltage.
Example: Incandescent lamps, electric heaters, etc.
Characteristics of resistive load
(i)
They consume only active power
(ii)
Voltage and current are in phase with each other.
(iii)
Power factor is unity
(ii) Inductive load
An inductive load is a type of load that stores electrical energy in the form of magnetic field.
Inductive loads (any load having magnetic coils) include devices such as motors, etc.
Motor converts electrical energy to magnetic energy further which is used as mechanical
energy. This load exhibit inductance.
- 41 -
Characteristics of inductive load
(i)
An inductive load causes the current to lag the voltage.
(ii)
The power factor of inductive load is lagging
(iii) Capacitive load
A capacitive load is a type of load that stores electrical energy in the form of electric field.
It can release stored energy when the power supply is disconnected.
Capacitors are typical examples of capacitive loads.
Characteristics of capacitive load
(i)
A capacitive load causes the current to lead the voltage.
(ii)
The power factor of capacitive load is leading
(iv) Combination load
Combination loads are loads that make use of various combinations of resistors, inductors
and capacitors to perform specific functions.
Example: Single-phase motors often use capacitors to aid the motor during starting and
running.
Types of electrical load according to the load consumer category
(i)
Residential load: Domestic load consists of lights, fans, refrigerators, heaters,
television, small motor for pumping water, etc.
(ii)
Commercial load: Commercial load consists of lighting for shops and electric
appliances used in restaurants, etc.
(iii)
Industrial load: Industrial load consists of load demand by industries.
The magnitude of industrial load depends upon the type of industry.
- 42 -
(iv)
Municipal load: Municipal load consists of street lighting, power required for
water supply and drainage purposes.
(v)
Irrigation load: This type of load is the electric power needed for pumps driven
by motors to supply water to fields.
(vi)
Traction load: This type of load includes tram cars, trolley buses, railways, etc.
Types of load according to the number of electrical load phases
(i) Single phase load: A single-phase load refers to an electrical load that can operate on
a single-phase power supply or, on a three-phase power supply by connecting the load
between one of the phases and neutral.
(ii) Three phase load: A three-phase load refers to an electrical load that operates on a
three-phase power supply.
- 43 -
Types of electrical load according to load / phase distribution
(i) Balanced load: A balanced three-phase load refers to a condition when all three
phases carry the same magnitude of current along with the same phase differences.
(ii) Unbalanced load: Unbalanced load refers to the condition when unequal currents are
carried by the three phases.
- 44 -
- 45 -
1.3.9. Single phase AC voltage to series circuits
(i) A.C. Circuit containing resistance only (Purely resistive a.c. circuit)
Consider a circuit containing a resistance R connected across an alternating voltage source.
Let the alternating voltage be given by the equation: 𝒗 = 𝑽𝒎 Sin wt
As a result of this voltage, an alternating current will flow in the circuit. i.e. 𝑖 = 𝑣 = 𝑉𝑚 Sin wt
𝑅
𝑅
where 𝑅 is the resistance of the circuit.
Resistance: It is the opposition offered by a circuit or a conductor to the flow of electric
current.
The value of 𝑖 will be maximum when Sin wt = 1, then 𝐼𝑚 =
𝑉𝑚
𝑅
hence 𝒊 = 𝑰𝒎 Sin wt
The alternating voltage and current are in phase with each other.
Its phasor diagram:
The voltage and current reach maximum and zero value at the same time.
- 46 -
(ii) A.C. Circuit containing inductance only (Purely inductive a.c. circuit)
Consider a circuit containing an inductance L connected across an alternating voltage source.
The instantaneous current is given by the equation: 𝑖 = 𝑉𝑚 Sin (wt - 𝜋)
𝐿𝑤
Current is maximum when Sin (wt -
𝜋)
2
= 1 then, 𝐼 =
𝑚
2
𝑉𝑚
𝐿𝑤
Hence the equation of the current becomes: 𝒊 = 𝑰𝒎 Sin (wt - 𝝅).
𝟐
The current 𝑖 lags the voltage 𝑣 by 90° or 𝜋 radian.
2
Here, 𝐿𝑤 plays the part of resistance. It is called the inductive reactance of the coil.
It is denoted by: 𝑿 or 𝑋 = 𝐿𝑤 and 𝒊 =
𝑳
𝐿
𝑽𝒎
Sin (wt - 𝝅)
𝑿𝑳
𝟐
where
𝑋𝐿 = inductive reactance in ohms (Ω)
𝐿 = inductance in henry (H)
w = angular velocity in radian/second
Inductive reactance: The opposition to the current flow in a.c. circuit caused by an
inductance or inductive device.
- 47 -
Current reaches its maximum or zero value later than the voltage.
Its phasor diagram:
(iii) A.C. Circuit containing capacitance only (Purely capacitive a.c. circuit)
Consider a circuit containing a capacitance C connected across an alternating voltage source.
The instantaneous current is given by: 𝒊 =
𝑽𝒎
Sin (wt + 𝝅) or 𝒊 = Cw 𝑽 Sin (wt + 𝝅)
𝟏
𝑪𝒘
Current is maximum when Sin (wt + 𝝅 ) = 1 then, 𝐼 =
𝑚
𝟐
𝒎
𝟐
𝑽𝒎
or 𝐼 = Cw 𝑉
𝑚
𝟏
𝑪𝒘
Hence the equation of the current becomes: 𝒊 = 𝑰𝒎 Sin (wt + 𝝅)
𝟐
The current 𝑖 leads the voltage 𝑣 by 90° or 𝝅 radian.
𝟐
Here,
𝟏
𝑪𝒘
plays the part of resistance. It is called the capacitive reactance.
It is denoted by: 𝑿 or 𝑿 =
𝑪
𝑪
𝟏
and 𝒊 =
𝑪𝒘
𝑽𝒎
Sin (wt + 𝝅)
𝑿𝑪
𝟐
Where:
𝑋𝐶 = capacitive reactance in ohms (Ω)
𝐶 = capacitance in farad (F)
- 48 -
𝑚
𝟐
w = angular velocity in radian/second
Capacitive reactance: The opposition to the current flow in a.c. circuit caused by a
capacitive device.
Current reaches its maximum or zero values earlier than the voltage.
Its phasor diagram:
(iv) A.C. through resistance and inductance in series
A resistance R and inductance L are connected in series as shown in the figure:
- 49 -
Let
V = r.m.s. value of applied voltage
𝐼 = r.m.s. value of current
𝑉𝑅= 𝐼R = Voltage drop across resistance R (in phase with current)
𝑉𝐿= 𝐼𝑋𝐿= Voltage drop across an inductance (at right angle to the current)
Voltage triangle
Vector OA represents voltage drop 𝑉𝑅 across R.
Vector AB represents voltage drop 𝑉𝐿 across L.
Vector OB represents vector sum of 𝑉𝑅 and 𝑉𝐿.
𝑉2 = 𝑉2 + 𝑉2
𝑅
𝐿
or 𝑉= √𝑉2 + 𝑉2 = √(𝐼R)2 + (𝐼𝑋 )2 = √𝐼2𝑅2 + 𝐼2𝑋
𝑅
𝐿
𝐿
2
𝐿
= 𝐼√𝑅2 + 𝑋2
𝐿
The quantity √𝑅2 + 𝑋2𝐿 is known as impedance of the circuit.
It is noted as 𝑍 and is measured in ohm (Ω).
𝒁 = √𝑹𝟐 + 𝑿𝟐𝑳 and 𝑽= 𝑰𝒁
Impedance: It is the total opposition (including resistance and reactance) offered by an a.c.
circuit to the flow of alternating current.
From the voltage triangle, it is clear that the current 𝐼 lags behind the voltage 𝑉 by an angle ɸ.
𝑉 = 𝑉𝐶𝑜𝑠 ɸ and 𝑉 = 𝑉𝑆𝑖𝑛 ɸ. Then,
𝑅
𝐿
𝑉𝐿
𝑉𝑅
= 𝑉 𝑆𝑖𝑛 ɸ = 𝒕𝒂𝒏 ɸ.
𝑉 𝐶𝑜𝑠 ɸ
𝑽𝑳
It means 𝒕𝒂𝒏 ɸ = 𝑽𝑳 or ɸ = 𝐭𝐚𝐧−𝟏 ( )
𝑽𝑹
𝑽𝑹
Hence if the applied voltage is given by: 𝑣 = 𝑉𝑚 Sin wt
The current is given by: 𝒊 = 𝑰𝒎 Sin (wt - ɸ) where 𝐼𝑚 =
- 50 -
𝑉𝑚
𝑍
Impedance triangle
From the impedance triangle, it is clear that: R = 𝑍𝐶𝑜𝑠 ɸ and 𝑋𝐿= 𝑍𝑆𝑖𝑛 ɸ.
Then, 𝑋𝐿 =
𝑅
𝑿
𝑍 𝑆𝑖𝑛 ɸ
= 𝒕𝒂𝒏 ɸ or 𝑡𝑎𝑛 ɸ = 𝑋𝐿 and ɸ = 𝐭𝐚𝐧−𝟏( 𝑳).
𝑍 𝐶𝑜𝑠 ɸ
𝑅
𝑹
Hence the applied voltage is given by: 𝑣 = 𝑉𝑚 Sin wt
The current is given by: 𝒊 = 𝑰𝒎 Sin (wt - ɸ) where 𝐼𝑚 =
𝑉𝑚
𝑍
(v) A.C. through resistance and capacitance in series
A resistance R and a capacitance C are connected in series as shown in the figure:
- 51 -
Let
V = r.m.s. value of applied voltage
𝐼 = r.m.s. value of current
𝑉𝑅 = 𝐼R = Voltage drop across resistance R (in phase with current)
𝑉𝐶 = 𝐼𝑋𝐶 = Voltage drop across a capacitor 𝐶 (at right angle to the current)
Voltage triangle
𝑉2 = 𝑉2 + 𝑉2
𝑅
𝐶
or 𝑉 = √𝑉2 + 𝑉2 = √(𝐼R)2 + (𝐼𝑋 )2 = √𝐼2𝑅2 + 𝐼2𝑋
𝑅
𝐶
𝐶
𝐶
2
= 𝐼√𝑅2 + 𝑋2
𝐶
√𝑹𝟐 + 𝑿𝟐𝑪 = 𝒁 and 𝑉= 𝐼𝑍 where 𝑍 is the impedance of the circuit.
From the voltage triangle, it is clear that the current 𝐼 leads the applied voltage 𝑉 by an angle
ɸ.
𝑉 = 𝑉𝐶𝑜𝑠 ɸ and 𝑉 = 𝑉𝑆𝑖𝑛 ɸ. Then, 𝑉𝐶 =
𝑅
𝐶
𝑉𝑅
𝑉 𝑆𝑖𝑛 ɸ
= 𝒕𝒂𝒏 ɸ
𝑉 𝐶𝑜𝑠 ɸ
𝑽
𝑪
It means 𝒕𝒂𝒏 ɸ = 𝑽𝑪 or ɸ = 𝐭𝐚𝐧−𝟏 ( )
𝑽𝑹
𝑽𝑹
If the applied voltage is given by: 𝑣 = 𝑉𝑚 Sin wt
the current is given by: 𝒊 = 𝑰𝒎 Sin (wt + ɸ) where 𝐼𝑚 =
𝑉𝑚
𝑍
Impedance triangle
From the impedance triangle, it is clear that: R = 𝑍𝐶𝑜𝑠 ɸ and 𝑋𝐶 = 𝑍𝑆𝑖𝑛 ɸ.
Then, 𝑋𝐶 =
𝑅
𝑿
𝑍 𝑆𝑖𝑛 ɸ
= 𝒕𝒂𝒏 ɸ or 𝑡𝑎𝑛 ɸ = 𝑋𝐶 and ɸ = 𝐭𝐚𝐧−𝟏( 𝑪).
𝑍 𝐶𝑜𝑠 ɸ
𝑅
𝑹
- 52 -
Hence the applied voltage is given by: 𝑣 = 𝑉𝑚 Sin wt
The current is given by: 𝒊 = 𝑰𝒎 Sin (wt + ɸ) where 𝐼𝑚 =
𝑉𝑚
𝑍
(vi) A.C. through resistance, inductance and capacitance in series
A resistance R, an inductance L and a capacitance C are connected in series as shown in the
figure:
Let
V = r.m.s. value of applied voltage
𝐼 = r.m.s. value of current
𝑉𝑅 = 𝐼R = Voltage drop across resistance R (in phase with current)
𝑉𝐿 = 𝐼𝑋𝐿 = Voltage drop across a capacitor 𝐿 (at right angle to the current)
𝑉𝐶 = 𝐼𝑋𝐶 = Voltage drop across a capacitor 𝐶 (at right angle to the current)
Voltage triangle
Here,
Vector OA represents voltage drop 𝑉𝑅 across R.
Vector AB represents voltage drop 𝑉𝐿 across L.
Vector AC represents voltage drop 𝑉𝐶 across C.
Vector AD represents 𝑉𝐿- 𝑉𝐶
- 53 -
Vector OD represents vector sum of 𝑉𝑅 and (𝑉𝐿- 𝑉𝐶)
𝑉2 = 𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶)2
or 𝑉 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶)2 = √(𝐼R)2 + (𝐼𝑋𝐿 − 𝐼𝑋𝐶)2
= √(𝐼R)2 + [𝐼 (𝑋𝐿 − 𝑋𝐶)]2 = √𝐼2R2 + 𝐼2(𝑋𝐿 − 𝑋𝐶)2 = 𝐼√R2 + (𝑋𝐿 − 𝑋𝐶)2
and 𝑍 = √R2 + (𝑋𝐿 − 𝑋𝐶)2
Impedance Triangle
From the impedance triangle, it is clear that: √R2 + (𝑋𝐿 − 𝑋𝐶)2
𝑋𝐿 − 𝑋𝐶 = 𝑋 or 𝑍 = √R2 + 𝑋2 𝑽 = 𝑰𝒁
Where:
𝑍 is the impedance
𝑋 is the reactance
Reactance (𝑿): It is the opposition to the current flow in a.c. circuit caused by inductance or
capacitance.
R = 𝑍𝐶𝑜𝑠 ɸ and 𝑋 - 𝑋 = 𝑍𝑆𝑖𝑛 ɸ or 𝑋𝐿 −
𝐿
𝐶
𝑅
𝑋𝐶
=
𝑍𝑆𝑖𝑛 ɸ
𝑍𝐶𝑜𝑠 ɸ
= 𝑡𝑎𝑛 ɸ or ɸ =
−𝟏 𝑿
𝐭𝐚𝐧 ( )
𝑹
Three cases of R-L-C series circuit:
The impedance of a R-L-C series circuit is given by: 𝑍 = √R2 + (𝑋𝐿 − 𝑋𝐶)2
(i)
When 𝑿𝑳 − 𝑿𝑪 is positive (i.e. 𝑿𝑳 > 𝑿𝑪), the circuit is inductive.
(ii)
When 𝑿𝑳 − 𝑿𝑪 is negative (i.e. 𝑿𝑳 < 𝑿𝑪), the circuit is capacitive.
(iii)
When 𝑿𝑳 − 𝑿𝑪 is zero (i.e. 𝑿𝑳 = 𝑿𝑪), the circuit is purely resistive.
If the equation of the applied voltage is 𝑣 = 𝑉𝑚 Sin wt, then the equation for the current will
be: 𝒊 = 𝑰𝒎 Sin (wt ± ɸ) where 𝑰𝒎 =
𝑉𝑚
𝑍
- 54 -
 When 𝑿𝑳 > 𝑿𝑪 𝑡ℎ𝑒𝑛 𝑽𝑳 > 𝑽𝑪: The negative sign (-) is used.
It means that the current lags the voltage by an angle ɸ.
 When 𝑿𝑳 < 𝑿𝑪 𝑡ℎ𝑒𝑛 𝑽𝑳 < 𝑽𝑪: The positive sign (+) is used.
It means that the current leads the voltage by an angle ɸ.
1.3.10. Power in AC circuits
(i) Active power (P)
It may be defined in one of the following ways:
(i)
The power which is actually consumed or utilized in electrical circuit.
It represents the rate of useful work done by a load.
(ii)
The power actually converted to another form of power in a.c. circuit.
(iii)
The power drawn by electrical resistance of a system doing useful work.
P = V I Cos Ɵ = 𝑰𝟐R = S Cos Ɵ
It is measured in watt (W) or kilowatt (kW) or MW.
(ii) Reactive power (Q)
It may be defined in one of the following ways:
(i)
It is the power needed to establish and maintain the magnetic fields in inductive
elements and the electric fields in capacitive elements.
(ii)
The power developed in the reactance of the circuit.
Reactive power exists in a.c. circuits when the voltage and current are not in phase.
Q = V I Sin Ɵ = 𝑰𝟐X = S Sin Ɵ
It is measured in volt-ampere reactive (VAR) or kilo volt-ampere reactive (kVAR) or
MVAR.
(iii) Apparent power (S)
It may be defined in one of the following ways:
(i)
It is the product of root mean square (RMS) value of applied voltage and current.
i.e., S = V I
(ii)
It is the vector sum of active and reactive power. i.e. S = √𝑷𝟐 + 𝑸𝟐
It is measured in volt-ampere (VA) or kilo volt-ampere (kVA) or MVA.
- 55 -
Phasor diagram for lagging p.f.
1.3.11. Energy in AC circuits
(i) Active energy (𝑾)
It is the product of active power and time. 𝑾 = 𝑷 𝑿 𝒕 = 𝑽𝑰𝒕 𝑪𝒐𝒔 ɸ
It is measured in Kwh. It is measured by active energy meter.
The active energy meter is used in domestic installations.
(ii) Reactive energy (𝑾𝒓)
It is the product of reactive power and time. 𝑊𝑟 = 𝑄 𝑋 𝑡 = 𝑉𝐼𝑡 𝑆𝑖𝑛 ɸ.
It is measured in KVarh. It is measured by reactive energy meter.
In industrial installations, two energy meters (active and reactive) are used.
These allow determining the power factor of the installation.
𝐶𝑜𝑠 ɸ =
𝑊
√𝑊 2
+ 𝑊𝑟2
1.3.12. Power factor (𝑪𝒐𝒔 𝜃)
It is the cosine of angle between voltage and current.
It may be calculated as follows:
(i)
Power factor = Cos Ɵ = 𝑅 = 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
(ii)
Power factor = Cos Ɵ = 𝑃 =
𝑍
𝑆
𝐼𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒
𝐴𝑐𝑡𝑖𝑣𝑒 𝑝𝑜𝑤𝑒𝑟
𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑝𝑜𝑤𝑒𝑟
It may be noted that power factor can never have a value greater than 1.
Reactive power required by inductive loads increases the amount of apparent power in the
distribution system.
Increasing the reactive and apparent power causes the power factor to decrease.
- 56 -
1.3.13. Active and reactive components of current
Active component of current
The current component, which is in phase with the circuit voltage and contributes to the
active or true power of the circuit, is called an active component or watt-full component or
in-phase component of current.
Reactive component of current
The component, which is in quadrature or 90° degrees out of phase to the circuit voltage and
contributes to the reactive power of the circuit, is called a reactive component of current.
Basic electrical quantities
Electrical quantity
Active
Reactive
Current (𝐼)
Active component of current
Reactive component of current
= 𝐼 𝐶𝑜𝑠 𝜃
= 𝐼 𝑆𝑖𝑛 𝜃
Voltage across resistance.
Voltage across inductive element
𝑉𝑅 = 𝑉 𝐶𝑜𝑠 𝜃
or voltage across capacitive
Voltage (𝑉)
element.
𝑉𝐿 or 𝑉𝐶 = 𝑉 𝑆𝑖𝑛 𝜃
Impedance (𝑍)
Resistance
Inductive reactance or capacitive
𝑅 = 𝑍 𝐶𝑜𝑠 𝜃
reactance.
𝑋𝐿 or 𝑋𝐶 = 𝑍 𝑆𝑖𝑛 𝜃
Power (𝑆)
Energy
Active power
Reactive power
𝑃 = 𝑆 𝐶𝑜𝑠 𝜃
𝑄 = 𝑆 𝑆𝑖𝑛 𝜃
Active energy = 𝑃 𝑡
Reactive energy = 𝑄 𝑡
- 57 -
Summary of results of series a.c. circuits
Type of
Value of impedance
impedance
Phase angle for
Power factor
current
Resistance only
𝑅
0°
1
Inductance only
𝑋𝐿 = 𝐿𝑤
90° lag
0
Capacitance
𝑋𝐶 =
90° lead
0
0° < Ɵ < 90° lag
1 > p.f. > 0 lag
0° < Ɵ < 90° lead
1 > p.f. > 0 lead
𝑍=
0° < Ɵ < 90° lag or
1 > p.f. > 0 lag or
√R2 + (𝑋𝐿 − 𝑋𝐶)2
lead
lead
1
𝑐𝑤
only
Resistance and
𝑍 = √𝑅2 + 𝑋2
𝐿
inductance
(𝑅 = 𝑍 𝐶𝑜𝑠 ɸ
and 𝑋𝐿 = 𝑍 𝑆𝑖𝑛 ɸ)
Resistance and
𝑍 = √𝑅2 + 𝑋2
𝐶
capacitance
(𝑅 = 𝑍 𝐶𝑜𝑠 ɸ
and 𝑋𝐶 = 𝑍 𝑆𝑖𝑛 ɸ)
Resistance,
inductance and
capacitance
(𝑅 = 𝑍 𝐶𝑜𝑠 ɸ
and
𝑋𝐿 − 𝑋𝐶 = 𝑍 𝑆𝑖𝑛 ɸ)
- 58 -
Type of
Circuit diagram
Phasor diagram
impedance
Equations of
instantaneous
values of applied
voltage and
current
𝑣 = 𝑉𝑚 Sin wt
Resistance only
𝑖 = 𝐼𝑚 Sin wt
𝑣 = 𝑉𝑚 Sin wt
Inductance only
𝜋
𝑖 = 𝐼 Sin (wt − )
𝑚
2
Capacitance
𝑣 = 𝑉𝑚 Sin wt
only
𝑖 = 𝐼 Sin (wt + 𝜋)
Resistance and
𝑣 = 𝑉𝑚 Sin wt
inductance
𝑖 = 𝐼𝑚 Sin (wt - ɸ)
Resistance and
𝑣 = 𝑉𝑚 Sin wt
capacitance
𝑖 = 𝐼𝑚 Sin (wt + ɸ)
Resistance,
𝑣 = 𝑉𝑚 Sin wt
inductance and
𝑖 = 𝐼𝑚 Sin (wt ± ɸ)
𝑚
capacitance
or
- 59 -
2
Type of
Applied voltage
Power
impedance
Resistance only
𝑉 = 𝑉𝑅
𝑃 = 𝑆 𝐶𝑜𝑠 0° = 𝑆
𝑄 = 𝑆 𝑆𝑖𝑛 0° = 0
Inductance only
𝑉 = 𝑉𝐿
𝑃 = 𝑆 𝐶𝑜𝑠 90° = 0
𝑄 = 𝑆 𝑆𝑖𝑛 90° = 𝑆
Capacitance
𝑉 = 𝑉𝐶
𝑃 = 𝑆 𝐶𝑜𝑠 90° = 0
𝑄 = 𝑆 𝑆𝑖𝑛 90° = 𝑆
only
Resistance and
𝑉 = √𝑉2 + 𝑉2
𝑅
𝑃 = 𝑆 𝐶𝑜𝑠 ɸ or 𝑃 = 𝑉𝐼𝐶𝑜𝑠 ɸ
𝐿
𝑄 = 𝑆 𝑆𝑖𝑛 ɸ or 𝑄 = 𝑉𝐼𝑆𝑖𝑛 ɸ
inductance
(𝑉𝑅 = 𝑉𝐶𝑜𝑠 ɸ and
𝑆 = √𝑃2 + 𝑄2 or 𝑆 = 𝑉𝐼
𝑉𝐿 = 𝑉𝑆𝑖𝑛 ɸ)
Resistance and
𝑉 = √𝑉2 + 𝑉2
𝑅
𝑃 = 𝑆 𝐶𝑜𝑠 ɸ or 𝑃 = 𝑉𝐼𝐶𝑜𝑠 ɸ
𝐶
𝑄 = 𝑆 𝑆𝑖𝑛 ɸ or 𝑄 = 𝑉𝐼𝑆𝑖𝑛 ɸ
capacitance
(𝑉𝑅 = 𝑉𝐶𝑜𝑠 ɸ and
𝑆 = √𝑃2 + 𝑄2 or 𝑆 = 𝑉𝐼
𝑉𝐶 = 𝑉𝑆𝑖𝑛 ɸ)
Resistance,
𝑉 = √𝑉2 + (𝑉𝐿 − 𝑉𝐶)2
𝑅
𝑄 = 𝑆 𝑆𝑖𝑛 ɸ or 𝑄 = 𝑉𝐼𝑆𝑖𝑛 ɸ
inductance and
capacitance
𝑃 = 𝑆 𝐶𝑜𝑠 ɸ or 𝑃 = 𝑉𝐼𝐶𝑜𝑠 ɸ
(𝑉𝑅 = 𝑉𝐶𝑜𝑠 ɸ
and 𝑉𝐿 - 𝑉𝐶 = 𝑉𝑆𝑖𝑛 ɸ)
- 60 -
𝑆 = √𝑃2 + 𝑄2 or 𝑆 = 𝑉𝐼
EXERCISES
1. A voltage 𝑣 = 142 sin 314 𝑡 is applied across a resistance of 20Ω, an inductance of
0.1H and a capacitance of 100µF in series.
Determine:
(i)
r.m.s. value of current
(ii)
Equation of instantaneous current
(iii)
Power factor
(iv)
Active, reactive and apparent power.
2. A resistance of 6Ω and an inductance of 0.03H are connected in series across 60V,
50Hz supply. Determine:
(i)
Impedance
(ii)
Current
(iii)
Phase angle between the applied voltage and the current
(iv)
Equations of voltage and current
(v)
Voltage drop across resistance and inductance
(vi)
Active, reactive and apparent power.
3. A resistance of 30Ω and a capacitance of 79.5µF are connected in series across 100V,
50 Hz supply. Find:
(i)
Impedance
(ii)
Current
(iii)
Phase angle between the applied voltage and the current
(iv)
Equation of instantaneous value of supply voltage and current
(v)
Voltage drop across resistance and capacitance.
(vi)
Active, reactive and apparent power.
4. A resistance of 20Ω, an inductance of 0.2H and a capacitance of 100µF are connected
in series across a 220V, 50Hz supply. Calculate:
(i)
Impedance
(ii)
Current
(iii)
Voltage drop across resistance, inductance and capacitance.
(iv)
Phase angle between voltage and current.
(v)
Equations of voltage and current
(vi)
Active, reactive and apparent power.
(vii)
Power factor
- 61 -
5. A purely inductive coil allows a current of 10 A to flow from a 230V, 50Hz supply.
Find:
(i)
Inductive reactance of the coil
(ii)
Inductance of the coil
(iii)
Equations of voltage and current
6. A voltage 𝑣 = 100 Sin 314 𝑡 is applied across a resistance and an inductance in series
and a current 𝑖 = 5 Sin (314 𝑡 - 𝜋) flows in the circuit. Determine:
3
(i)
Resistance
(ii)
Reactance
(iii)
Inductance of the circuit
(iv)
Voltage drop across resistance and inductance
(v)
Active, reactive and apparent power
7. A 100V, 60W lamp is to be operated on 240V, 50Hz supply.
Find the value of:
(i)
Pure resistance
(ii)
Pure inductance
(iii)
Capacitance; in series with the lamp so that the lamp may take their rated
power and voltage.
8. A capacitance of 8µF takes a current of 1A when an alternating voltage is applied to
it, is 250V. Calculate:
(i)
The frequency of the applied voltage
(ii)
The resistance to be connected in series with the capacitance to reduce the
current in the circuit to 0.5A at the same frequency.
9. When a certain inductive coil is supplied at 240V, 50 Hz, the circuit current is 6.45A.
When the frequency is changed to 40 Hz at 240V, the current taken is 7.48A.
Calculate the inductance and resistance of the coil.
1.3.14. Single phase AC voltage to parallel circuits
As in dc circuits, the voltage across all branches is the same in parallel ac circuits.
But current in any branch depends upon the impedance of that branch.
The total line current supplied to the circuit the phasor sum of the branch currents.
The parallel circuit is the exact opposite to the series circuit.
This time instead of the current being common to the circuit components (i.e.𝑅, 𝐿 and 𝐶), the
- 62 -
applied voltage is now common to all, so we need to find the individual branch currents
through each element or branch.
Solving parallel AC circuits
The methods of solving parallel ac circuits are:
(i)
By phasor diagram (Methods of components)
(ii)
By phasor algebra (Complex method)
(iii)
Parallelogram method (for two branches)
Phasor diagram
In this method, we draw the phasor diagram of the parallel circuit taking voltage as the
reference phasor. The circuit or line current is determined by the method of components.
Example 1
Consider a parallel circuit consisting of two branches and connected to an alternating voltage
of V volts (r.m.s.) as shown in the figure:
- 63 -
Branch 1
𝑍 = √𝑅
𝑋 = 2 𝜋𝑓𝐿
𝐿
1
2
1
+ 𝑋
𝐼 =
1
2
𝐿
𝜃 = 𝑡𝑎𝑛−1
𝑉
1
𝑍1
𝑋𝐿
𝑅1
The current 𝐼1 lags the applied voltage 𝑉 by 𝜃1
Branch 2
1
𝑍2 = √𝑅2 2 + 𝑋𝐶 2
𝑋𝐶 = 2𝜋𝑓𝐶
𝐼2 =
𝑉
𝜃2 = 𝑡𝑎𝑛−1
𝑍2
𝑋𝐶
𝑅2
The current 𝐼2 leads the applied voltage 𝑉 by 𝜃2
Total circuit
The line current 𝐼 is the phasor sum of 𝐼1 and 𝐼2.
Suppose its phase angle is 𝜃.
The values of 𝐼 and 𝜃 can be determined by resolving the currents into two rectangular
components.
𝐼 𝐶𝑜𝑠 𝜃 = 𝐼1 𝐶𝑜𝑠 𝜃1 + 𝐼2 𝐶𝑜𝑠 𝜃2
𝐼 𝑆𝑖𝑛 𝜃 = - 𝐼1 𝑆𝑖𝑛 𝜃1 + 𝐼2 𝑆𝑖𝑛 𝜃2
𝐼 = √(𝐼 𝐶𝑜𝑠 𝜃)2 + (𝐼 𝑆𝑖𝑛 𝜃)2
tan Ɵ =
𝐼 𝑆𝑖𝑛 𝜃
𝐼 𝐶𝑜𝑠 𝜃
=
− 𝐼1 𝑆𝑖𝑛 𝜃1 + 𝐼2 𝑆𝑖𝑛 𝜃2
𝐼1 𝐶𝑜𝑠 𝜃1 + 𝐼2 𝐶𝑜𝑠 𝜃2
−1 𝐼 𝑆𝑖𝑛 𝜃
Ɵ = 𝑡𝑎𝑛 (
=
𝑇𝑜𝑡𝑎𝑙 𝑌−𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡
𝑇𝑜𝑡𝑎𝑙 𝑋−𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡
)
𝐼 𝐶𝑜𝑠 𝜃
Circuit power factor, 𝐶𝑜𝑠 𝜃 =
𝐼 𝐶𝑜𝑠 𝜃
𝐼
=
𝐼1 𝐶𝑜𝑠 𝜃1 + 𝐼2 𝐶𝑜𝑠 𝜃2
𝐼
- 64 -
=
𝑇𝑜𝑡𝑎𝑙 𝑋−𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡
𝐼
Example 2
Example 3
Consider the circuit below;
𝐼𝑟 𝐶𝑜𝑠 ɸ𝑟 = 𝐼1 𝐶𝑜𝑠 ɸ1 + 𝐼2 𝐶𝑜𝑠 ɸ2
𝐼𝑟 𝑆𝑖𝑛 ɸ𝑟 = - 𝐼1 𝑆𝑖𝑛 ɸ1 - 𝐼2 𝑆𝑖𝑛 ɸ2
𝐼𝑟 = √ (𝐼𝑟 𝐶𝑜𝑠 ɸ𝑟)2 + (𝐼𝑟 𝑆𝑖𝑛 ɸ𝑟)2
−1
ɸ𝑟 = tan (
𝐶𝑜𝑠 ɸ𝑟 =
𝐼𝑟 𝑆𝑖𝑛 ɸ𝑟
)
𝐼𝑟 𝐶𝑜𝑠 ɸ𝑟
𝐼𝑟 𝐶𝑜𝑠 ɸ𝑟
𝐼𝑟
Example 4
- 65 -
Law of parallelogram methods (for two branches)
Alternating current may be added in the same way as forces are added, using law of
parallelogram.
Formula of parallelogram:
𝐼𝑡 = √𝐼2 + 𝐼2 + 2 𝐼1 𝐼2 𝐶𝑜𝑠 𝛼
1
2
α = Angle between 𝐼1 and 𝐼2
EXERCISES
1. A 240V, 50 Hz source supplies four loads connected in parallel.
Load 1: 10 kW at 0.8 power factor lagging
Load 2: 50 A at 0.707 power factor lagging
Load 3: 5 kW at unity power factor.
Load 4: 8 kVA at 0.6 power factor leading.
Determine the total kW, kVAR, kVA, current and overall power factor.
2. The following figure shows three loads connected in parallel across a 1000 V at 50 Hz
single phase supply.
Load 1: 125 kVA at 0.3 power factor lagging
Load 2: 10 kW and 40 kVAR, leading power factor
Load 3: Resistive load, 15 kW.
Determine the total kW, kVAR, kVA, current and overall power factor.
- 66 -
3. The given figure shows the source suppling two loads. Load 1 consumes 34 kVA at
0.3 p.f. leading and load 2 consumes 42 kVA at 0.5 p.f. lagging.
Find the kW, kVAR, kVA and power factor of combined load.
4. A 120V, 60 Hz source supplies two loads connected in parallel as shown in the figure.
Find the kW, kVAR, kVA and power factor of combined load.
EXERCISES
For the following circuits, find:
(i)
Maximum value of supply voltage.
(ii)
Equation of instantaneous values of supply voltage.
(iii)
Instantaneous value of supply voltage when time is 4 milli seconds.
(iv)
Impedance for each branch.
(v)
Branch currents.
(vi)
Phase angles between supply voltage and branch currents.
(vii)
Total current
(viii)
Phase angle between supply voltage and total current
(ix)
Circuit power factor
(x)
Equations of instantaneous values of branch currents.
(xi)
Equations of instantaneous values of total current.
(xii)
Total active, reactive and apparent power.
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1.
2.
3.
4.
- 68 -
5.
6.
7.
8.
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9.
10.
11.
- 70 -