EE22151_AC Circuits Analysisst dsdcvssvd

B.TECH – INFORMATION TECHNOLOGY
EE22151 – BASIC ELECTRICAL AND ELECTRONICS
ENGINEERING
(UNIT- 1)
ELECTRICAL CIRCUITS
(PART-2: AC CIRCUIT ANALYSIS)
COURSE INSTRUCTOR
THAMIZMANI S
AP/EEE,SVCE

UNIT I
ELECTRICAL CIRCUITS
• Ohm’s Law – Kirchhoff’s Laws – Steady State Solution of DC
Circuits using Mesh Analysis
• Introduction to AC Circuits – Waveforms and RMS Value – Power
and Power factor, Single Phase and Three Phase AC Balanced
Circuits.

01-11-2023 EE18151-BEEE-UNIT_1 Electrical Circuits and Measurements 3
 Previously you learned that DC sources have ﬁxed polarities and constant
magnitudes and thus produce currents with constant value and unchanging
direction
AC Fundamentals
 In contrast, the voltages of ac sources alternate in polarity and vary in
magnitude and thus produce currents that vary in magnitude and alternate in
direction.

 Sinusoidal ac Voltage
One complete variation is referred to as a cycle.
Starting at zero,
the voltage increases to a positive peak
amplitude, decreases to zero,
changes polarity,
increases to a negative peak amplitude,
then returns again to zero.
Since the waveform repeats itself at regular intervals, it is called a periodic signal.
 Symbol for an ac Voltage Source
Lowercase letter e or v is used
to indicate that the voltage varies with time.
AC Fundamentals

 During the ﬁrst half-cycle,
the source voltage is positive
 Therefore, the current is in
the clockwise direction.
 During the second half-cycle,
the voltage polarity reverses
 Therefore, the current is in the
counterclockwise direction.
 Since current is
proportional to
voltage, its shape is
also sinusoidal
A voltage which changes its polarity at regular intervals of time is called an alternating
voltage. When an alternating voltage is applied in a circuit, the current flows first in one
direction and then in the opposite direction; the direction of current at any instant depends
upon the polarity of the voltage.
Alternating Voltage and Current

Generating ac Voltages
 One way to generate an ac voltage is to rotate a coil of wire at
constant angular velocity in a fixed magnetic field
 The magnitude of the resulting voltage is proportional to the rate at which flux
lines are cut
 its polarity is dependent on the direction the coil sides move through the field.

Sinusoidal Alternating Voltage and Current
The sinusoidal alternating voltage can be expressed by the equation
A sinusoidal current can be expressed in the same way as voltage

Important A.C. Terminology
Waveform:
 The shape of the curve obtained by plotting the instantaneous values of voltage or
current as ordinate against time as abcissa is called its waveform or waveshape.
Fig. shows the waveform of an alternating voltage varying sinusoidally

Instantaneous value:
 The value of an alternating quantity at any instant is called instantaneous value.
 The instantaneous values of alternating voltage and current are represented by v and
i respectively.
 As an example, the instantaneous values of voltage (See Fig.) at 0º, 90º and 270º are
0, + Vm, −Vm respectively
Cycle:
 One complete set of positive and negative values
of an alternating quantity is known as a cycle.
 Fig. shows one cycle of an alternating voltage.
 One cycle corresponds to 360º or 2π radians

Frequency:
 The number of cycles that occur in one second is called the frequency (f) of the
alternating quantity.
 It is measured in cycles/sec (C/s) or Hertz (Hz).
 One Hertz is equal to 1C/s.

Time period:
The time taken in seconds to complete one cycle of an alternating quantity is called its
time period. It is generally represented by T.
It is the inverse of frequency.

Amplitude , Peak-Value, and Peak-to-Peak Value
 The maximum value (positive or negative)
attained by an alternating quantity is called its
amplitude or peak value.
 The amplitude of an alternating voltage or current
is designated by Vm (or Em) or Im
Amplitude (Vm or Em):
It is measured between minimum and maximum peaks.
Peak-to-Peak Value (Vp-p or Ep-p):
Peak Value (Vp or Ep)
The peak value of a voltage or current is its
maximum value with respect to zero.
In this figure : Peak voltage = E + Em

Angular velocity and frequency:
 The coil is rotating with an angular velocity of ω rad/sec in a uniform magnetic field.
 In one revolution of the coil, the angle turned is 2π radians and the voltage wave
completes 1 cycle.
 The time taken to complete one cycle is the time period T of the alternating voltage.

The standard form of an alternating voltage
is given by
Relationship between
radians
Let t =T
rad
/ T rad/sec
rad/sec
Different Forms of Alternating Voltage

In a d.c. system, the voltage and current are constant so that there is no
problem of specifying their magnitudes.
However, an alternating voltage or current varies from instant to instant. A
natural question arises how to express the magnitude of an alternating voltage
or current.
There are four ways of expressing it, namely ;
(i) Peak value
(ii) Average value or mean value
(iii)R.M.S. value or effective value
(iv) Peak-to-peak value
Values of Alternating Voltage and Current

 It is the maximum value attained by an alternating quantity.
 The peak or maximum value of an alternating voltage or current is represented
by Vm or Im.
 The knowledge of peak value is important in case of testing materials
Peak Value

 The average value of a waveform is the average of all its values over a
period of time.
 In performing such a computation, we regard the area above the time axis
as positive area and area below the time axis as negative area.
 The algebraic signs of the areas must be taken into account when
computing the total (net) area.
 The time interval over which the net area is computed is the period T of the
waveform.
Average Value

The equation of an alternating current varying sinusoidally is given by
𝐚𝐯
∫ 𝒊 𝒅𝜽
𝝅
𝟎
𝝅
∫ 𝑰𝒎 𝜽 𝒅𝜽
𝝅
𝟎
𝝅
𝐚𝐯
𝒎
𝟎
𝝅
𝐚𝐯
𝟐𝑰𝒎
𝝅
= 0.637 𝒎
Average Value, 𝐚𝐯
𝐀𝐫𝐞𝐚 𝐨𝐟 𝐨𝐧𝐞 𝐚𝐥𝐭𝐞𝐫𝐚𝐭𝐢𝐨𝐧
𝐛𝐚𝐬𝐞 𝐥𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝐨𝐧𝐞 𝐚𝐥𝐭𝐞𝐫𝐚𝐭𝐢𝐨𝐧
𝐚𝐯
𝟐𝑽𝒎
𝝅
= 0.637 𝒎
Average Value of Sinusoidal Current
Similarly for Voltage

 The effective or r.m.s. value of an alternating current is that steady current (d.c.)
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.
For example, when we say that the r.m.s. or effective value of an alternating current
is 5A, it means that the alternating current will do work (or produce heat) at the same
rate as 5A direct current under similar conditions.
R.M.S. or Effective Value

The r.m.s. value of symmetrical wave can also be expressed as :
The r.m.s. or effective value of an alternating voltage can similarly be
expressed as :
R.M.S. or Effective Value

The equation of an alternating current varying sinusoidally is given by
𝒓𝒎𝒔
∫ 𝒊𝟐 𝒅𝜽
𝝅
𝟎
𝝅
∫ 𝑰𝒎
𝟐 𝒔𝒊𝒏𝟐𝜽 𝒅𝜽
𝝅
𝟎
𝝅
𝒓𝒎𝒔
𝑰𝒎
𝝅
𝟏 𝟐𝜽
𝟐
𝝅
𝟎
=
𝑰𝒎
𝟐 𝝅
𝜽
𝟐 𝟎
𝝅
=
𝑰𝒎
𝟐 𝝅
𝒓𝒎𝒔
𝑰𝒎
𝟐
= 0.707 𝒎
R.M.S value, 𝒓𝒎𝒔
𝐀𝐫𝐞𝐚 𝐨𝐟 𝐡𝐚𝐥𝐟 𝐜𝐲𝐜𝐥𝐞 𝐨𝐟 𝐬𝐪𝐮𝐚𝐫𝐞 𝐰𝐚𝐯𝐞
𝐡𝐚𝐥𝐟 𝐜𝐲𝐜𝐥𝐞 𝐛𝐚𝐬𝐞
𝒓𝒎𝒔
𝑽𝒎
𝟐
= 0.707 𝒎
R.M.S. Value of Sinusoidal Current
Similarly for Voltage

Form factor:
 The ratio of r.m.s. value to the average value of an alternating quantity is known as
form factor
Form Factor and Peak Factor

Peak factor:
 The ratio of maximum value to the r.m.s. value of an alternating quantity is known
as peak factor
 Peak factor is also called crest factor or amplitude factor.
Form Factor and Peak Factor

Voltages and Currents with Phase Shifts
Leading w.r.t Reference wave
 If a sine wave does not pass through zero at t =0 s, it has a phase shift.
 Waveforms may be shifted to the left or to the right
Lagging w.r.t Reference wave

Phasor Difference
 Phase difference refers to the angular displacement between different waveforms of the same
frequency.
 The terms lead and lag can be understood in terms of phasors. If you observe phasors rotating as in
Figure, the one that you see passing first is leading and the other is lagging.

AC Series circuits
1. A.C. Circuit Containing Resistance (R) Only
2. A.C. Circuit Containing Inductance (L) Only
3. A.C. Circuit Containing Capacitance (C) Only
4. RL series A.C. circuit
5. RC series A.C. circuit
6. RLC series A.C. circuit

Let the alternating voltage be given by the equation
𝑊ℎ𝑒𝑟𝑒, 𝐼 =
𝑉
𝑅
Circuit diagram
Phasor diagram
𝑉 = 𝑖𝑍 𝑜𝑟 𝑖 =
𝑣
𝑍
=
𝑣
𝑅
𝒊 =
𝑽𝒎
𝑹
𝐬𝐢𝐧 𝝎𝒕
𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕
𝑽𝒓𝒎𝒔 =
𝑽𝒎
𝟐
𝑰𝒓𝒎𝒔 =
𝑰𝒎
𝟐
A.C. Circuit Containing Resistance Only

Phase angle:
The phase angle = 𝟎𝟎
Instantaneous Power:
In any circuit, electric power consumed at any instant is the product
of voltage and current at that instant
The applied voltage and the circuit current are in phase with each
other for pure resistive circuit
Voltage & Current waveform
Power waveform, p = vi
𝐀𝐯𝐞𝐫𝐚𝐠𝐞 𝐏𝐨𝐰𝐞𝐫, 𝑷𝐚𝐯 𝒐𝒓 (𝑷) =
𝟏
𝟐𝝅
𝑽𝒎 𝑰𝒎𝒔𝒊𝒏𝟐
𝝎𝒕 𝒅𝝎𝒕
𝟐𝝅
𝟎
𝑷 =
𝑽𝒎𝑰𝒎
𝟐
=
𝑽𝒎
𝟐
𝑰𝒎
𝟐
= 𝑽 𝑰
A.C. Circuit Containing Resistance Only

Circuit diagram
Phasor diagram
Where,
Inductive Reactance, 𝑿𝑳 = 𝝎𝑳
𝑿𝑳= 𝟐𝝅𝒇𝑳
𝒊 =
𝑽𝒎
𝑿𝑳
𝐬𝐢𝐧 𝝎𝒕 − 𝝅
𝟐
𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 − 𝝅
𝟐
A.C. Circuit Containing Pure Inductance Only

Phase angle:
Power:
The phase angle = 𝟗𝟎𝟎
( i lags v by 𝟗𝟎𝟎
)
Hence power absorbed in pure inductance is zero
The current lags behind the voltage by π/2 radians or 90º. Hence in
a pure inductance, current lags the voltage by 90
A.C. Circuit Containing Pure Inductive Only

Circuit diagram
Phasor diagram
Where,
Capacitive Reactance, 𝑪
𝟏
𝝎𝑪
𝟏
𝟐𝝅𝒇𝑪
𝒊 =
𝑽𝒎
𝟏
𝝎𝑪
𝐬𝐢𝐧 𝝎𝒕 + 𝝅
𝟐 =
𝑽𝒎
𝑿𝑪
𝐬𝐢𝐧 𝝎𝒕 + 𝝅
𝟐
𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 + 𝝅
𝟐
A.C. Circuit Containing Pure Capacitor Only

Phase angle:
Power:
The phase angle = 𝟗𝟎𝟎
( i leads v by 𝟗𝟎𝟎
)
Hence power absorbed in a pure capacitance is zero.
The current leads the voltage by π/2 radians or 90º. Hence in a
pure capacitance, current leads the voltage by 90º
A.C. Circuit Containing Pure Capacitance Only

Formulas for AC circuits for R or L or C
Electrical Parameter Pure Resistive
Circuit
Pure Inductive
Circuit
Pure Capacitive Circuit
Instantaneous Voltage (v) 𝒗 = 𝑽𝒎 𝐬𝐢𝐧 𝝎𝒕 𝒗 = 𝑽𝒎 𝐬𝐢𝐧 𝝎𝒕 𝒗 = 𝑽𝒎 𝐬𝐢𝐧 𝝎𝒕
Instantaneous Current (i) 𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 − 𝝅
𝟐 𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 + 𝝅
𝟐
RMS voltage (V)
𝑽 = I R =
𝑽𝒎
𝟐
𝑽 = I 𝑿𝑳 =
𝑽𝒎
𝟐
𝑽 = I 𝑿𝑪 =
𝑽𝒎
𝟐
RMS Current (I)
𝑰 =
𝑽
𝑹
=
𝑰𝒎
𝟐
𝑰 =
𝑽
𝑿𝑳
=
𝑰𝒎
𝟐
𝑰 =
𝑽
𝑿𝑪
=
𝑰𝒎
𝟐
Impedance (Z) Z = R 𝒁 = 𝑿𝑳 = 𝝎𝑳 = 𝟐𝝅𝒇𝑳 𝒁 = 𝑿𝑪 = 𝟏
𝝎𝑪 = 𝟏
𝟐𝝅𝒇𝑪
Maximum Current 𝑰𝒎
𝑰𝒎=
𝑽𝒎
𝑹
𝑰𝒎=
𝑽𝒎
𝑿𝑳
𝑰𝒎=
𝑽𝒎
𝑿𝑪
Maximum Voltage 𝑽𝒎 𝑽𝒎= 𝑰𝒎 R 𝑽𝒎= 𝑰𝒎 𝑿𝑳 𝑽𝒎= 𝑰𝒎 𝑿𝑪
Phase angle 0 90 90
Instantaneous power p 𝒑 = 𝑽𝒎𝑰𝒎𝐬𝐢𝐧𝟐
𝝎𝒕 𝒑 = −𝑽𝑰 𝐬𝐢𝐧 𝟐𝝎𝒕 𝒑 = 𝑽𝑰 𝐬𝐢𝐧 𝟐𝝎𝒕
Average Power P = V I 0 0

1. An a.c. circuit consists of a pure resistance of 10 Ω and is connected
across an a.c. supply of 230 V, 50 Hz. Calculate (i) current (ii) power
consumed and (iii) equations for voltage and current.
Solution:
𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕
(iii)The voltage and current equation are
𝒗 = 𝑽𝒎 𝐬𝐢𝐧 𝝎𝒕
Problems on Resistive circuit
Given Data: R=10 Ω, V = 230 V and f = 50 Hz

2. A Voltage of 100 is applied to a 10 ohm resistor. Find the
Voltage, Current, Instantaneous current, Instantaneous power and
Average Power.
Solution:
𝒊 =
𝒗
𝑹
=
𝟏𝟎𝟎 𝐬𝐢𝐧 𝝎𝒕
𝟏𝟎
= 𝟏𝟎 𝐬𝐢𝐧 𝝎𝒕
The instantaneous current equation is
Problems on Resistive circuit
Given Data: R=10 Ω and
𝑽 =
𝑽𝒎
𝟐
=
𝟏𝟎𝟎
𝟐
= 𝟕𝟎. 𝟕𝟐 𝑽
The RMS Voltage is
The RMS Current is
𝑰 =
𝑰𝒎
𝟐
=
𝟏𝟎
𝟐
= 𝟕. 𝟎𝟕𝟐 𝑨
The Average Power is
P = V I = 70.72 x 7.072 = 500 Watts
Instantaneous power
𝒑 = 𝒗𝒊 = 𝟏𝟎𝟎 𝐬𝐢𝐧 𝝎𝒕 × 𝟏𝟎 𝐬𝐢𝐧 𝝎𝒕
= 1000 𝐬𝐢𝐧𝟐
𝝎𝒕

3. A pure inductive coil allows a current of 10 A to flow from a 230 V, 50
Hz supply. Find (i) inductive reactance (ii) inductance of the coil (iii)
power absorbed. Write down the equations for voltage and current
Solution: (iv)The voltage and current equation are
𝒗 = 𝑽𝒎 𝐬𝐢𝐧 𝝎𝒕 𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 − 𝝅
𝟐
Problems on Inductive circuit
Given Data: I=10 A, V = 230 V and f = 50 Hz

4. A pure inductance L = 0.2 H has an applied voltage . Find
the instantaneous current, instantaneous and average powers, inductive
reactance and the RMS current
Solution:
𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 − 𝝅
𝟐
= 𝟏. 𝟓𝟗𝟐 𝐬𝐢𝐧 𝟑𝟏𝟒𝒕 − 𝟗𝟎°
Problems on Inductive circuit
𝑳 = 62.8 ohm
Inductive Reactance:
𝒎
𝑽𝒎
𝑿𝑳
𝟏𝟎𝟎
𝟔𝟐.𝟖
= 1.592 A
The instantaneous current equation is:
Average Power P = 0 Watts
I
𝑰𝒎
𝟐
=
𝟏.𝟓𝟗𝟐
𝟐
RMS Current:
Instantaneous Power:
=
𝑽𝒎𝑰𝒎
𝟐
𝟏𝟎𝟎×𝟏.𝟓𝟗𝟐
𝟐
79.6

5. A 318 µF capacitor is connected across a 230 V, 50 Hz system. Determine (i)
the capacitive reactance (ii) r.m.s. value of current and (iii) equations for
voltage and current.
Solution: (iii)The voltage and current equation are
𝒗 = 𝑽𝒎 𝐬𝐢𝐧 𝝎𝒕 𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 + 𝝅
𝟐
Problems on Capacitive circuit
Given Data: C=318 µF, V = 230 V and f = 50 Hz

R-L Series A.C. Circuit
V = r.m.s. applied voltage
I = r.m.s. value of the circuit current
𝑉 = I R ... where 𝑉 is in phase with I
𝑉 = I 𝑋 ... where 𝑉 leads I by 90°
Let the applied voltage is
Now the current equation
for RL circuit
𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 − ∅
𝑰𝒎=
𝑽𝒎
𝒁
Where
j = +90°
𝑉 = 𝑉 + 𝑉
= 𝐼𝑅 + 𝑗𝐼𝑋
𝑉 = 𝐼(𝑅 + 𝑗𝑋 )
𝑉
𝐼
= 𝑅 + 𝑗𝑋
𝒁 = 𝑹 + 𝒋𝑿𝑳
By KVL
𝒁 = 𝑹 + 𝒋𝑿𝑳
𝐙 = 𝑹𝟐 + 𝑿𝑳
𝟐

Impedance 𝐙 = 𝑹𝟐 + 𝑿𝑳
𝟐
𝑽𝟐
= 𝑽𝑹
𝟐
+ 𝑽𝑳
𝟐
Power Factor
𝑽𝑳
𝑽𝑹
𝑰𝑹
𝑰
𝑹
𝒘𝒉𝒆𝒓𝒆 𝑿𝑳 = 𝝎𝑳 = 𝟐𝝅𝒇𝑳
𝑳= Inductive Reactance
𝑽
𝑹𝟐 𝑿𝑳
𝟐
𝑽
𝒁
R-L Series A.C. Circuit
𝑺𝟐
= 𝑷𝟐
+ 𝑸𝟐
𝑹𝒆𝒂𝒍 𝑷𝒐𝒘𝒆𝒓, 𝑷 = 𝑽𝑰 𝐜𝐨𝐬 ∅ (Watts)
𝑹𝒆𝒂𝒄𝒕𝒊𝒗𝒆 𝒑𝒐𝒘𝒆𝒓, 𝑸 = 𝑽𝑰 𝒔𝒊𝒏 ∅
(VAR)
𝑨𝒑𝒑𝒂𝒓𝒆𝒏𝒕 𝑷𝒐𝒘𝒆𝒓, 𝑺 = 𝑽𝑰 (VA)
Power:

R-C Series A.C. Circuit
Now the current equation
for RC circuit
𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 + ∅
𝑰𝒎=
𝑽𝒎
𝒁
Where
-j = -90°
𝑉 = 𝑉 + 𝑉
= 𝐼𝑅 − 𝑗𝐼𝑋
𝑉 = 𝐼(𝑅 − 𝑗𝑋 )
𝑉
𝐼
= 𝑅 − 𝑗𝑋
𝒁 = 𝑹 − 𝒋𝑿𝑪
By KVL
𝒁 = 𝑹 − 𝒋𝑿𝑪
V = r.m.s. applied voltage
I = r.m.s. value of the circuit current
𝑉 = I 𝑋 ... where 𝑉 lags I by 90°
𝐙 = 𝑹𝟐 + 𝑿𝑪
𝟐

Impedance 𝐙 = 𝑹𝟐 + 𝑿𝑪
𝟐
𝑽𝟐
= 𝑽𝑹
𝟐
+ −𝑽𝑪
𝟐
Power Factor:
𝑽𝑪
𝑽𝑹
𝑰
𝑰𝑹 𝑹
𝑪
𝟏
𝝎𝑪
𝟏
𝟐𝝅𝒇𝑪
𝑪 Capacitive Reactance
𝑽
𝑹𝟐 𝑿𝑪
𝟐
𝑽
𝒁
R-C Series A.C. Circuit
𝑺𝟐
= 𝑷𝟐
+ 𝑸𝟐
(VAR)
𝑨𝒑𝒑𝒂𝒓𝒆𝒏𝒕 𝑷𝒐𝒘𝒆, 𝑺 = 𝑽𝑰 (VA)
Power:

𝑉 = I 𝑋 ... where 𝑉 lags I by 90°
𝑉 = I 𝑋 ... where 𝑉 leads I by 90°
R-L-C Series A.C. Circuit
𝒁 = 𝑹 + 𝒋(𝑿𝑳 − 𝑿𝑪)
𝑉 = 𝑉 +𝑉 +𝑉
= 𝐼𝑅 + 𝑗𝐼𝑋 − 𝑗𝐼𝑋
𝑉 = 𝐼 𝑅 + 𝑗(𝑋 − 𝑋 )
𝑉
𝐼
= 𝑅 + 𝑗(𝑋 − 𝑋 )
𝒁 = 𝑹 + 𝒋(𝑿𝑳 − 𝑿𝑪)
By KVL
𝟐
𝑳 𝑪
𝟐

𝟐
𝑳 𝑪
𝟐
Power Factor:
𝑽𝑳 𝑽𝑪
𝑽𝑹
𝑿𝑳
𝑹
𝑪
𝟏
𝝎𝑪
𝟏
𝟐𝝅𝒇𝑪
𝑳
𝑽
𝑹𝟐 𝑿𝑳 𝑿𝑪
𝟐
𝑽
𝒁
(Kvar)
𝑨𝒑𝒑𝒂𝒓𝒆𝒏𝒕 𝑷𝒐𝒘𝒆𝒓, 𝑺 = 𝑽𝑰 (VA)
Power:
𝟐
𝑳 𝑪
𝟐
R-L-C Series A.C. Circuit

Formulas for AC circuits for RL, RC and RLC Series Circuit
Electrical Parameter RL Circuit RC Circuit RLC Circuit
Instantaneous Voltage (v) 𝒗 = 𝑽𝒎 𝐬𝐢𝐧 𝝎𝒕 𝒗 = 𝑽𝒎 𝐬𝐢𝐧 𝝎𝒕 𝒗 = 𝑽𝒎 𝐬𝐢𝐧 𝝎𝒕
Instantaneous Current (i) 𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 − ∅ 𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 + ∅ 𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 ± ∅
RMS Voltage
𝐕 = 𝑽𝑹
𝟐
+ 𝑽𝑳
𝟐
𝐕 = 𝑽𝑹
𝟐
+ 𝑽𝑪
𝟐
𝐕 = 𝑽𝑹
𝟐
+ 𝑽𝑳 − 𝑽𝑪
𝟐
Impedance (Z)
𝐙 = 𝑹𝟐 + 𝑿𝑳
𝟐
𝐙 = 𝑹𝟐 + 𝑿𝑪
𝟐 𝒁 = 𝑹𝟐 + 𝑿𝑳 − 𝑿𝑪
𝟐
RMS Current (I)
𝐈 =
𝑽
𝒁
𝐈 =
𝑽
𝒁
𝐈 =
𝑽
𝒁
Power Factor (cos ∅)
𝐜𝐨𝐬 ∅ =
𝑹
𝒁
𝒍𝒂𝒈 𝐜𝐨𝐬 ∅ =
𝑹
𝒁
𝒍𝒆𝒂𝒅 𝐜𝐨𝐬 ∅ =
𝑹
𝒁
Real Power (P) 𝑷 = 𝑽𝑰 𝐜𝐨𝐬 ∅ 𝑷 = 𝑽𝑰 𝐜𝐨𝐬 ∅ 𝑷 = 𝑽𝑰 𝐜𝐨𝐬 ∅
Reactive Power (Q) 𝑸 = 𝑽𝑰 𝒔𝒊𝒏 ∅ 𝑸 = 𝑽𝑰 𝒔𝒊𝒏 ∅ 𝑸 = 𝑽𝑰 𝒔𝒊𝒏 ∅
Apparent Power (S) 𝑺 = 𝑽𝑰 𝑺 = 𝑽𝑰 𝑺 = 𝑽𝑰
Phase angle 𝐭𝐚𝐧 ∅ =
𝑽𝑳
𝑽𝑹
𝑹
𝑿𝑳
𝐭𝐚𝐧 ∅ =
𝑽𝑪
𝑽𝑹
𝑿𝑪
𝑹
𝐭𝐚𝐧 ∅ =
𝑽𝑳 𝑽𝑪
𝑽𝑹
𝑿𝑳 𝑿𝑪
𝑹

Three cases of R-L-C series circuit.
𝑳 𝑪
𝒎
𝑳 𝑪
𝒎
𝑳 𝑪
𝒎
𝒎
𝑹 = I R : 𝑪 = I 𝑪 : 𝑳 = I 𝑳

1. A coil having a resistance of 7 Ω and an inductance of 31·8 mH is
connected to 230 V, 50 Hz supply. Calculate (i) the circuit current (ii)
phase angle (iii) power factor (iv) power consumed and (v) voltage
drop across resistor and inductor
Solution:
Problems on RL circuit
Given Data: R=7 Ω , L = 31.8mH , V= 230 V
and f = 50 Hz lag
= = 𝟎

2. A capacitor of capacitance 79·5 µ F is connected in series with a
non-inductive resistance of 30 Ω across 100 V, 50 Hz supply. Find (i)
impedance (ii) current (iii) phase angle and (iv) equation for the
instantaneous value of current.
Solution:
Problems on RC circuit
Given Data: R=30 Ω , C = 79.5 µ F , V= 100 V
and f = 50 Hz
=
=
.
𝟎
lead
iv) The current equation for RC circuit is
𝒎

3. A 230 V, 50 Hz a.c. supply is applied to a coil of 0.06 H inductance
and 2.5 Ω resistance connected in series with a 6·8 µF capacitor.
Calculate (i) impedance (ii) current (iii) phase angle between current
and voltage (iv) power factor and (v) power consumed.
Solution:
Problems on RLC circuit
Given Data: R=2.5 Ω , L = 0.06 H, C = 6.8 µ F ,
V= 230 V and f = 50 Hz
=
𝟎
lead
𝑿𝑳 − 𝑿𝑪 < 𝟎 (-ve) then the circuit behave like RC circuit
lead

Generation of three phase voltages
• The 3-Phase voltage can be produced in a stationary armature with rotating field or in
a rotating armature with a stationary field.
• Three separate but identical set of coils placed 120 degree electrically apart.
• Hence voltage generated in them are 120 degree apart.
• Three voltages are of the same magnitude and frequency
• The voltages are assumed to be sinusoidal
• Counting the time from the instant when the voltage in R phase is zero
the equations for the instantaneous voltages of the 3 phases are expressed as
𝑹𝑹 𝒎
𝒀𝒀 𝒎
𝑩𝑩 𝒎
𝒎
Three phase circuits
𝑹𝑹
𝒀𝒀
𝑩𝑩

Generation of three phase voltages
𝑹𝑹
𝒀𝒀
𝑩𝑩

3 phase coil Arrangement Generated EMF
𝒂𝒏 𝑹𝑹 𝒎
𝒃𝒏 𝒀𝒀 𝒎
𝒄𝒏 𝑩𝑩 𝒎
𝒎
Instantaneous Three phase voltage equations are

• At any given instant, the algebraic sum of the three voltages is zero.
• The vector sum of ER + Ey +EB = 0
Phasor diagram
𝑹𝑹 𝒎
𝒀𝒀 𝒎
𝑩𝑩 𝒎
𝒎
Instantaneous Three phase voltage equations are
𝑹𝑹
𝒀𝒀
𝑩𝑩

• The order in which the alternating quantity attain their maximum values is
called the phase sequence
RYB or (+) sequence RBY (-) sequence
Phase sequence
R
Y
B
R
B
Y
𝑹𝑹
𝒀𝒀
𝑩𝑩
𝑹𝑹
𝑩𝑩
𝒀𝒀

Star or wye connection Delta or mesh connection
Interconnection of the phases
𝑉 = 𝑉 = 𝑉 = 𝑉 = 𝑃ℎ𝑎𝑠𝑒 𝑉𝑜𝑙𝑎𝑡𝑔𝑒𝑠
𝑉 = 𝑉 = 𝑉 = 𝑉 = 𝐿𝑖𝑛𝑒 𝑉𝑜𝑙𝑎𝑡𝑔𝑒𝑠
𝑳
𝑷
𝑳 𝑷
𝑳 𝑷
𝑳
𝑷

Balanced star connected voltage source
Notation Defined:
𝑷
𝑳
𝑳 𝑷

Balanced star connected voltage source
𝑷𝒐𝒘𝒆𝒓 𝒄𝒐𝒏𝒔𝒖𝒎𝒆𝒅 𝒊𝒏 𝒐𝒏𝒆 𝒑𝒉𝒂𝒔𝒆, 𝑷 = 𝑬𝑷 𝑰𝑷 𝐜𝐨𝐬 ∅
𝑷𝒐𝒘𝒆𝒓 𝒄𝒐𝒏𝒔𝒖𝒎𝒆𝒅 𝒊𝒏 𝑻𝒉𝒓𝒆𝒆 𝒑𝒉𝒂𝒔𝒆, 𝑷 = 𝟑 𝑬𝑷 𝑰𝑷 𝐜𝐨𝐬 ∅
𝑹𝒆𝒂𝒄𝒕𝒊𝒗𝒆 𝒑𝒐𝒘𝒆𝒓, 𝑸 = 𝟑 𝑽𝑳 𝑰𝑳 𝒔𝒊𝒏 ∅
(Kvar)
𝑨𝒑𝒑𝒂𝒓𝒆𝒏𝒕 𝑷𝒐𝒘𝒆, 𝑺 = 𝟑 𝑽𝑳 𝑰𝑳(VA)
Power Relations:
Let Consider be the power factor of the system
𝑷 = 𝟑 𝑽𝑳 𝑰𝑳 𝐜𝐨𝐬 ∅ (Watts)
𝑳 𝑷

Balanced Delta connected voltage source
Notation Defined:
𝑳 𝑷 𝑳
𝑷

𝑷𝒐𝒘𝒆𝒓 𝒄𝒐𝒏𝒔𝒖𝒎𝒆𝒅 𝒊𝒏 𝒐𝒏𝒆 𝒑𝒉𝒂𝒔𝒆, 𝑷 = 𝑬𝑷 𝑰𝑷 𝐜𝐨𝐬 ∅
𝑷𝒐𝒘𝒆𝒓 𝒄𝒐𝒏𝒔𝒖𝒎𝒆𝒅 𝒊𝒏 𝑻𝒉𝒓𝒆𝒆 𝒑𝒉𝒂𝒔𝒆, 𝑷 = 𝟑 𝑬𝑷 𝑰𝑷 𝐜𝐨𝐬 ∅
𝑹𝒆𝒂𝒄𝒕𝒊𝒗𝒆 𝒑𝒐𝒘𝒆𝒓, 𝑸 = 𝟑 𝑽𝑳 𝑰𝑳 𝒔𝒊𝒏 ∅
(Var)
𝑨𝒑𝒑𝒂𝒓𝒆𝒏𝒕 𝑷𝒐𝒘𝒆, 𝑺 = 𝟑 𝑽𝑳 𝑰𝑳(VA)
Power Relations:
Let Consider be the power factor of the system
𝑷 = 𝟑 𝑽𝑳 𝑰𝑳 𝐜𝐨𝐬 ∅ (Watts)
Balanced Delta connected voltage source
𝑳 𝑷

Formulas for balanced Star and Delta Circuit
Electrical Parameter Star / Wye Delta / Mesh
Phase Voltage
𝑽𝑷 =
𝑽𝑳
𝟑
𝑽𝑳 = 𝑽𝑷
Phase Current
𝑰𝑷 =
𝑽𝑷
𝒁
𝑰𝑷 =
𝑽𝑷
𝒁
Line Current 𝑰𝑳 = 𝑰𝑷 𝑰𝑳 = 𝟑𝑰𝑷
Power Factor (cos ∅)
𝐜𝐨𝐬 ∅ =
𝑹
𝒁
𝐜𝐨𝐬 ∅ =
𝑹
𝒁
Real Power (P) 𝑷 = 𝟑 𝑽𝑳 𝑰𝑳 𝒄𝒐𝒔 ∅ 𝑷 = 𝟑 𝑽𝑳 𝑰𝑳 𝒄𝒐𝒔∅
Reactive Power (Q) 𝑸 = 𝟑 𝑽𝑳 𝑰𝑳 𝒔𝒊𝒏 ∅ 𝑸 = 𝟑 𝑽𝑳 𝑰𝑳 𝒔𝒊𝒏 ∅
Apparent Power (S) 𝑺 = 𝟑 𝑽𝑳 𝑰𝑳 𝑺 = 𝟑 𝑽𝑳 𝑰𝑳

1. Three coils, each having a resistance of 20 Ω and an inductive reactance of 15
Ω, are connected in star to a 400 V, 3-phase, 50 Hz supply. Calculate (i) the line
current (ii) power factor and (iii) power supplied
Solution:
Problems in balanced Star circuit
Given Data: R=20 Ω , 𝑿𝑳= 𝟏𝟓 , 𝑽𝑳= 400 V and
f = 50 Hz
lag
Phase Impedance
= Ω
Phase Voltage = 230.94V
Phase Current
.
i) Line Current
9.237 A
iii) Power supplied
= x 400 x 9.237 x 0.8
= 5119.66 Watts

2. Three similar coils each having a resistance of 5 Ω and an inductance of 0.02H
are connected in delta to a 440V, 3-phase, 50Hz supply. Calculate the line current
and total power absorbed.
Solution:
Problems in balanced Delta circuit
Given Data: R=5 Ω , L = 0.02 H , 𝑽𝑳= 440 V
and f = 50 Hz
.
lag
Phase Impedance
= Ω
Phase Current
.
Phase Voltage = 440V
Power observed
= x 440 x 94.94 x 0.622
= 45004.18 Watts
= 6.28 Ω
Line Current = x
= 94.94 A

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