AC FUNDAMENTALS
Periodic waves
A periodic wave is a wave in which the particles of the medium oscillate
continuously repeating their vibratory motion regularly at fixed intervals of time.
A wave whose displacement has a periodic variation with time or distance, or both.
A number of parameters can be defined to describe a periodic wave:
Frequency - One of the most important properties of waveform is to identify the
number of complete cycles it goes through in a fixed period of time. For standard
measurements, the period of time is one second, so the frequency of the wave is
commonly measured in cycles per second (cycles/sec) and, in normal usage, is
expressed in units of Hertz (Hz). It is represented by the letter ‘f ’.
Period - Sometimes we need to know the amount of time required to complete
one cycle of the waveform, rather than the number of cycles per second of time.
This is logically the reciprocal of frequency. Thus, period is the time duration of
one cycle of the waveform, and is measured in seconds/cycles or seconds.
Wavelength (λ)- A waveform moves physically as well as changing in time,
sometimes we need to know how far it moves in one cycle of the wave, rather than
how long that cycle takes to complete. This of course depends on how fast the
wave is moving as well. The Greek letter (lambda) is used to represent wavelength
in mathematical expressions. And, λ = c/f. As shown in the figure to the above,
wavelength can be measured from any part of one cycle to the equivalent point in
the next cycle. Wavelength is very similar to period except that wavelength is
measured in distance (or length) per cycle while period is measured in time per
cycle.
Amplitude (A) - the maximum distance a particle gets from its undisturbed
position.
Types of Periodic Waveform
AC Waveform
Direct Current or D.C. as it is more commonly called, is a form of current that
flows around an electrical circuit in one direction only, making it a
“Unidirectional” supply.
DC Circuit and Waveform
AC Waveform on the other hand is defined as one that varies in both magnitude
and direction in more or less an even manner with respect to time making it a “Bidirectional” waveform.
Generating AC Voltages
Thus the expression for induced e.m.f. generated in each conductor is given by
e.m.f. generated = Blv sinθ volt
The total e.m.f. generated in the loop = 2Blv sinθ volt
where, B= Flux density of the magnetic field in wb/m2
l= active length of the conductor in meter
v = linear velocity of the conductor in meter/sec.
ω = θ / t = angular velocity of the conductor in radian / sec.
Now Em = 2Blv volts = Maximum value of the induced e.m.f. in the conductor.
This is achieved at θ=90°.
Hence the equation giving the instantaneous e.m.f. induced in the conductor is
given by
e = Em sinθ volts.
Alternating Current or Voltage
• A sinusoidal ac waveform starts at zero
– Increases to a positive maximum
– Decreases to zero
– Changes polarity
– Increases to a negative maximum
– Returns to zero.
Alternating Waveform
In the figure above, Em represents the maximum value of the e.m.f. and e is the
value of the e.m.f. after the loop or the conductors has been rotated through an
angle θ from the position of zero e.m.f.
AC Waveform Characteristics
1. Period
• The time required for an ac waveform to complete one full cycle is called
the period (T).
• A cycle consists of one complete positive, and one complete negative
alternation.
The period of a given ac wave is the same for each cycle
Measurement of the period
The period of a sine wave can be measured between any two corresponding points
on the waveform.
2. Frequency
• Frequency ( f ) is the number of cycles that an ac wave completes in
one second.
– The more cycles completed in one second, the higher the
Frequency.
– Frequency is measured in hertz (Hz)
• Relationship between frequency ( f ) and period (T) is: f = 1/T
Illustration of frequency
T = 333mS
F = 3Hz
(3 Cycles/Sec)
T = 200mS
F = 5Hz
(5 Cycles/Second)
3. Amplitude
• The Amplitude is the magnitude or intensity of the signal waveform
measured in volts or amps.
• Amplitude, better known as its Maximum or Peak value represented
by the terms, Vmax (Vp) for voltage or Imax (Ip) for current. The peak
value of an ac wave is the value of voltage or current at the positive or
negative maximum with respect to zero.
★ Peak-to-Peak Values
✓ The peak-to-peak value of an ac wave is the voltage or current from the
positive peak to the negative peak.
✓ The peak-to-peak value is twice the actual voltage value
✓ Not Often Used
✓ The peak-to-peak values are represented as:Vpp and Ipp where: Vpp =
2Vp and Ipp = 2Ip
8V Peak (Actual Value) = 16V Peak-To-Peak
4. Average Value
• The average value of an AC current or voltage is the average of all the
instantaneous values during one alternation. They are actually DC values.
Considering the general case of a voltage waveform which cannot be
represented by a simple mathematical expression. Figure below shows this
waveform.
Average Value of an AC Waveform
Where: n equals the actual number of mid-ordinates taken over either the positive
or the negative half cycle.
Or alternatively
Vaverage = area enclosed over half cycle / length of the base over half cycle.
Similarly if it is a current waveform similar in nature of that of a voltage waveform
as shown above,
Where: n equals the actual number of mid-ordinates used.
Or alternatively
Iaverage =
area enclosed over half cycle / length of the base over half cycle.
5. R.M.S. Value
Fig. RMS Value of ac waveform
If the current (a) represented in above Fig. is passed through a resistor having
resistance R ohms, the heating effect of i1 is i12 R, that of i2 is i 22R, etc. as shown in
Fig. (b).The variation of the heating effect during the second half-cycle is exactly
the same as that during the first half-cycle.
Average heating effect = (i12R + i22R + ………..in2R) / n
Let I to be the value of direct current through the same resistance R to produce a
heating effect equal to the average heating effect of the alternating current, then
= square root of the mean of the squares of the
current
= root-mean-square (r.m.s.) value of current
✓ The r.m.s (root mean square) value, or effective value, of an alternating
current is defined as that value of steady current which would produce the
same heating effect in the same resistance.
✓ It is sometimes called the resistive DC equivalent value.
✓ Most AC sources are specified with the RMS Value.
Consider the two circuits shown below:
If the bulbs light with the same brightness (that is, they are working at the same
power), then it would be logical to regard the current Iac as being equivalent to the
current Idc.
The power dissipated by the direct current is P = I2dcR and the power dissipated
by the alternating current is P = I2acR = I2rmsR.
When the same amount of heat is being produced by the resistor in both
setups, the ac voltage has an rms value equal to the dc voltage.
Alternatively, the average heating effect or mean power may be expressed as :
Average heating effect over half cycle = area enclosed by i²R curve over half
cycle / length of the base.
Form Factor and Crest Factor
Form Factor is the ratio between the average value and the RMS value and is
given as:
For a pure sinusoidal waveform the Form Factor will always be equal to 1.11.
Crest Factor is the ratio between the R.M.S. value and the Peak value of the
waveform and is given as:
For a pure sinusoidal waveform the Crest Factor will always be equal to 1.414.
Average and R.M.S. values of Sinusoidal Waveform
Average and R.M.S. value of Sinusoidal current
For a very small interval dθ radians, the area of the shaded strip is i · dθ ampere
radians. The use of the unit ‘ampere radian’ avoids converting the scale on the
horizontal axis from radians to seconds.
Therefore total area enclosed by the current wave over half-cycle is
=
From the above
Average value of current over half cycle = Iav = 2Im ampere radian / π radian
=0.637ampere.
If the current is passed through a resistor having resistance R ohms, instantaneous
heating effect = i 2R watts. The variation of i 2R during a complete cycle is shown
in Fig.(b). During interval dθ radians, heat generated is i 2R · dθ watt radians and is
represented by the area of the shaded strip. Hence
Heat generated during the first half-cycle = area enclosed by the i 2R curve
The average heating effect = (π / 2)Im2R watt radian / π radian = (Im2R / 2) watts.
If I is the value of direct current through the same resistance to produce the same
heating effect,
I2R = (Im2R / 2) and I = Im / √2 = 0.707Im
Representation of an Alternating quantity by a phasor
A vector is a quantity having direction as well as magnitude, especially as
determining the position of one point in space relative to another.
Phasor
• A rotating vector, simply called a “Phasor” is a scaled line whose
length represents an AC quantity that has both magnitude (“peak
amplitude”) and direction (“phase”) which is “frozen” at some point in
time.
• A phasor is a vector that has an arrow head at one end which signifies
partly the maximum value of the vector quantity ( V or I ) and partly the
end of the vector that rotates.
• Generally, vectors are assumed to pivot at one end around a fixed zero
point known as the “point of origin” while the arrowed end representing
the quantity, freely rotates in an anti-clockwise direction at an angular
velocity, ( ω ) of one full revolution for every cycle. This anti-clockwise
rotation of the vector is considered to be a positive rotation. Likewise, a
clockwise rotation is considered to be a negative rotation.
• The projection of the phasor onto the vertical axis represents the
instantaneous value of the quantity it represents.
• Phasor apply only to sinusoidally varying quantities.
• A sinusoidal waveform can be produced by plotting vertical projection
of a phasor that rotates in the counterclockwise direction at a constant
angular velocity.
Phasor representation of an alternating quantity
Fig. shows OA when it has rotated through an angle θ from the position occupied
when the current was passing through its zero value. If AB and AC are drawn
perpendicular to the horizontal and vertical axes respectively :
OC = AB = OA sin θ = Im sin θ = i, namely the value of the current at that instant.
Hence the projection of OA on the vertical axis represents to scale the
instantaneous value of the current. Thus when θ = 90 ° the projection is OA itself;
when θ = 180°, the projection is zero and corresponds to the current passing
through zero from a positive to a negative value; when θ = 210° the phasor is in
position OA1 and the projection = OD = 0.5 OA1 = 0.5Im; and when θ = 360°, the
projection is again zero and corresponds to the current passing zero from a
negative to a positive value. It follows that OA rotates through one revolution or 2π
radians in one cycle of the current wave.
Angular Velocity
• Rate at which the generator coil rotates with respect to time,  (Greek letter
omega)
If f is the frequency in hertz, then OA rotates through f revolutions or 2π f
radians in 1 second. Hence the angular velocity of OA is 2π f radians/second
i.e.  =2π f radians/second.
If the time taken by OA to rotate through an angle θ radians be t seconds in the
figure above, then
θ = angular velocity x time = t = 2π f t radians
Therefore the instantaneous value of the current can be expressed as
i = Im sin θ = Im sint = Im sin2π f t.
Phase Angle
The phase angle of a single wave is the angle from the zero point on the wave to
the value at the point from which time is reckoned. Thus i = Im sin(t + )
represents a sine wave of current with a phase angle . The angle  is the phase
angle of the current with respect to the point where i = 0 as a reference. The figure
below illustrates this principle.
Voltages and Currents with Phase Shifts
• If a sine wave does not pass through zero at t = 0, it has a phase shift
• For a waveform shifted left
i = Im sin(t + )
• For a waveform shifted right
i = Im sin(t - )
Shifted Sine Waves
Phase Difference
• Phase difference is angular displacement between waveforms of same
frequency
• If angular displacement is 0°
– Waveforms are in phase
• If angular displacement is not 0o, they are out of phase by amount of
displacement
• May be determined by drawing two waves as phasors.
Phasor representations of alternating quantities differing in phase
The phase angle of a sine wave can be used to describe the relationship of one sine
wave to another by using the terms “Leading” and “Lagging” to indicate the
relationship between two sinusoidal waveforms of the same frequency, plotted
onto the same reference axis.
The current waveform is in phase with the voltage waveform, as shown by the
graph and the corresponding phasor diagram.
The current waveform leads the voltage waveform by an angle π/2, as shown
by the graph and the corresponding phasor diagram.
The current waveform lags the voltage waveform by an angle π/2, as shown by
the graph and the corresponding phasor diagram.
• If v1 = 5 sin(100t) and v2 = 3 sin(100t - 30°), v1 leads v2 by 30°
Phasor Diagram of Sinusoidal(Alternating)Waveform
A diagram representing alternating current and alternating voltage (of same
frequency) as vectors (phasors) with the phase angle between them is called a
phasor diagram.
Five Rules for Drawing Phasor Diagrams
Rule 1. The length of the phasor is directly proportional to the amplitude of the
wave depicted.
Rule 2. In circuits which have combinations of L, C & R in SERIES
it is customary to draw the phasor representing CURRENT horizontally, and
call this the REFERENCE phasor. This is because the current in a series circuit
is common to all the components.
Rule 3. In parallel circuits, where L, C and R are connected in parallel the phasor
representing the SUPPLY VOLTAGE is always drawn in the REFERENCE
direction. This is because in a parallel circuit it is the supply voltage that is
common to all components.
Rule 4. The direction of rotation of all phasors is considered to be
ANTICLOCKWISE.
Rule 5. In any one diagram, the same type of value (RMS, peak etc.) is used for
all phasors, not a mixture of values.
The phasor diagram is drawn corresponding to time zero ( t = 0 ) on the horizontal
axis. The length of the phasor is directly proportional to the amplitude of the wave
depicted.
In the example above the two waveforms are out-of-phase by 30o so one can say
that v2 lags v1 or v1 leads v2 by 30o.
Phasor Diagram Summary
•
Phasors and Phasor Diagrams only apply to sinusoidal AC waveforms.
•
A Phasor Diagram can be used to represent two or more stationary
sinusoidal quantities at any instant in time.
•
Generally the reference phasor is drawn along the horizontal axis and at
that instant in time the other phasors are drawn. All phasors are drawn
referenced to the horizontal zero axis.
•
Phasor diagrams can be drawn to represent more than two sinusoids. They
can be either voltage, current or some other alternating quantity but the
frequency of all of them must be the same.
•
All phasors are drawn rotating in an anticlockwise direction. All the
phasors ahead of the reference phasor are said to be “leading” while all the
phasors behind the reference phasor are said to be “lagging”.
•
Generally, the length of a phasor represents the R.M.S. value of the
sinusoidal quantity rather than its maximum value.
•
Sinusoids of different frequencies cannot be represented on the same
phasor diagram due to the different speed of the vectors. At any instant in time
the phase angle between them will be different.
•
Two or more vectors can be added or subtracted together and become a
single vector, called a Resultant Vector.
•
The horizontal side of a vector is equal to the real or x vector. The vertical
side of a vector is equal to the imaginary or y vector. The hypotenuse of the
resultant right angled triangle is equivalent to the r vector.
•
In a three-phase balanced system each individual phasor is displaced by
120o.
Find out the rms values of the following waveforms as shown
below:
Figure 1.
2.
Figure 2
3.
Figure 3
4.
Figure 4.
5.
Figure 5.
Note:
Lower-case letters are used to represent instantaneous value and
upper-case letters represent definite values such as maximum,
average or r.m.s. values. In alternating current circuits, capital V
and I without any subscript represent r.m.s. values.
AC Circuit
AC Circuit Containing Resistance only
Resistance is the opposition that an element offers to the flow of electric current.
It is represented by the uppercase letter R. The standard unit of resistance is
the ohm, sometimes written out as a word, and sometimes symbolized by the
uppercase Greek letter omega: Ω. The behavior of an ideal resistor is dictated by
the relationship specified by Ohm's law.
Resistor is passive element. Passive element is an electrical component that does
not generate power, but instead dissipates, stores, and/or releases it.
Consider a circuit consisting of an ac source and a resistor. The instantaneous
voltage across the resistor
vR = v = Vmax sin wt
From Ohm's law
i = v / R = (Vmax sin wt) / R= Imax sin wt, the instantaneous current flowing through
the resistor .
If Vmax and Imax be the maximum values of the voltage and current respectively, it
follows that: Imax = Vmax / R……… (1)
But the r.m.s. value of a sine wave is 0.707 times the maximum value so that:
r.m.s. value of voltage = V = 0.707 Vmax
r.m.s. value of current = I =0.707 Imax
Substituting for Vmax and Imax in equation (1)
I/0.707 = V / 0.707R, and I = V / R.
Hence Ohm’s law can be applied without any modification to an a.c. circuit
possessing resistance only.
Phase Relationship between applied voltage and current in Resistor
 The graph shows the current through and the voltage across the resistor.
 The current and the voltage reach their maximum values at the same time.
 The current and the voltage are said to be in phase.
 The direction of the current has no effect on the behavior of the resistor.
 Resistors behave essentially the same way in both DC and AC circuits.
This “in-phase” effect can also be represented by a phasor diagram. Therefore, as
the voltage and current are both in-phase with each other, there will be no phase
difference ( θ = 0 ) between them, so the vectors of each quantity are drawn superimposed upon one another along the same reference axis.
Power in a resistive circuit
The instantaneous power in a resistive circuit is given by the product of
instantaneous voltage and instantaneous current. The instantaneous power is given
by
p = vi = Vmax sin ωt * Imax sin ωt
Writing
Vmax= Vm and Imax= Im
The average power consumed in the circuit over a complete cycle is given by
Paverage =
=
=
= Vr.m.s.Ir.m.s.
= VI
In summary Power to a Resistive Load
• p is always positive. From the above equation it is clear that whatever may
be the value of ωt the value of cos2ωt cannot be greater than 1 hence the
value of p cannot be negative. The value of p is always positive irrespective
of the instantaneous direction of voltage v and current i.
• Power flows only from source to load and p is the rate of energy
consumption by the load
– All of the power delivered by the source is absorbed by the load.
• This power is known as active power. Power to a pure resistance consists of
active power only.
• Average value of power is halfway between zero and peak value of VmIm
• P = VmIm/2
• If V and I are in RMS values
– Then P = VI
• Also, P = I2R and P = V2/R
• Active power relationships for resistive circuits are the same for ac as for dc.
AC Circuit Containing Inductance only
Inductance is the property of an electrical conductor by which a change
in current through it induces an electromotive force in the conductor. It consists of
a conductor such as a wire, usually wound into a coil.
An "ideal inductor" has inductance, but no resistance or capacitance.
When the current flowing through an inductor changes, the time-varying magnetic
field induces an “e.m.f.” (eL) in the coil, according to Faraday’s law of
electromagnetic induction. According to Lenz's law the direction of induced
"e.m.f." is always such that it opposes the change in current that created it. As a
result, inductors always oppose a change in current. The instantaneous value of the
induced e.m.f. is given by
Since the resistance of the coil is assumed to be negligible, the whole of the
applied voltage is absorbed in neutralizing the induced e.m.f.,
=
where
Imax= Vmax/ωL.
Thus the current in an inductor lags the applied voltage by an angle π/2 or 90°.
Also from the expression it follows that maximum value of the current is V max/ωL,
i.e. Imax= Vmax/ωL, so that Vmax/ Imax = ωL= 2πfL.
If V and I are the r.m.s. values, then V/I = 0.707Vmax/0.707Imax = 2πfL
= inductive
reactance. The term inductive reactance is denoted by the symbol X L. Hence I =
V/2πfL = V/ XL = V/ ωL. This is similar to I = V / R.
The inductive reactance is the opposition that an inductor (or coil) offers to the
alternating current. Therefore, ωL plays the same role as that of a resistor. The
inductor impedes the flow of alternating current in the circuit. Unit of X L is also
ohm.
To have a large reactance the coil
(i) Should have many turn as L
(ii) Should have an iron-core as L
N.
μrμo.
(iii)
Length and area of the coil as L
area / length.
(iv) Also the frequency of a.c should be high.
 This is consistent with Faraday’s Law:
 The larger the rate of change of the current in the inductor, the
larger the back emf, giving an increase in the reactance and a
decrease in the current.
iv) XL in case of DC (direct current), is zero.
The inductive reactance of an inductor increases as the frequency increases. Also
as the frequency increases the current flowing through the inductor also reduces in
value.
The effect of very low and very high frequencies on the reactance of a pure AC
Inductance as follows:
In an AC circuit containing pure inductance the following formula applies:
Phase Relationship between applied voltage and current in Inductor
Phasor Diagram for an Inductor
Power in a Inductive circuit
The instantaneous power delivered to the purely inductive circuit is obtained by
In the above expression, it is found that the power is flowing in alternative
directions. From 0 o to 90o it will have negative half cycle, from 90 o to 180o it
will have positive half cycle, from 180 oto 270o it will have again negative half
cycle and from 270o to 360o it will have again positive half cycle. Therefore this
power is alternating in nature with a frequency, double of supply frequency. As
the power is flowing in alternating direction i.e. from source to load in one
quater cycle and from load to source in next half cycle the average value of this
power is zero. The implication is that the inductive element receives energy
from the source during one-quarter of a cycle of the applied voltage and returns
exactly the same amount of energy to the driving source during the next quarter
of a cycle. Therefore this power does not do any useful work.
The power associated with an inductance is reactive power.
Energy Stored in an Inductor
If the circuit is purely inductive, energy will be stored in the magnetic field
during quarter of a cycle and is obtained by integrating power wave p between
limits of t = T/4 and t = T/2,
=
If L in henrys and Im in amperes respectively, WL is given in joule.
AC Circuit Containing Capacitance only
A capacitor (originally known as a condenser) is a passive two-terminal electrical
component used to store electrical energy temporarily in an electric field. The
forms of practical capacitors vary widely, but all contain at least two electrical
conductors (plates) separated by a dielectric (i.e. an insulator that can store
energy).
An ideal capacitor is wholly characterized by a constant capacitance C, defined as
the ratio of charge Q on each conductor to the voltage V between them.
The circuit contains a capacitor and an AC source. An inductor opposes a change
in current. A capacitor does the opposite. It opposes a change in voltage. Pure
capacitor has zero resistance. When an alternating voltage applied across the
capacitor, the capacitor first charged in one direction and then in another direction.
The charge q is given by
q = Cv = CVmaxsinωt
The flow of electrons “through” a capacitor (i.e. the charging current) is directly
proportional to the rate of change of voltage across the capacitor.
Expressed mathematically, the relationship between the current “through” the
capacitor and rate of voltage change across the capacitor is as such:
=
Thus the current in a pure capacitor leads the applied voltage by π/2 radian or 90°.
From the above expression it follows that the maximum value of the current is
ωCVmax or 2πfCVmax.
Vmax / Imax = 1 / 2πfC. If V and I are the r.m.s. values of voltage and current
then
= capacitive reactance. The impeding effect of a
capacitor on the current in an AC circuit is called the capacitive reactance. The
capacitive reactance is expressed in ohms and is represented by XC.
★ Capacitive reactance decreases with increasing frequency. In other words,
the higher the frequency, the less it opposes (the more it “conducts”) the AC
flow of electrons and the current increases.
★ As the frequency approaches zero, XC approaches infinity and the current
approaches zero.
o The capacitor would act as an open circuit, and that is why capacitor
blocks DC.
Capacitive Reactance against Frequency
The effect of very low and very high frequencies on the reactance of a pure AC
Capacitance as follows:
Phase Relationship of applied voltage and current in Capacitor
Pure capacitive circuit waveforms
Phasor Diagram for AC Capacitance
Power in a purely Capacitive circuit
In the first-quarter cycle both v and i are positive, therefore the power is also
positive (since p = vi, at any instant). In the second quarter-cycle v stays positive
while i has gone negative, therefore p is negative. In the third-quadrant
both i and v are negative and so p is positive. Finally, in the fourth-quadrant i is
positive and v is still negative resulting in p being negative. The power wave is
thus a series of identical positive and negative pulses whose average value over
an half-cycle of voltage is zero, also note that its frequency is twice the
frequency of the voltage.
During the first and third quarter-cycles the power is positive meaning that power
is supplied by the circuit to charge the capacitor. In the second and fourth quartercycles the capacitor is discharging and thus supplies the energy stored in it back to
the circuit, thus p has a negative value. The minus or plus signs simply indicate the
direction in which the power is flowing. Since this interchange of energy dissipates
no average power no heating will occur and no power is lost.
The capacitive power does not do any useful work. This power is also a reactive
power.
Energy Stored in an Capacitor
The amount of energy received by the capacitor during quarter of a cycle and is
obtained by integrating power wave p between limits of t = 0 and t = T/4,
Since Im=ωCVm,
Summary
Resistance, Reactance
The following is a summary of the relationship between voltage and current in
circuits:
★ Resistance is the special case when φ = 0.
★ Reactance the special case when φ = ± 90°.
Component
Resistor
Difference of
Phase between
Voltage and
Current
Ohm’s Law
Voltage and
Current are in
phase
Inductor
Current lags
behind Voltage
by π/2
Capacitor
Voltage lags
behind Current by
π/2
R =V / I
XL = V / I = ωL
Xc =V / I = 1 /ωC
Memory Aid for Passive Elements in AC
An old, but very effective, way to remember the phase differences for inductors
and capacitors is:
“E L I” the “I C E”
E.m.f E is before current I in inductors L;
Emf E is after current I in capacitors C.
Series RL, RC and RLC Circuit
Series Resistance-Inductance Circuit
The circuit above consists of a pure non-inductive resistance, R is connected in
series with a pure inductance, L. In this RL series circuit above the current is
common to both the resistance and the inductance while the voltage is made up of
the two component voltages, vR and vL. A sinusoidal driving voltage, v = Vm sinωt
is applied to a series combination of a resistive element (R is pure resistor) and an
inductive element (L is pure inductor), the voltage balance equation is
v = vR + vL ……….(1)
This is Kirchoff’s voltage law applied to instantaneous voltages. It states that the
instantaneous voltage drop across the resistive element plus the instantaneous
voltage drop across the inductive element equals the instantaneous voltage drop
across the RL branch.
If it is assumed that the current i = Im sinωt, flows through a series branch
consisting of a resistive element, R and an inductive element, L, then
Ri + L
= voltage applied = v
Or
R Im sinωt + ωL Im cosωt = v ………..(2)
In the phasor diagram a reference or common component is the current as the same
current flows through the resistance and the inductance. The individual vector
diagrams for a pure resistance and a pure inductance are given as:
Individual phasor Diagram
Phasor Diagram of R-L series circuit
From the phasor diagram above, it is seen that the red line is the horizontal current
reference and VR is the voltage across the resistive component which is in-phase
with the current. The inductive voltage VL which is 90o in front of the current
therefore it can l be seen that the current lags the purely inductive voltage by 90o.
Blue Line gives the resulting supply voltage. Then:

V equals the r.m.s value of the applied voltage.

I equals the r.m.s. value of the series current.

VR equals the I.R voltage drop across the resistance which is in-phase
with the current.

VL equals the I.XL voltage drop across the inductance which leads the
current by 90o.
From equation (2) the RIm and ωLIm components may be related as shown in
the diagram below:
If both sides of the equation(2) are divided by
, the equation takes
the following form
Im
Then
Im[sinωt cosφ + cosωt sinφ] =
From which
v=
It is thus shown that
Im sin(ωt+φ) or v = Im Z sin(ωt+φ) = Vm sin(ωt+φ)
 Z=
= Vm / Im = √2V/ √2I = V/I
 φ = tan-1(ωL / R)
 v leads i in RL branch by φ°.
∠ tan (ωL / R)
 Z=
-1
 Impedance depends upon the frequency, ω of the circuit as this affects the
circuits reactive components.
If R is expressed in ohms, ω is in rad/sec. and L is in henry in which case
is given in ohms.
Variation of Impedance and Phase Angle with Frequency
For a series RL circuit; as frequency increases:
– R remains constant
– XL increases
–Z increases
– φ increases
Alternative solution
Applying Pythagoras theorem to mathematically find the value of the resultant
voltage across the resistor/inductor ( RL ) circuit.
As VR = I.R and VL = I.XL the applied voltage will be the vector sum of the two as
follows:
The quantity
represents the impedance, Z of the circuit.
Impedance, Z is the “TOTAL” opposition to current flowing in an AC circuit that
contains both Resistance, ( the real part ) and Reactance ( the imaginary part ).
Impedance also has the units of Ohms, Ω‘s. Impedance can also be represented by
a complex number, Z = R + jXL
The Impedance Triangle
Then: ( Impedance )2 = ( Resistance )2 + ( Reactance )2. This means that the
positive phase angle, φ between the voltage and current is given as.
Phase Angle
Power
The instantaneous power or the instantaneous volt-amperes, delivered to the RL
branch may be obtained as
p = vi = [Vm sin(ωt+φ)][Im sinωt]
=VmIm sinωt[sinωt cosφ + cosωt sinφ]
= VmIm sin2ωt cosφ + VmIm (sinωt cosωt) sinφ
= (VmIm/2) cosφ - (VmIm/2) [cos2ωt]cosφ + (VmIm/2)[ sin2ωt] sinφ
= (VmIm/2) cosφ - (VmIm/2) cos(2ωt-φ)
So the instantaneous power in a single phase circuit varies sinusoid ally. The
instantaneous power, p = constant term + sinusoidal oscillating term.
The expression of instantaneous power has two components:
(i)
(VmIm/2) cosφ, which contains no reference to ωt and therefore remains
constant in value.
(ii)
(VmIm/2) cos(2ωt-φ), the term 2ωt indicating that it varies twice the
supply frequency; thus it is seen that the power undergoes two cycles of
variation for one cycle of variation of voltage wave. Furthermore, since
the average value of a cosine curve over a complete cycle is zero, it
follows that this component does not contribute anything towards the
average value of the power taken from the alternator.
The average value of the power over complete cycle is given by
Pav = (VmIm/2) cosφ
or
Pav =
=
The waveform and power curve of the RL Series Circuit is shown below
The various points on the power curve are obtained by the product of voltage and
current. If you analyze the curve carefully, it is seen that the power is negative
between angle 0 and φ and between 180 degrees and (180 + φ) and during the rest
of the cycle the power is positive.
It is seen that the instantaneous power is negative for a small time and the negative
area (below horizontal axis and above p curve) represents energy returned from the
circuit to the source (alternator). The positive area (above horizontal-axis and
below p curve) during the time interval represents energy supplied from the
alternator to the circuit and the difference between the two areas represents the
energy absorbed by the circuit over half cycle. By dividing this net area by the time
interval T the average power (P) is obtained.
Series Resistance-Capacitance Circuit
The series RC circuit above consists of an pure resistance, R is connected in series
with a pure capacitance, C. The current flowing into the circuit is common to both
the resistance and capacitance, while the voltage is made up of the two component
voltages, vR and vC. The resulting voltage of these two components can be found
mathematically but since vectors vR and vC are 90o out-of-phase, they can be added
vectorially by constructing a phasor diagram.
If it is assumed that the current i = Im sinωt, flows through a series branch
consisting of a resistive element, R and an capacitive element, C, then
vR = Ri = RIm sinωt, and vC = q / C =
The voltage applied to the branch is, physically, the sum of the two component
voltages. In the form of equation
RIm sinωt
……………(3)
In a series AC circuit the current is common and can therefore be used as the
reference source because the same current flows through the resistance and into the
capacitance. The individual phasor diagrams for a pure resistance and a pure
capacitance are given as:
Phasor Diagrams for the Two Pure Components
Series RC circuit phasor diagram
From equation (3) the RIm and (-Im / ωC) components may be related as shown
in the diagram below:
If both sides of the equation (3) are divided by
takes the following form
, the equation
Im
Then
Im [sinωt cosφ - cosωt sinφ] =
From which
Im sin(ωt+θ) or v = Im Z sin(ωt-φ) = Vm sin(ωt-φ)
v=
It is thus shown that
 Z=
= Vm / Im = √2 V /√2 I = V / I
 φ = tan-1((1/ωC) / R)
 i leads v in RC branch by φ°.
 Z=
∠ tan ((1/ωC) / R)
-1
 Impedance depends upon the frequency, ω of the circuit as this affects the
circuits reactive components.
If R is expressed in ohms, ω is in rad/sec. and C is in farad in which case
is given in ohms.
Alternative Solution
Applying Pythagoras theorem to mathematically find the value of the resultant
voltage across the resistor-capacitor ( RC ) circuit.
As VR = I.R and Vc = I.XC the applied voltage will be the vector sum of the two as
follows:
The quantity
represents the impedance, Z of the circuit. Impedance, Z
which has the units of Ohms, Ω’s is the “TOTAL” opposition to current flowing in
an AC circuit that contains both Resistance, ( the real part ) and Reactance ( the
imaginary part ).
The Impedance Triangle
Then: ( Impedance )2 = ( Resistance )2 + ( Reactance )2. This means then by
using Pythagoras theorem the negative phase angle, φ between the voltage and
current is calculated as.
Phase Angle
Power in RC Series Circuit
If the alternating voltage applied across the circuit is given by the equation
v = Vm sin(ωt-φ) and I = Imsinωt
Therefore, the instantaneous power is given by p = vi
Putting the value of v and i
p = Vm sin(ωt-φ) Imsinωt
= VmIm sin(ωt-φ) sinωt
=VmIm sinωt[sinωt cosφ - cosωt sinφ]
= VmIm sin2ωt cosφ - VmIm (sinωt cosωt) sinφ
= (VmIm/2) cosφ - (VmIm/2) [cos2ωt]cosφ + (VmIm/2)[ sin2ωt] sinφ
= (VmIm/2) [cosθ – cos(2ωt + φ)]
The Average power consumed in the circuit over a complete cycle is given by
Pav or P = average of (VmIm/2) cosφ – average of (VmIm/2) cos(2ωt + φ)
= (VmIm/2) cosφ – zero = (VmIm/√2√2)cosφ = VIcosφ
Where, cosφ is called the power factor of the circuit.
cosφ = R/Z
Waveform and Power Curve of the RC Series Circuit
The waveform and power curve of the RC Circuit is shown below
The various points on the power curve is obtained from the product of the
instantaneous value of voltage and current. The power is negative between the
angle (180◦ – ϕ) and 180◦ and between (360◦ -ϕ) and 360◦ and in the rest of the
cycle the power is positive. Since the area under the positive loops is greater than
that under the negative loops, therefore the net power over a complete cycle is
positive.
Series RLC Circuit
The series RLC circuit above has a single loop with the instantaneous current, i,
flowing through the loop being the same for each circuit element. Since the
inductive and capacitive reactance’s XL and XC are a function of the supply
frequency, the sinusoidal response of a series RLC circuit will therefore vary with
frequency, ƒ. Then the individual voltage drops across each circuit element
of R, L and C element defined by:
 i = Im sinωt
 The instantaneous voltage across a pure resistor, vR is “in-phase” with the
current.
 The instantaneous voltage across a pure inductor, vL “leads” the current by
90o
 The instantaneous voltage across a pure capacitor, vC “lags” the current by
90o
 Therefore, vL and vC are 180o “out-of-phase” and in opposition to each
other.
For the series RLC circuit above, this can be shown as:
The amplitude of the source voltage across all three components in a series RLC
circuit is made up of the three individual component voltages, vR, vL and vC with
the current common to all three components. The vector diagrams will therefore
have the current vector as their reference with the three voltage vectors being
plotted with respect to this reference as shown below:
Individual Voltage Phasors
To find the supply voltage, v , the Phasor Sum of the three component voltages
combined together vectorially.
Kirchoff’s voltage law (KVL) states that around any closed loop the sum of
voltage drops around the loop equals the sum of the EMF’s. Then applying this law
to the these three voltages will give the amplitude of the source voltage, v as
vR+ vL + vC = v
Now, vR = Ri = RIm sinωt,
vL= L(di/dt) = ωL Im cosωt and vC= q/C =
RIm sinωt + ωL Im cosωt
=v
RIm sinωt + (ωL -
Im cosωt = v
Phasor Diagram for a Series RLC Circuit
Alternative solution
Using Pythagoras’s theorem to obtain the value of v as shown
The instantaneous voltage across each of the three circuit elements can be
expressed as
VRmax is the maximum voltage across the resistor and VRmax = ImaxR
VLmax is the maximum voltage across the inductor and VLmax = ImaxXL.
VCmax is the maximum voltage across the capacitor and VCmax = ImaxXC.
The r.m.s. value of the voltages across each element
VR is the r.m.s. voltage across the resistor and VR = IR
VL is the r.m.s. voltage across the inductor and VL = IXL.
VC is the maximum voltage across the capacitor and VC = IXC.
The sum of these voltages must equal the voltage of the AC source.
By substituting the above values into Pythagoras’s equation
Z is called the impedance of the circuit and it plays the role of resistance in the
circuit, where
Z  R 2   X L  XC 
2
 Impedance has units of ohms.
 Impedance of the circuit which ultimately depends upon the
resistance and the inductive and capacitive reactance’s.
 The Impedance Triangle for a Series RLC Circuit
Phase Angle
The phase angle, φ between the source voltage, V and the current, I is the same as
for the angle between Z and R in the impedance triangle. This phase angle may be
positive or negative in value depending on whether the source voltage leads or lags
the circuit current and can be calculated mathematically from the ohmic values of
the impedance triangle as:
cosφ = R / Z , sinφ = (XL- XC) / Z and tanφ = (XL- XC) / R.
Determining the Nature of the Circuit
If φ is positive
 XL> XC (which occurs at high frequencies)
 The current lags the applied voltage.
 The circuit is more inductive than capacitive.
VL is greater than VC so the circuit behaves like an inductor/inductive circuit,
and current lags the applied voltage.
If φ is negative
 XL< XC (which occurs at low frequencies)
 The current leads the applied voltage.
 The circuit is more capacitive than inductive.
When VC is larger than VL the circuit is capacitive and current leads the
applied voltage.
If φ is zero
 XL= XC
 The circuit is purely resistive.
When VL and VC are equal the circuit is purely resistive.
Series Circuit with R, L and C: Effect of variation of Frequency
Figure above shows the effect of frequency upon the inductive and capacitive
reactances and upon the resultant reactance and the impedance of a circuit having
R, L and C. In general it is convenient to take inductive reactance as positive and
capacitive reactance as negative. These positive and negative signs are merely
conventions. It is only the phase of the current that is affected.
For frequency at point A, the inductive reactance AB and the capacitive reactance
AC are equal in magnitude so that their resultant is zero. Consequently the
impedance is then same as resistance (R) AD of the circuit. Furthermore, as the
frequency is reduced below point A or increased above point A, the impedance
increases and therefore the current decreases.
Figure below shows the behavior of the series RLC circuit graphically below and
above the value of frequency at point A. The frequency at point A is denoted by fo.
Series Resonance
This is the characteristics of the circuit if a supply voltage of fixed
amplitude but of different frequencies was applied to the circuit.
In a series RLC circuit there becomes a frequency point where the inductive
reactance of the inductor becomes equal in value to the capacitive reactance of the
capacitor. In other words, XL = XC. The point at which this occurs is called
the Resonant Frequency point, ( ƒr ) of the circuit, and this resonance frequency
produces a Series Resonance.
Series RLC Circuit
From the above equation for inductive reactance, if either the Frequency or
the Inductance is increased the overall inductive reactance value of the inductor
would also increase. The same is also true for the capacitive reactance formula
above but in reverse. If either the Frequency or the Capacitance is increased the
overall capacitive reactance would decrease. As the frequency approaches infinity
the capacitors reactance would reduce to zero causing the circuit element to act like
a perfect conductor of 0Ω’s.
But as the frequency approaches zero or DC level, the capacitors reactance would
rapidly increase up to infinity causing it to act like a very large resistance acting
like an open circuit condition. This means then that capacitive reactance is
“Inversely proportional” to frequency for any given value of capacitance.
At a higher frequency XL is high and at a low frequency XC is high. Then there
must be a frequency point were the value of XL is the same as the value of XC.
Series Resonance Frequency
where: ƒr is in Hertz, L is in Henry and C is in Farads.
Electrical resonance occurs in an AC circuit when the two reactances which are
opposite and equal cancel each other out as XL = XC. In a series resonant circuit,
the resonant frequency, ƒr point can be calculated as follows:
Since XL = XC, XL − XC = 0 so that
Since the impedance at resonance Z equals the resistance R, the impedance is a
minimum. With minimum impedance, the circuit has maximum current determined
by I = V / R. The resonant circuit has a phase angle equal to 0° so that the power
factor is unity.
At frequencies below the resonant frequency (Fig. above), XC is greater than XL so
the circuit consists of resistance and capacitive reactance. However, at frequencies
above the resonant frequency (Fig. above), XL is greater than XC so the circuit
consists of resistance and inductive reactance.
At resonance in series RLC circuit, two reactances become equal and cancel each
other. So in resonant series RLC circuit, the opposition to the flow of current is due
to resistance only. At resonance, the total impedance of series RLC circuit is equal
to resistance i.e Z = R, impedance has only real part but no imaginary part and
this impedance at resonant frequency is called dynamic impedance and this
dynamic impedance is always less than impedance of series RLC circuit.
Before series resonance i.e. before frequency, fr capacitive reactance dominates
and after resonance, inductive reactance dominates and at resonance the circuit acts
purely as resistive circuit causing a large amount of current to circulate through the
circuit.
Impedance in a Series Resonance Circuit
Then in a series resonance circuit as VL = VC the resulting reactive voltages are
zero
and
all
the
supply
voltage
is
dropped
Therefore, VR = Vsupply.
Series RLC Circuit at Resonance
Series Circuit Current at Resonance
The current in series resonance circuit is given by
across
the
resistor.
Variation of current with frequency in series resonance circuit
The frequency response curve of a series resonance circuit shows that the
magnitude of the current is a function of frequency and plotting this onto a graph
shows us that the response starts at near to zero at zero frequency, reaches
maximum value at the resonance frequency when IMAX = IR and then drops again
to nearly zero as ƒ becomes infinite. The result of this is that the magnitudes of the
voltages across the inductor, L and the capacitor, C can become many times larger
than the supply voltage, even at resonance but as they are equal and at opposition
they cancel each other out.
Series Circuit Voltage at Resonance
The voltages across the inductor (VL ) and the capacitor (VC) are 180° out of phase
with each other. They are both 90° out of phase with the voltage across the resistor.
The current I and VR are always in phase. Figure shows the phasor diagram of the
voltages in the series RLC circuit below and above the resonant frequency and at
the resonant frequency fr.
Phase Angle of a Series Resonance Circuit
Figure below shows the overall variation of phase as the frequency is varied.
The phase angle is positive for frequencies above ƒr and negative for frequencies
below ƒr and this can be proven by,
Effect of resistance on current variation in series resonance
Phasor diagram of series resonance circuit
Quality Factor (Q)
The Q, quality factor, of a resonant circuit is a measure of the “goodness” or
quality of a resonant circuit.
The voltage applied to the series RLC circuit is V, and the current at resonance is I,
then the voltage across L is VL= IXL = (V/R) ωL
Similarly, the voltage across C
VC = IXC= (V/R) (1/ ωC)
Q is termed the Q factor or voltage magnification, because VC equals Q multiplied
by the source voltage V:
Fig. Voltage magnification Q in series resonant circuit
The capacitive and inductive reactance’s store energy that oscillates between them,
the energy being at one moment stored as electrostatic energy in the capacitor, and
a quarter of a cycle later as magnetic energy in the inductor. At the resonant
frequency, when the capacitive and inductive reactances are equal, they transfer
equal energy, and the circuit appears resistive. The maximum magnetic energy
stored in L at any instant is (1/2)LIm2 joules, where Im is the maximum value of
current in the inductor, and the maximum electrostatic energy in C is (1/2)CVm2
joules, where Vm represents the maximum value of the voltage across the capacitor.
However, energy is dissipated as I2R losses in the resistance of the circuit as the
energy is passed backwards and forwards between L and C.
This leads to a more general definition of Q factor. It is defined as the ratio of
the reactive power, of either the capacitor or the inductor to the power
dissipated in the resistor at resonance:
For inductive reactance XL at resonance:
For capacitive reactance XC at resonance:
Bandwidth of series resonance circuit
Variation of current with frequency in series resonance circuit
The bandwidth of a circuit is defined as the frequency range between the halfpower points when I = Imax/√2. This is illustrated in Figure above.
The value of current that is 70.7% of its maximum resonant value which is defined
as: 0.707IMAX.
The bandwidth, BW, equals ωH- ωL, where the frequencies ωH and ωL are referred
to as half-power points or frequencies. They are also referred to as cut-off
frequencies. The term half-power frequency can be justified by consideration of
the conditions for maximum and half-power for the series RLC
circuit.
At maximum power, when ω = ωr,
For ω = ωL, and
ω = ωH the current I = =0.707 Imax = 0.707(V/R)
……………….(1)
The point corresponding to the lower frequency is called the “lower cut-off
frequency”, labelled ƒL with the point corresponding to the upper frequency being
called the “upper cut-off frequency”, labelled ƒH. The distance between these two
points, i.e. ( ƒH – ƒL ) is called the Bandwidth, (BW).
Half Power Frequency
Half power frequencies is the frequency when the magnitude of the current is
decrease by the factor of 1/√2 from its maximum value.
The equation of current is I =
The value of current I is
(1/√2)(V/R) =
Or 2R2 =
Solving for ω
and
for ω = ωL and ω = ωH
,
R=±
for ω = ωL and ωH
The bandwidth, BW = ωH - ωL = R / L.
Slectivity
The sharpness of the resonance curve depends on the Q factor. The bandwidth, the
range of frequencies for which the power half-power, is narrower, the higher Q is.
A circuit is said to be selective if the response has a sharp peak and narrow
bandwidth and is achieved with a high Q factor. Q is therefore a measure of
selectivity.
Q = ωr / Δω = ωr L / R
 Δω is the width of the curve, measured between the two values of ω
for which Pavg has half of its maximum value.
 These points are called the half-power points.
A high-Q circuit responds only to a narrow range of frequencies.
 Narrow peak
In order to obtain higher selectivity, Q must be large.
The effect of Q on Imax and on the bandwidth (BW)
thus, for high selectivity, R must be small. This means that the total series
resistance of the circuit including the source resistance must be small. Therefore a
series tuned circuit must be driven by a voltage source having a low internal
resistance if it is to exhibit a resonance peak and be selective.
Power in a Series Resonant Circuit
The average power dissipated in a series resonant circuit can be expressed in terms
of the rms voltage and current as follows:
Where
Substitution now gives the expression for average power as a function of
frequency.
This expression shows that at resonance, when ω = ωr, the average power is a
maximum and the value of the average power is
since at the resonant frequency ωr the reactive parts cancel so that the circuit
appears as just the resistance R.
The above Figure is a plot of average power versus frequency for two values of R
in a series RLC circuit. As the resistance is made smaller, the curve becomes
sharper in the vicinity of the resonance frequency.
Magnification in Resonance
The voltage applied to the series RLC circuit is V, and the current at resonance is I,
then the voltage across L is VL=IXL= (V/R) ωL
Similarly, the voltage across C
VC = IXC= (V/R) (1/ ωC)
Since Q = 1/ωrCR = ωr L/R
Where ωr is the frequency at resonance.
Therefore
VL = VQ
VC = VQ
The ratio of voltage across either L or C to the voltage applied at resonance can be
defined as magnification.
Magnification = Q = VL / V or VC / V
Quality factor is the voltage magnification that circuit produces at resonance.
Series resonance by varying inductance (L) only
When L is varied to produce resonance, the equations for current and voltage drops
are given by
For any value of current the drop across resistance
Similarly the drop across the inductance and capacitance are respectively
And
It is noted that
of
becomes a maximum at resonance whereas the maximum value
across inductance occurs after resonance. Since
and
is
constant, the maximum drop across condenser will occur when the current is
maximum. In case of
, both
are increasing before resonance and
the product must be increasing. At resonance, I is not changing but
is increasing
and hence drop is increasing. The drop is increasing until the reduction in current
offsets the increase in
. This point can be determined from d
.
Differentiating the equation and setting the result equal to zero yield
The variation in phase angle between V and I as L varied can be seen to vary from
(a negative angle) when L is zero to +900 when L becomes ∞. Hence the
power factor varies from
(when L is 0) to 0 (when L becomes infinite).
Series resonance by varying capacitance (C) only
When C is varied to produce resonance, The equations for current and voltage
drops are given by
For any value of current the drop across resistance
Similarly the drop across the inductance and capacitance are respectively
And
Here the drop across the inductance is maximum when the current in the circuit is
maximum, since XL is constant. The maximum drop across the condenser occurs
before resonance. At resonance, XC is decreasing whereas the current is not
changing (slope being zero). The drop IXC must, therefore, be decreasing.
Consequently, the drop must have been a maximum before resonance. At
resonance the drops across the inductance and capacitance are equal and opposite.
The conditions for maximum VC may be determined analytically by setting the first
derivative with respect to C or XC equal to zero.
For capacitance the capacitive reactance is infinite and the current is therefore zero.
For infinite capacitance the capacitive reactance is zero and the current
is
The power factor varies from
, when C is infinite, to zero
when C is zero.
Application of Series RLC Resonant Circuit
★
The receiving circuit of a radio is an important application of a resonant
circuit. One tunes the radio to a particular station (which transmits a specific
electromagnetic wave or signal) by varying a capacitor, which changes the
resonant frequency of the receiving circuit. When the resonance frequency
of the circuit matches with that of the incoming electromagnetic wave, the
current in the receiving circuit increases. This signal caused by the incoming
wave is then amplified and fed to a speaker. Because many signals are often
present over a range of frequencies, it is important to design a high-Q circuit
to eliminate unwanted signals. In this manner, stations whose frequencies
are near but not equal to the resonance frequency give signals at the receiver
that are negligibly small relative to the signal that matches the resonance
frequency.
★ Metal Detectors in Airports
An airport metal detector is essentially a resonant circuit. The portal you step
through is an inductor (a large loop of conducting wire) that is part of the
circuit. The frequency of the circuit is tuned to the resonant frequency of the
circuit when there is no metal in the inductor. Any metal on your body
increases the effective inductance of the loop and changes the current in it.
When you walk through with metal in your pocket, you change the effective
inductance of the resonance circuit, resulting in a change in the current in the
circuit. This change in current is detected, and an electronic circuit causes a
sound to be emitted as an alarm.
Summary
For a series RLC circuit at certain frequency called resonant frequency, the
following points must be remembered. So at resonance:
1. Inductive reactance X L is equal to capacitive reactance XC.
2.
Total impedance of circuit becomes minimum which is equal to R i.e Z = R.
3.
Circuit current becomes maximum as impedance reduces, I = V / R.
4.
Voltage across inductor and capacitor cancels each other, so voltage across
resistor Vr = V, supply voltage.
5.
Since net reactance is zero, circuit becomes purely resistive circuit and
hence the voltage and the current are in same phase, so the phase angle
between them is zero.
6.
Power factor is unity.
7.
Frequency at which resonance in series RLC circuit occurs is given by
Parallel Circuit
Parallel RL Circuit
In RL parallel circuit resistor and inductor are connected in parallel with each
other and this combination is supplied by a voltage source, Vin. Since the resistor
and inductor are connected in parallel, the currents flowing in resistor and inductor
are different.
IT = the total electric current flowing from voltage source in amperes.
IR = the electric current flowing in the resistor branch in amperes.
IL = the electric current flowing in the inductor branch in amperes.
θ = the angle between IR and IL.
From the Kirchhoff’s current law
IT = IR + IL = Vin/ ZR + Vin/ ZL = Vin (1/ ZR +1/ ZL) = Vin (YR +YL) = VinY
where the symbol Y represents the reciprocal of impedance and is called
admittance. The resultant current in parallel RL branch is the product of the input
voltage and the sum of the reciprocals of branch impedances.
Phasor Diagram of Parallel RL circuit
The total electric current IT can also be calculated from the phasor diagram as
and the phase angle is given by
= tan-1 ((V/XL) / (V / R)) = tan-1 (R/XL)
= tan-1(R/ωL)
Also the phase angle again be given by, θ = cos-1 (IR / IT) = cos-1(Z/R)
The total phase angle of a parallel RL circuit always lies between 0° to -90°. It is
0° for pure resistive circuit and -90° for pure inductive circuit.
In complex form the currents are written as,
Where G is the conductance and BL is the inductive suseptance of a parallel RL
circuit. Thus if the admittance of the circuit is (0.2-j0.1) mho, such a circuit can be
represented as a resistance of 5 Ω(ohm) in parallel with an inductive reactance of
10Ω; whereas if the impedance of a circuit is (5+j10)Ω, such a circuit can be
represented as a resistance of 5Ω in series with an inductive reactance of 10Ω.
In parallel circuit, the voltage across inductor and resistor remains the same so,
Impedance of Parallel RL Circuit
Let,
Z = total impedance of the circuit in ohms.
R = resistance of circuit in ohms.
L = inductor of circuit in henry, XL = inductive reactance in ohms.
Since resistance and inductor are connected in parallel, the total impedance of the
circuit is given by
In order to remove "j" from the denominator multiply and divide numerator and
denominator by (R - j XL),
Convert a parallel RL circuit into its equivalent series RL circuit
The parallel RL circuit is shown in the figure below
An equivalent series version of the circuit can be made using the values of
resistance and inductive reactance.
The equivalent circuit will behave identically to the parallel version with respect to
points A and B of the circuit
The relationships for calculating the equivalent series resistance and inductive
reactance can be used to create the equivalent series circuit
The value of RS and XS are given by the equations
And
Example:
Calculate the series resistance and series inductive reactance for the equivalent
series circuit. Rp= 47Ω and Lp = 22mh. Frequency = 60hz.
Rp = 47Ω and Xp = 2πfLp= 8.294Ω
= 1.419Ω ,
= 8.044Ω. From which LS
21.337mh
Convert a series RL circuit into its equivalent parallel RL circuit
,
Y=
=
upon rationalizing,
Equating real and imaginary part
The above equations are parallel equivalent of series impedance ,
Parallel RC Circuit
In RC parallel circuit resistor and capacitor are connected in parallel with each
other and this combination is supplied by a voltage source, Vin. Since the resistor
and capacitor are connected in parallel, the currents flowing in resistor and
capacitor are different.
From the Kirchhoff’s current law
IT = IR + IC = Vin/ ZR + Vin/ ZC = Vin (1/ ZR +1/ ZC) = Vin (YR +YC) = VinY
In complex from the current is given by
The phasor diagram of the circuit
Phasor diagram of parallel RC circuit
From the above phasor diagram
The phase angle is given by
Impedance of Parallel RC Circuit
R × (- jXC)
Z=
R - jXC
Thus, we can multiply the parallel expression by (R+jXC)/(R+jXC) and get the following result:
Z=
=
R × (- jXC)
R - jXC
×
R + jXC
R + jXC
(-jRXC)(R + jXC)
(R - jXC)(R + jXC)
= -jR²XC - j²RXC²
R² + XC²
RXC² - jR²XC
=
=
–j
R² + XC²
Note:
For inductive circuit, impedance = Z = R+jXL (imaginary component +ve)
admittance = Y = G - jBL (imaginary component –ve)
For capacitive circuit, impedance = Z = R - jXC (imaginary component -ve)
admittance = Y = G + jBC (imaginary component +ve)
Thus Z = 3 – j 5 indicates impedance of a capacitive circuit.
while Y = 3 – j5 indicates admittance of a inductive circuit.
Parallel RLC Circuit
In parallel
RLC
Circuit the resistor, inductor and capacitor are
connected
in
parallel across a voltage supply. The parallel RLC circuit is exactly opposite to
the series RLC circuit. The applied voltage remains the same across all
components whilst the supply current IS consists of three parts. The current flowing
through the resistor, is IR, the current flowing through the inductor is IL and the
current through the capacitor is IC.
But the current flowing through each branch and therefore each component will be
different to each other and to the supply current, IS. The total current drawn from
the supply will not be the mathematical sum of the three individual branch currents
but their vector sum.
Since the voltage across the circuit is common to all three circuit elements, the
current through each branch can be found using Kirchoff’s Current Law, (KCL).
Phasor Diagram for a Parallel RLC Circuit
IR is the electric current flowing in the resistor, R in amps.
IC is the electric current flowing in the capacitor, C in amps.
IL is the electric current flowing in the inductor, L in amps.
Is is the supply electric current in amps.
In the parallel RLC circuit, all the components are connected in parallel; so
the voltage across each element is same. Therefore, for drawing phasor diagram,
take voltage as reference vector and all the other currents i.e IR , IC, IL are drawn
relative to this voltage vector.
Using Pythagoras’s theorem
Impedance of a Parallel RLC Circuit
The admittance of the circuit and the impedance with respect to admittance as:
Determining the Nature of the Circuit
If θ is negative
 BL> BC (which occurs at lower frequencies)
 The current lags the applied voltage.
 The circuit is more inductive than capacitive.
IL is greater than I C so the circuit behaves like an inductor/inductive circuit,
and current lags the applied voltage.
If θ is positive
 BL< BC (which occurs at higher frequencies)
 The current leads the applied voltage.
 The circuit is more capacitive than inductive.
IC is larger than IL the circuit is capacitive and current leads the applied
voltage.
If φ is zero
 BL= BC or IL = IC
Parallel Impedance Network
Fig. (a) Circuit Diagram
Consider the circuit shown in Fig.(a) above in which two series circuits are
connected in parallel. To analyze the arrangement, the phasor diagrams for each
branch have been drawn as shown in Figs (b) and (c). In each branch the current
has been taken as reference; however, when the branches are in parallel, it is easier
to take the supply voltage as reference, hence Figs (b) and (c) have been drawn
separately and then superimposed on one another to give Fig.(d). The current
phasors may then be added to give the total current in correct phase relation to the
voltage. The analysis of the diagram is carried out in the manner noted above.
Figure (b) Phasor diagram of branch 1, (c) Phasor diagram of branch 2 and
(d) Phasor diagram of complete circuit.
The phase angle for the network shown in Fig.(a) is a lagging angle if I1 sin 1 >
I2 sin2 and is a leading angle if I1 sin1 < I2 sin2.
Parallel Resonance
Resonance
The circuit is said to be in resonance if
 The current is in phase with the applied voltage.
 The impedance (or admittance) is completely real when this condition exists.
 Power factor of the circuit at resonance is unity.
 The circuit behaves like a resistive circuit.
A parallel resonance circuit is exactly the same as the series resonance circuit.
Both are 3-element networks that contain two reactive components, both are
influenced by variations in the supply frequency and both have a frequency point
where their two reactive components cancel each other out influencing the
characteristics of the circuit. Both circuits have a resonant frequency point.
The difference this time however, is that a parallel resonance circuit is influenced
by the currents flowing through each parallel branch. Consider the parallel RLC
circuit below.
The total Admittance of the circuit
Where
A
parallel
circuit
containing
a
resistance, R,
an
inductance, L and
a
capacitance, C will produce a parallel resonance circuit when the resultant current
through the parallel combination is in phase with the supply voltage. At resonance
there will be a large circulating current between the inductor and the capacitor due
to the energy of the oscillations, then parallel circuits produce current resonance.
Susceptance at Resonance
From above, the inductive susceptance, BL is inversely proportional to the
frequency
as
represented
by
the
hyperbolic
curve.
The capacitive
susceptance, BC is directly proportional to the frequency and is therefore
represented by a straight line. The final curve shows the plot of total susceptance
of the parallel resonance circuit versus the frequency and is the difference between
the two susceptance’s.
At the resonant frequency point where it crosses the horizontal axis the total circuit
susceptance is zero. Below the resonant frequency point, the inductive susceptance
dominates the circuit producing a “lagging” power factor, whereas above the
resonant frequency point the capacitive susceptance dominates producing a
“leading” power factor.
Series resonance takes place when VL = VC and this situation occurs when the
two reactances are equal, XL = XC. Parallel Resonance occurs when the imaginary
parts of Y become zero. Then
Also at resonance the parallel LC circuit acts like an open circuit with the
circuit current being determined by the resistor, R only.
At resonance, the impedance of the parallel circuit is at its maximum value and
equal to the resistance of the circuit. Also at resonance, as the impedance of the
circuit is now that of resistance only, the total circuit current, I will be “in-phase”
with the supply voltage, V.
The circuit’s response can be changed by changing the value of this resistance.
Changing the value of R affects the amount of current that flows through the
circuit at resonance, if both L and C remain constant. Then the impedance of the
circuit at resonance Z = RMAX is called the “dynamic impedance” of the
circuit.
Impedance in a Parallel Resonance Circuit
The parallel circuit’s impedance is at its maximum at resonance then
consequently, the circuits admittance must be at its minimum and one of the
characteristics of a parallel resonance circuit is that admittance is very low limiting
the circuits current. At the resonant frequency, ƒr the admittance of the circuit is
equal to the conductance, G given by 1/R because in a parallel resonance circuit
the imaginary part of admittance, i.e. the susceptance, B is zero because BL = BC.
Current in a Parallel Resonance Circuit
As the total susceptance is zero at the resonant frequency, the admittance is at its
minimum and is equal to the conductance, G. Therefore at resonance the current
flowing through the circuit must also be at its minimum as the inductive and
capacitive branch currents are equal ( IL = IC ) and are 180o out of phase.
The total current flowing in a parallel RLC circuit is equal to the vector sum of the
individual branch currents and for a given frequency is calculated as:
At resonance, currents IL and IC are equal and cancelling giving a net reactive
current equal to zero. Then at resonance the above equation becomes.
Since the current flowing through a parallel resonance circuit is the voltage divided
by impedance, at resonance the impedance, Z is at its maximum value, ( =R ).
Therefore, the circuit current at this frequency will be at its minimum value
of V/R and the graph of current against frequency for a parallel resonance circuit is
given as.
A parallel resonant circuit stores the circuit energy in the magnetic field of the
inductor and the electric field of the capacitor. This energy is constantly being
transferred back and forth between the inductor and the capacitor which results in
zero current and energy being drawn from the supply. This is because the
corresponding instantaneous values of IL and IC will always be equal and opposite
and therefore the current drawn from the supply is the vector addition of these two
currents and the current flowing in IR.
The frequency response curve of a parallel resonance circuit shows that the
magnitude of the current is a function of frequency and plotting this onto a graph
shows us that the response starts at its maximum value, reaches its minimum value
at the resonance frequency when IMIN = IR and then increases again to maximum
as ƒ becomes infinite.
The result of this is that the magnitude of the current flowing through the
inductor, L and the capacitor, C circuit can become many times larger than the
supply current, even at resonance but as they are equal and at opposition (
180o out-of-phase ) they effectively cancel each other out.
As a parallel resonance circuit only functions on resonant frequency, this type of
circuit is also known as a Rejecter Circuit because at resonance, the impedance of
the circuit is at its maximum thereby suppressing or rejecting the current whose
frequency is equal to its resonant frequency. The effect of resonance in a parallel
circuit is also called “current resonance”.
Magnification at Resonance
The voltage applied to the parallel RLC circuit is V, and the current at resonance is
IR.
The current flowing through the inductor is IL and is given by
IL = V / XL. At resonant frequency ωr , IL = V / ωrL
Since, at resonance, V = I/G
IL = I / G ωrL = QI, where Q is the current magnification
where B is the inductive or capacitive susceptance and X is the inductive or
capacitive reactance.
By substituting By substituting ωr = 1/√(LC) into equation
Again by Q is defined as the ratio of the reactive power, of either the capacitor
or the inductor to the power dissipated in the resistor at resonance:
Q has the same inherent definition for both parallel and series circuits. It may
appear, at first glance, that the expressions for Q for a series and a parallel resonant
circuit are quite different. It will be shown that they are the same. Meanwhile, it
should be remembered that R and X in equation are parallel circuit components –
unlike R and X in the series circuit.
Parameters in Parallel Resonant Circuit
Salient Features of Series RLC Circuit and Parallel RLC Circuit at
Resonance
SL
RLC SERIES CIRCUIT
RLC PARALLEL CIRCUIT
No.
Resistor, inductor and capacitor Resistor, inductor and capacitor
1.
2.
3.
4.
5.
6.
7.
8.
are connected in series
Current is same in
element
are connected in parallel.
each Current is different in all elements
and the total current is equal to
vector sum of each branch of
current.
Voltage across all the elements Voltage across each element
is different and the total voltage remains the same.
is equal to the vector sum of
voltages
across
each
component.
For drawing phasor diagram, For drawing phasor diagram,
current is taken as reference voltage is taken as reference
vector.
vector.
Voltage across each element is Current in each element is given
given by : VR= IR, VL = I XL, VC by: IR = V / R , IC = V / XC , IL = V
= I XC
/ XL
Its more convenient to use Its more convenient to use
impedance for calculations.
admittance for calculations.
At resonance , when XL = XC,
At resonance, when XL = XC, the
the
circuit
has
minimum
circuit has maximum impedance.
impedance.
An acceptor circuit
A rejector circuit.
Series and Parallel circuits – Difference and similarities
Impedance and Admittance Power, VA, VAR and Power Factor
The current that flows in a circuit as result a result of applying sinusoidal voltage is
governed in magnitude and phase by the circuit parameters (resistance R, selfinductance L, Capacitance C and mutual inductance M) and the angular velocity or
frequency of the applied voltage. If the circuit parameters are constant, the current
that flows will be of sinusoidal waveform but will, differ in phase from the
sinusoidal applied voltage.
Mathematically a particular type of function is required to relate voltage and
current in a-c circuit. The one generally employed is called the impedance
function or simply the impedance of the circuit. The impedance function tells two
important facts: (1) the ratio of voltage to current i.e. Vmax to Imax or V to I, and (2)
the phase angle between the waves of voltage and current. A special type of
notation is required to signify the two properties of the impedance and the notation
is
Z∠ angle
Z
is the magnitude of the impedance and in particular case is represented by a
certain number of ohms. The angle associated with Z, if it is positive, defines the
lead of voltage with respect to current and specifies the number of degrees or
radians by which the current lags the voltage.
For pure resistive a-c circuit
Applied voltage, v = Vmax sin ωt
i = v / R = (Vmax sin wt) / R= Imax sin ωt
From the above equation it is evident that V max / Imax = R or
and that
the current wave is in time phase with the voltage wave.
So when using pure resistors in AC circuits the term Impedance, ZR =R∠0°ohm.
For pure inductive circuit
Applied voltage, v = Vmax sin ωt
i
, where Imax= Vmax/ωL.
From the above equation it is evident that Vmax / Imax = ωL or
and
that the current wave lags by one-quarter of a cycle or 90° from the voltage wave.
So when using pure inductance in AC circuits the term Impedance, ZL = ωL
∠90° = XL ∠90° ohm.
For pure capacitive circui
Applied voltage, v = Vmax sin ωt
where Imax= ωCVmax or 2πfCVmax .
i=
From the above expression it follows that the maximum value of the current is
ωCVmax or 2πfCVmax.
From the above equation it is evident that V max / Imax = 1/ ωC or
and
that the current wave leads by one-quarter of a cycle or 90° from the voltage wave.
So when using pure capacitance in AC circuits the term Impedance, ZC =
(1/ωC)
∠-90° = XC∠-90°ohm.
ADMITTANCE
Electrical impedance is the measure of the opposition that a circuit presents to
a current when a voltage is applied. The effective impedance of an electric circuit
to alternating current is arising from the combined effects of resistance and
reactance. Impedance is a vector (two-dimensional) quantity consisting of two
independent quantities (one-dimensional): resistance (R) and reactance (X).
Admittance is a measure of how easily a circuit or device will allow a current to
flow. It is defined as the inverse of impedance.
Admittance is defined as
Y = 1 / Z = Z-1
where
Y is the admittance, measured in mho and the symbol ℧. Admittance is a vector
(two-dimensional) quantity consisting of two independent quantities (onedimensional): conductance (G) which is inverse of resistance (R) and
suseptance (B) which is inverse of reactance (X).
Z is the impedance, measured in ohms.
Where 1/XL = BL = inductive suseptance and 1/ XC = BC = capacitive suseptance.
The j Operator
The symbol j, when applied to a phasor, alters its direction by 90º in an
anticlockwise direction, without altering its length, and is consequently referred to
as an operator.
Significance of the j operator
Start with phasor A in phase with the X-axis, then jA represents a phasor of the
same length upwards along the Y-axis. Again apply the operator j to jA , we turn
the phasor anticlockwise through another 90º, thus giving jjA or j2A. The symbol
j2 signifies that the operator j is used two times successively, thereby rotating the
phasor through 180º. This reversal of the phasor is equivalent to multiplying by -1,
i.e. j2A = -A. So that j2 may be regarded as being numerically equal to -1 and
j =√-1.
Thus
Phasor Rotation of the j-operator
So by multiplying a phasor by j2 will rotate the phasor by 180o anticlockwise,
multiplying by j3 rotates it 270o and by j4 rotates it 360o or back to its original
position. Multiplication by j10 or by j30 will causethe vector to rotate anticlockwise
by the appropriate amount. In each successive rotation, the magnitude of the
phasor always remains the same.
Now the expression A=3 + j4 could be represented graphically by plotting the 3
units of real number in the X-axis and the 4 units of imaginary number along the
Y-axis as shown in the figure below.
Representation of 3 + j4
This type of number, combining real and imaginary number is termed as
complex number. Since the real component of a complex number is drawn along
the reference axis, namely the X axis, and the imaginary component is drawn at
right-angles to that axis, these components are sometimes referred to as the
inphase and quadrature components respectively.
The various ways of representing the complex number algebraically:
A = a +jb (Cartesian or rectangular notation)
= A(cosθ +jsinθ) (trigonometric notation)
= A∠θ (polar notation)
Complex Numbers using the Rectangular Form
A complex number is represented by a real part and an imaginary part that takes
the generalized form of:





Where:
Z - is the Complex Number representing the Vector
x - is the Real part
y - is the Imaginary part
j - is defined by √-1
Complex Numbers using the Complex plane
Conjugate Complex Numbers
Complex Numbers using Polar Form
The Polar Form of a complex number is written in terms of its magnitude and
angle. Thus, a polar form vector is presented as: Z = A ∠±θ, where: Z is the
complex number in polar form, A is the magnitude or modulo of the vector and θ
is its angle or argument of A which can be either positive or negative.
Polar Form Representation of a Complex Number
Converting Polar Form into Rectangular Form, ( P→R )
Converting Rectangular Form into Polar Form, ( R→P )
Impedance is represented as a complex quantity Z and is given by Z= R ± jX
where the real part of impedance is the resistance R and the imaginary part is
the reactance X.
For pure Resistive Circuit
For pure Inductive Circuit
For pure Capacitive Circuit
Admittance, just like impedance, is a complex number, made up of a real part (the
conductance, G), and an imaginary part (the susceptance, B), thus:
Y = G + jB
where G (conductance) and B (susceptance). Now
Power in AC Circuits
• Power in an electric circuit is the rate at which electrical energy is generated
or absorbed or it is defined as the rate at which electrical energy is
transferred by an electric circuit.
• Instantaneous power to a load is p = v • i
• In an ac circuit
– p may be positive sometimes and negative other times
• Average value of the power, P
– Real power
• Average value of instantaneous power, real power, active power, and
average power mean the same thing.
Active Power
 The power associated with energy transfer from the electrical system to
another system such as heat, light or mechanical drives are termed active
power.
 The portion of power that, averaged over a complete cycle of the AC
waveform, results in net transfer of energy in one direction is active power
(sometimes also called real power). The active power is the power that is
dissipated in the resistance of the load.
 It uses the same formula used for DC (V & I are the magnitudes, not the
phasors).
V2
PI R
R
2

[watts, W]
True power is a function of a circuit’s dissipative elements, usually
resistances (R).
 Actually do some “work”.
 Measured in watt.
Reactive Power
 The portion of power due to stored energy, which returns to the source in
each cycle, is known as reactive power. It doesn’t do any useful “work”.
 The reactive power is the power that is exchanged between reactive
components (inductors and capacitors).
 The formulas look similar to those used by the active power, but use
reactance instead of resistances.
 Units: Volts-Amps-Reactive (VAR)
V2
QI X 
X
2
[VAR]
 Reactive power is a function of a circuit’s reactance (X).
 Reactive power is present when applied voltage and current are not in phase.
 One waveform leads the other.
 This can happen for inductive or capacitive loads.
 Average value of reactive power is zero.
Apparent Power
 The apparent power is the power that is “appears” to flow to the load.
 The magnitude of apparent power can be calculated using similar formulas
to those for active or reactive power:
V2
S  VI  I Z 
Z
2
[VA]
 Units: Volts-Amps (VA)
 Apparent power is a function of a circuit’s total impedance (Z).
 V & I are the magnitudes of effective voltage and effective current.
 If load has both resistance and reactance
 Product is neither the real power nor the reactive power, but a
combination of both.
 Resistive load only:

The true (active) power, reactive power and apparent power for resistive load

Reactive( Pure inductive) load only:
True power, reactive power, and apparent power for a purely reactive load.
 Resistive/reactive load:
True power, reactive power, and apparent power for a resistive/reactive load.
Power Triangle
 The power triangle graphically shows the relationship between real (P),
reactive (Q) and apparent power (S).
 AC Impedance is a complex quantity made up of real resistance and
imaginary reactance.
Z  R  jX
( )
 AC Apparent Power is a complex quantity made up of real active power and
imaginary reactive power:
S  P  jQ
(VA)
Active and Reactive Power Equations
• P = S cos  = VI cos 
• Q = S sin  = VI sin 
• V and I are RMS or Effective values
•  is the phase angle between V and I
Power to a Resistive Load
 In ac circuits, voltage and current are functions of time.
 Power at a particular instant in time is given by
p  vi  (Vm sin t )( I m sin t )  Vm I m sin 2 t 
Vm I m
1  cos 2t 
2
This is called instantaneous power.
 p is always positive
 All of the power delivered by the source is absorbed by the load.
Average power P = VmIm / 2
 Using RMS values V and I
VRMS 
Vm
I RMS 
Im
P
2
2
rms value of voltage
rms value of current
Vm I m  Vm  I m 

 VRMS I RMS


2
 2  2 
(watts)
 Active power is the average value of instantaneous power.
Power to an Inductive Load
 Consider the following circuit where
v = Vm sinωt and i = Im sin(ωt – π/2)
 p is equally positive and negative.
 All of the power delivered by the source is returned.
Average power PL = 0 W. It contributes nothing to average power.
• QL = I2XL = V2/XL
• Unit is VAR.
 The power that flows into and out of a pure inductor is reactive power only.
Power to a Capacitive Load
 Consider the following circuit where
V = Vm sinωt and i = Im sin(ωt + π/2)
 p is equally positive and negative
 All of the power delivered by the source is returned (no power losses with a
pure reactive load).
Average power PC = 0 W
• QC = I2XC = V2/XC
• Unit is VAR.
• Power that flows into and out of a pure capacitance is reactive power only.
Power Factor
• Ratio of real power (P) to apparent power (S) is called the power factor, Fp
• Power factor (FP) tells us what portion of the apparent power (S) is actually
real power (P).
• Fp = P/S = cos 
• Power factor angle  is given by  = cos-1(P / S). For a pure resistance,  =
0º. For a pure inductance,  = 90º. For a pure capacitance,  = -90º. For a
circuit containing a mixture,  is somewhere between 0° and 90°.
• Power factor is a dimensionless quantity is expressed as a number between 0
to 1.0.
• The power factor is one when the voltage and current are in phase. When the
power factor is 1, all the energy supplied by the source is consumed by the
load.
• The power factor is zero when the current leads or lags the voltage by 90
degrees. When power factor is equal to 0, the energy flow is entirely reactive
and stored energy in the load returns to the source on each cycle. Therefore
there is no real power consumed by the load.
• Whether the current is leading or lagging the power factor is termed as
leading or lagging power factor correspondingly.
The power factor of the circuit may also be defined in any one of the following
ways:
(i) Power factor is defined as the cosine of the angle between voltage and current in
an A.C. circuit.
(ii) Power factor is defined as the ratio of resistance to impedance of an A.C.
circuit.
cos θ = R/Z
Unity power factor (FP = 1)
 Implies that apparent power is real power (S = P).
 If FP = 1, then  = 0º.
 It could also be said that the load looks purely resistive. Load
current and voltage are in phase.
Lagging power factor
For load containing resistance and inductance
 Power factor will be less than one and lagging
 The load current lags load voltage.
 Implies that the load looks inductive.
Leading power factor
For load containing resistance and capacitance
 Fp (power factor) is less than one and is leading.
 The load current leads load voltage
 Implies that the load looks capacitive
The importance of power factor
cos  = Real power (watts) / Apparent power (volt-ampere).

To transfer a given amount of power at certain voltage, the electrical current
is inversely proportional to cos. Hence higher the power factor lower will be
the current flowing. A small current flow requires less cross sectional area of
conductor and thus it saves conductor and money.

Power factors below 1.0 require a utility to generate more than the minimum
volt-amperes necessary to supply the real power (watts). This increases
generation and transmission costs. For example, if the load power factor were
as low as 0.7, the apparent power would be 1.4 times the real power used by the
load. Line current in the circuit would also be 1.4 times the current required at
1.0 power factor, so the losses in the circuit would be doubled (since they are
proportional to the square of the current). Alternatively all components of the
system such as generators, conductors, transformers would be increased in size
(and cost) to carry the extra current.

Further the KVA rating of machines is also reduced by having higher power
factor as,
Hence, the size and cost of machine also reduced. So, electrical power factor
should be maintained close to unity.
 Increasing the Power Factor:
As the power factor (i.e. cos θ) increases, the ratio of real power to apparent
power (which = cos θ), increases and approaches unity, while the angle θ
decreases and the reactive power decreases. [As cos θ → 1, its maximum
possible value, θ → 0 and so Q → 0, the load becomes less reactive and
more purely resistive].
 Decreasing the Power Factor:
 As the power factor decreases, the ratio of real power to apparent power also
decreases, as the angle θ increases and reactive power increase.
 If an a.c. generator is rated to give, say, 2000 A at a voltage of 400 V, it
means that these are the highest current and voltage values the machine can
give without the temperature exceeding a safe value. Consequently the rating
of the generator is given as 400 2000/1000 800 kVA. The phase
difference between the voltage and the current depends upon the nature of
the load and not upon the generator. Thus if the power factor of the load is
unity, the 800 kVA are also 800 kW, and the engine driving the generator
has to be capable of developing this power together with the losses in the
generator. But if the power factor of the load is, say, 0.5, the power is only
400 kW, so that the engine is developing only about one-half of the power of
which it is capable, though the generator is supplying its rated output of 800
kVA.
Similarly, the conductors connecting the generator to the load have to be
capable of carrying 2000 A without excessive temperature rise.
Consequently they can transmit 800 kW if the power factor is unity, but
only 400 kW at 0.5 power factor for the same rise of temperature.
It is therefore evident that the higher the power factor of the load, the greater
is the active power that can be generated by a given generator and
transmitted by a given conductor. The matter may be put another way by
saying that, for a given power, the lower the power factor, the larger must be
the size of the source to generate that power and the greater must be the
cross-sectional area of the conductor to transmit it; in other words, the
greater is the cost of generation and transmission of the electrical energy.
BALANCED THREE-PHASE CIRCUITS
Three-phase voltages are generated in the same way as single-phase voltages. A
three-phase system is simply three single-phase systems, which are displaced in timephase from one another. The single-phase systems that form the three-phase systems are
generally interconnected in some way.
Generation of Three-Phase Voltages
a
b'
c'
N
120o
120o
S
120o
c
b
Fig.1.1
a'
ea
e
ec
eb
180
0
60
Consider the coil aa’ on the armature
of a two-pole machine. When the
poles are in the position as shown in
Fig.1.1, the emf of conductor ‘a’ of
the coil aa’ is maximum and its
direction is away from the reader. If a
conductor is placed 120o away in
space from ‘a’ at position ‘b’, then it
will have maximum emf in a direction
away from the reader when the north
pole will be at ‘b’, in other words 120o
later than the time instant when the
north pole was at ‘a’. In the same
manner, the maximum emf in the
direction away from the reader for the
conductor at ‘c’ would occur 120o
later than that at ‘b’ and 240o later
than that ‘a’.
120
t
300
240
360
Fig.1.2
Thus the coils aa’, bb’ and cc’ would have emfs that are 120 o out of phase in time as
shown in the wave diagram of Fig.1.2. This system is called three-phase because there
are three waves of different time phase. From the above discussion, it is apparent that any
number of phases could be developed through properly spacing the coils. In general, the
electrical displacement between phases for a balanced n-phase system is (360/n)
electrical degrees. However, for two-phase system, it is 90 electrical degrees.
1
The equation of the phase emfs for a balanced three-phase system are as follows.
ea = Em sin t
eb = Em sin (t – 120o)
ec = Em sin (t – 240o)
where, Em is the peak value of the voltages in the three phases.
Their phasor representation is as follows.
(1.1)
Ec
120o
Ea
120o
120o
Fig.1.3
Eb
The sum of the above-mentioned three emfs is




ea  eb  ec  Em sin t  sin t  120 o  sin t  240 o



 Em sin t  2. sin t  180 o cos 60 o

 Em sin t  2.sin t. 1
0
2



(1.2)
This is also evident from the wave-diagram, wherefrom it can be seen that the sum of the
ordinates of the three emfs at any instant is always zero. The same can also be proved
from the phasor diagram. In other words, at no time instant all the three emfs are of same
polarity.
Phase Sequence
Phase sequence means the order in which the three phases attain their peak values.
In the above discussion, clockwise rotation of the field system was assumed. This
assumption made the emf of phase b to lag behind that of a by 120 o and the emf of c to
lag behind that of b by 120 o. Hence, the order in which the phase emfs attain their peak
values is ‘abc’. It is called the phase sequences ‘abc’.
If now the rotation of the field structure is reversed, then the emf of phase c would
lag behind that of a by 120o and the emf of b would lag behind that of c by 120o. Then the
order in which the three phases would attain their peak values would be reversed. The
phase sequence would then be ‘acb’.
Obviously, in a three-phase system, there are only two possible sequences, i.e.
‘abc’ and ‘acb’, as shown in Fig.1.4.
2
Eb
Ec
Ea
Ea
Fig.1.4
Eb
a-b-c
a-c-b
Ec
Numbering of the Phases
The three phases may be numbered 1, 2 and 3 or a, b and c. They may be given
three colours. The colours used commercially are Red (R), Yellow (Y) and Blue (B).
Double-subscript Notation
While analyzing three-phase circuits, it is imperative that directions in which the
circuit is being traced be noted and recorded. For such cases, double subscript notation is
very convenient. The order in which the subscripts are written denotes the direction in
which the circuit is being traced. Thus the emf from a to b is designated as E ab and that
Eab= E 60o
a
E
b
E
c
d
Fig.1.5
60o
Ecd = E 0o
from c to d as Ecd as shown in Fig.1.5. If d is connected to a as shown in Fig.1.6, then the
emf from c to b is determined by adding all the emfs in the direction encountered as the
circuit is traced from c to b. Hence,
Ecb = Ecd + Eab
a
E
b
Ecb
Eab
30o
c
E
d
Fig.1.6
30o
Ecd
Ecb = 2E cos 30 = 3E
o
3
Interconnection of Three Phases
If the three armature coils of the generator discussed above are not interconnected
but are kept separate as shown in Fig.1.7, then each phase would need two conductors for
transfer of power and the total number of conductors in that case becomes six. Such a
system may be called a six-wire, three-phase system. Such a generator can be loaded with
Load a
a
Ea'a
a'
Load b
b
Eb'b
three independent single-phase loads.
However, the use of six conductors makes
such a system very complicated and
expensive. Hence, the three phases are
generally interconnected which results in
substantial savings in cost. The two most
popular methods of interconnection are
i)
Star or Y connection and
ii)
Mesh or  connection.
b'
Load c
c
Ec'c
c'
Fig.1.7
Star Connection
In this method of interconnection, the similar ends, i.e. a, b and c or a’, b’ and c’,
are joined together at the point n which is known as star point or neutral point as shown
in Fig.1.8. The three conductors meeting at the point n are replaced by a single conductor
known as neutral conductor. Such a system is known as 4-wire, 3-phase system.
a
a
Ia
c'
Ic'c
Ina
a'
n
n'
In'n = Ia + Ib + Ic
Inc
b'
Ib'b
c
Ia
Fig.1.8
Ia'a
b
Ib
Ic
Inb
c
b
Ib
Ic
While considering the distribution of currents in a three-phase system it is to be
borne in mind that the arrows placed alongside the currents I a, Ib, Ic etc indicate the
directions of currents when they are assumed to be positive in magnitude and not the
directions at a particular instant.
4
If balanced three-phase voltages are applied across a balanced three-phase load,
the three currents Ia, Ib and Ic are equal in magnitude and are 120o out of time phase. Then
the neutral current
In’n = Ia + Ib + Ic = I0o + I – 120o + I – 240o = 0.
(1.3)
Hence, in that case, the neutral wire may be omitted and such a system is then known as
3-wire, 3-phase system as shown in Fig.1.9. Since all the currents are not positive or
negative at any time instant, hence any one or two line conductors offer the return path
for current at every instant of time.
a
Ia
Ina
Fig.1.9
n
Inc
Inb
c
Ib
b
Ic
Line and Phase Voltages in Star connection
a
Fig.1.10
Ena
Enc
Eba
n
Enb
c
b
Ecb
Eac
The voltage induced in each coil is
called the phase voltage. In other
words, the voltage between the
neutral point and any one of the lines
is called the phase voltage. On the
other hand, voltage between any pair
of terminals or lines is called the line
voltage as shown in Fig.1.10.
Considering ‘abc’ phase sequence, for a balanced three-phase system:
Ena = Eph 0o, Enb = Eph – 120o and Enc = Eph – 240o
(1.4)
Then the line voltages are as follows and are shown in Fig.1.11 in phasor form.
Eba = Ebn + Ena = Ena – Enb
Ecb = Ecn + Enb = Enb – Enc
and
Eac = Ean + Enc = Enc – Ena
(1.5)
5
Ebn
Eba
Enc
Eac
30o
30o
Ena
Ean
Ecn
Enb
Fig.1.11
Ecb
From the phasor diagram shown in Fig.1.11
Eba 
E
2
ph
2
 E ph
 2.E ph .E ph cos 60 o   3.E ph
(1.6)
and Eba leads Ena by 30o
Similarly, |Eba| = |Ecb| = |Eac| = EL = 3.Eph and
Eba, Ecb and Eac lead Ena, Enb and Enc respectively by 30o.
Line and Phase Currents in Star connection
a
Ia
Ina
Fig.1.12
n
Inc
Inb
c
Ib
b
The current in each winding is
known as phase current while the
current flowing in each line is called
the line current, as shown in
Fig.1.12.
For balanced 3-phase system
Ic
|Ina| = |Inb| = |Inc| = Iph.
In phasor form (considering a lagging phase angle),
Ina = Iph – , Inb = Iph – (120o + ), Inc = Iph – (240o + )
(1.7)
6
From the circuit it is obvious that
Ia = Ina ,
Ib = Inb
and Ic = Inc
 |Ia| = |Ib| = |Ic| = IL = Iph.
(1.8)
Power in Star connection
For balanced three-phase system
Total power = 3  Phase power.
Now, Phase power = Eph. Iph. cos

Total Power = P = 3. Eph. Iph. cos
Again, E ph 
EL
and I ph  I L
3
Therefore, in terms of line quantities
E
(1.9)
P  3  L  I L  cos  3E L .I L .cos
3
In Eqn.(1.9), it is to be noted that  is not the angle between EL and IL, but is the angle
between Eph and Iph.
Delta Connection
In this form of interconnection, the dissimilar ends of the three phase windings are
joined together. In other words, the three windings are joined in series to form a closed
mesh as shown in Fig.1.13.
Inb
n
n
Inc
c
Ic = Inc - Inb
c
Fig.1.13
Inb
b
Ina
n
Inc
n
Inc
Inb
b
a
Ina
Ina
n
Ina
Inc
n
Ia = Ina - Inc
a
Ib = Inb - Ina
Inb
From the circuit diagram, it appears that the delta connection results in shortcircuiting the three windings in a closed loop. But, for balanced three-phase system
Ena+Enb+Enc = 0 and, hence, no current of fundamental frequency will flow in the closed
loop. The delta connection results in 3-wire, 3-phase system.
7
Ic
Enc = Eac
Inc
Fig.1.14
Ibn
Ian
Ib

30o
30o
30o
30o

Inb

30o
Ena = Eba
Ina
30o
Icn
Ia
Enb = Ecb
Line and Phase Voltages in Delta Connection
| Ena | = | Enb | = | Enc | = Eph.
From the circuit arrangement shown in Fig.1.13, it is obvious that
Eba = Ena , Ecb = Enb and Eac = Enc.

| Eba | = | Ecb | = |Eac | = EL = Eph.
(1.10)
Line and Phase Currents in Delta Connection
| Ina | = | Inb | = | IIc | = Iph.
From the circuit diagram shown in Fig.1.13, the line currents are as follows:
Ia = Ina – Inc = Ina + Icn
Ib = Inb – Ina = Inb + Ian
and
Ic = Inc – Inb = Inc + Ibn.
(1.11)
From the phasor diagram shown in Fig.1.14,
8
I
Ia 
2
ph

2
 I ph
 2.I ph .I ph . cos 60 o  3.I ph
I a leads I na by 30 o
and
Similarly , | Ia | = | Ib | = | Ic | = IL = 3. Iph and Ia, Ib and Ic leads Ina, Inb and Inc
respectively by 30o.
Power in Delta Connection
Total power = P = 3. Eph. Iph. cos
Now, Eph = EL and Iph = IL / 3.
Therefore, in terms of line quantities
P  3  EL 
IL
3
 cos   3.E L .I L . cos 
Balanced / Conversion
IL
a
Fig.1.15
IL
ZY
a
EL
EL
Z
Z
ZY
ZY
IL
c
Z
IL
b
c
EL
b
EL
IL
IL
When a balanced star-connected load is equivalent to a balanced delta-connected
load as shown in Fig.1.15, the line voltages and currents must have the same values in
both the cases.
For balanced -load:
E ph 
 ZY 
EL
and
3
E ph
I ph

I ph  I L
1 EL
.
3 IL
9
For balanced  - load:
E ph  E L and I ph 
 Z 
E ph
I ph
 3.
IL
3
 1 E 
EL
 3. . L   3.Z Y
IL
 3 IL 
Hence, for Y to  conversion : Z   3Z Y
1
: Z Y  .Z 
3
and for  to Y conversion
Alternate Proof
For balanced Y – load:
Za = Zb = Zc = ZY
and for balanced  - load:
Zab = Zbc = Zca = Z
For Y to  conversion:
Z ab  Z a  Z b 
or, Z   Z Y  Z Y 
Z a .Z b
Zc
Z Y .Z Y
 3.Z Y
ZY
For  to Y conversion :
Za 
or, Z Y 
Z ab .Z ca
Z ab  Z bc  Z ca
Z 2
1
 .Z 
3.Z  3
Power-factor of balanced Three-phase System
The power-factor of a balanced three-phase system, when the wave-forms of the
voltage and current are sinusoidal, is defined as the cosine of the angle between phase
voltage and phase current irrespective of whether the connection is Y or . In other
words, it is the ratio of phase active power (W) to phase apparent power (VA).
10
Comparison of Copper requirement of Three-phase with Single-phase System
Three-phase system can be compared with the single-phase system on the basis of
a fixed amount of power transmitted over a fixed distance with the same amount of
power loss. In all the cases, the total weight of copper will be directly proportional to the
number of wires since the distance is fixed and inversely proportional to the resistance of
each wire.
Same Voltage between the Lines
For Single-phase system:
For Three-phase system:
P1 = V.I1 cos
P3 = 3.V.I3 cos
Since, P1 = P3 , V.I1 cos = 3.V. I3.cos
or, I1 = 3.I3
Again,
Loss1 = Loss3
or,
I 12 .R1  2  I 32 .R3  3
or,
3I 32
R1 3I 32
1
 2 

2
R3 2 I 1
2  3I 3 2

R
No of wires of three - phase
Copper for three - phase

 1
Copper for single - phase No of wires of single - phase
R3
3 1
 
2 2
3
  0.75
4
Same Voltage to Neutral
The voltage to neutral in single-phase system is half the voltage between lines.
For single-phase system :
P1 = 2. Vn.I1 cos
For three-phase system:
P3 = 3.Vn I3 cos
Since, P1 = P3 , 2.Vn I1 cos = 3. Vn. I3 cos
3
or, I1  .I 3
2
Again,
Loss1 = Loss3
11
or,
2.I 12 .R1  3I 32 R3
3I 32
R1 3I 32
2
 2 

9
R3 2 I 1
3
2  I 32
4
Copper for three - phase
No of wires in three - phase R 1



Copper for single - phase No of wires in single - phase R 3
or,

3 2
 1
2 3
Example – 1
Calculate the active and reactive current components in each phase of a starconnected 10kV, three-phase alternator supplying 5000kW at a power-factor 0.8(lag). If
the total current remains the same when the load power-factor is raised to 0.9(lag), find
the new power output.
Solution
Given,
VL = 10,000V ,
cos = 0.8

5000  103 = 3  10,000  IL  0.8

IL = 360.8A

IL = 360.8 cos-1 0.8 = 360.8 36.87o = (288.6 – j216.5)A.

Active component = 288.6A and Reactive component = -216.5A.
Now, VL = 10,000V, IL = 360.8A and cos = 0.9
 P = 3  10,000  360.8  0.9 = 5624315 W
= 5624.3 kW
Example – 2
A balanced load of (8+j6) per phase is connected to a three-phase, 230V supply.
Find the line current, power-factor, power, reactive VA and total VA when the load is i)
star connected and ii) delta connected.
Solution
Zph = 8 + j6 = 1036.87o .
12
Star Connection:
Vph = 230/3 = 132.8 V
 I ph 
Vph
Z ph

132.8
 13.28 A.
10
IL = Iph = 13.28A.
cos = cos 36.87o
= 0.8 (lag).
P = 3  230  13.28  0.8
= 4232.3 W
VAR = 3  230  13.28  0.6
= 3174.2 VAR
VA = 3  230  13.28
= 5290.4 VA
Delta Connection:
Vph = 230 V.
 I ph 
230
 23 A
10
 IL = 3. Iph = 3  23 = 39.8 A.
cos = 0.8 (lag)
 P = 3  230  39.8  0.8
= 12684.1 W
VAR = 3  230  39.8  0.6
= 9513.1 VAR
VA = 3  230  39.8
= 15855.2 VA.
Example – 3
A balanced three-phase, star-connected load of 150 kW takes a leading current of
100A with a line voltage of 1100V at 50Hz. Find the circuit constants of the load per
phase.
Solution
P = 150 kW = 150,000 W
Given,
VL = 1100 V and IL = 100 A.

150,000 = 3  1100  100  cos
 cos = 0.787 (lead)
13
Now,
E ph 
1100
 635.1 V and I ph  100 A.
3
635.1

 6.35 
100
 Z ph
 Rph = 6.35  0.787 = 5 
Xcph = 6.35  sin (cos-1 0.787) = 6.35  0.617 = 3.917 
and
 C ph 
1
 812.6 F .
2  50  3.917
Example – 4
A balanced star-connected load is supplied from a symmetrical three-phase, 400V
system. The current in each phase is 30A and lags 30 o behind the phase voltage.
Calculate the phase voltage and the total power.
Solution
VL
= 400 V
 E ph 
Given,
400
 230.94 V
3
cos = cos 30o = 0.866 (lag)
Iph = 30 A  IL = 30 A.
 P = 3  400  30  0.866 = 18000 W.
Example – 5
A symmetrical three-phase, 400V system supplies a balanced delta-connected
load. The current in each branch circuit is 20A and the phase angle is 40 o (lagging). Find
the line current and the total power.
Solution
Given , Iph = 20 A and VL = 400 V
IL = 3  20 = 34.64 A
cos = cos 40o = 0.766 (lag)
P = 3  400  34.64  0.766 = 18383.5 W.
14
Example – 6
Between any two terminals of a three-phase balanced load the voltage is 415V
and the resistance is 3.0. The current in each of the three lines is 100A. Find the powerfactor of the load. Find also the resistance and reactance per phase of the load with (a)
star connection, (b) delta connection.
Solution
Star Connection:
Let, resistance / phase = Rph
 Resistance between two terminals = Rph + Rph = 2Rph
Now, 2Rph = 3
or, Rph = 3/2 = 1.5 
Again,
E ph 
 Z ph
415
 239.6V and I ph  100A
3
239.6

 2.396 
100
 X ph 
2.396
Power factor 
2

 1.5 2  1.868 
1.5
 0.626
2.396
Delta Connection
Resistance between tw o terminals 
Again,
R ph  2R ph
R ph  2R ph

2
R ph
3
2
3 3
 R ph  3
or, R ph 
 4.5 
3
2
E ph  415 V and I L  100 A
 I ph 
 Z pn
100
 57.73 A
3
415

 7.19 
57.73
 X pn 
7.19
2

 4.5 2  5.6 
 Power factor 
4.5
 0.626.
7.19
15
Example – 7
Three star-connected impedances Z1 = 20 + j37.7  per phase are in parallel with
three delta-connected impedances Z2 = 30 – j159.3  per phase. The line voltage is
398V. Find the line current, power-factor, power and reactive VA taken by the
combination.
Solution
 to  conversion:
1
30  j159.3  10  j53.1   54.03  79.3o 
3
Z1  20  j37.7  42.6762.05 o 
Z2 
 Z ph 
Z1 .Z2
54.03  79.3o  42.6762.05 o

20  j37.7   10  j53.1
Z1  Z2
2305.46  17.25 o

30  j15.4
2305.46  17.25 o

 68.379.92 o 
o
33.72  27.17
Given,
VL  398 V,
 Vph 
398
3
 229.8V
229.8
 3.36 A.
68.37
 I L  I ph  3.36 A.
 I ph 
Power-factor = cos(9.92o) = 0.985 (lag)

Power = 3  398  3.36  0.985 = 2281.5 W
VAR = 3  398  3.36  sin 9.92o = 399 VAR
Example – 8
A three-phase, star-connected alternator feeds a 2000 hp delta-connected
induction motor having a power-factor of 0.85(lag) and an efficiency of 93%. Calculate
the current and its active and reactive components in (a) each alternator phase, (b) each
motor phase. The line voltage is 2200V.
16
Solution
a
a'
IL
2200V
c'
Fig.1.16
b
c
b'
Motor output  2000 hp  2000  746  1492,000 W
1492,000
 Motor input 
 1604301 W
0.93
Given, VL  2200 V and cos  0.85
1604301
 IL 
and
3  2200  0.85
-1
  cos 0.85  31.79 o
 495.3 A
 I L  495.3  31.79 o A
Alternator :
I ph  I L  495.3  31.79 o
 421  j260.93 A
Motor :
I ph 
IL

495.3
  31.79 o
3
3
 285.96  31.79 o
 243  j150.65  A.
17
Transformer
Definition of Transformer:
It is a static device, which transfer electric powers from one circuit to another without any
change in frequency, but with a change in voltage and corresponding current level also.
So in brief transformer is a device that –
I).
ii).
iii).
iv).
v).
Transfer electrical power from one circuit to another.
It does so without a change in frequency.
It accomplishes this by electromagnetic induction.
It transforms voltage level corresponding change in current level.
Where the electrical circuits (primary & secondary) are in mutual
inductive influence to each other.
Ideal Transformer:
1)
Winding resistance are negligible (i.e. purely inductive coil), So, I2 R loss negligible.
2)
All the flux set up by primary links the secondary winding. i.e. no leakage flux.
3)
Core losses are negligible (Hysterisis & eddy current loss).
4)
The core has constant permeability, i.e. the magnetization curve for the core is linear.
Considering an ideal transformer whose secondary is open & whose primary is connected to a
Sinusoidal alternating voltage V1. Under this condition, the primary draws current from the
source to build up a counter electromotive force equal & opposite to the applied voltage. Since
primary coil is purely inductive & there is no output, the primary draws a magnetizing current Iø
only to magnetise the core. This current produces an alternating flux Ø which is proportional to
the current & hence in phase with it. The changing flux is linked with both the windings. So, it
produces self induced emf in the primary. This emf e1, at any instant, equal to & in opposition to
V1.
Similarly in the secondary winding, an induced elf e2 is produced. This emf is in phase opposition
with V1 & its magnitude is proportional to the rate of change of flux & the no. of secondary turns.
E.m. f. equation of a Ideal transformer (single phase) on no load
V1  Primary supply voltage
N1  Primary turn
E1  Primary Winding. Induced e m f
N2  Secondary turn
V2  Secondary terminal voltage
E2  Secondary Winding. Induced e m f
Ø  Flux = Ø max Sin t (If the supply voltage is sinusoidal)
  2  f where f = frequency
Induced e m f e1 = - N1 . d Ø / dt.
= - N1  Ø max Cos t.
= N1  Ø max Sin (t –/2)
= E1 max Sin (t –//2)
E1 = (E1 max ) / 2
= 2 f N1 Ø max / 2
= 2 f N1 Ø max = 4.44 f N1 Ø max .
v1 = - e1 = N1 d Ø / dt (As per Lenz’s Law, oppose I)
So, V1 = - E1
Similarly, e2 = - N2 dØ / dt = E2 max. Sin (t –//2)
E2 = E2 max /2 = 4.44 f N2 Ø max
So, (E1 ) / (E2 ) = (N1 ) / (N2 ) = (V1 ) / (V2 ) = (I2 ) / (I1 )
I1 = primary load current.
I2 = Secondary load current.
Magnetic Leakage
In an ideal transformer, it is assumed that all the flux linked with the primary winding
also links the secondary winding. But, in practice, it is impossible to realize this condition
as magnetic flux cannot be confined. The greater portion of Flux (mutual flux) flows in
the core while a small proportion called leakage flux links one or other winding, but not
both. On account of this leakage flux, both the primary & secondary windings have
leakage reactance. So, emf e x1=-j Ie X1
v1=e1+I1(R1+jX1)
Similarly, in the secondary winding, an emf of self induction I 2X2 is developed.
To reduce the leakage flux, primary & secondary coils are placed concentrically.
Voltage transformer ratio (K)
We know, E2/E1=N2/N1=K
a) If K>1, i.e. N2>N1, Transformer is called as Step-up Transformer
b) If K<1, i.e. N2<N1, Transformer is called as Step-down Transformer
For Ideal transformer, Input VA=Output VA, V1I1=V2I2 , I2/I1=V1/V2=1/K
Transformer on No-Load
The Primary input current under no-load condition has to supply i) Iron loss in the core i.e
Hysterisis Loss & Eddy current loss ii) a very small amount of copper-loss in primary.
No load primary input power W0= V1 I0 Cosφ
Ic = I0 Cosφ, (Core loss component + Cu loss component)
Iø = I0 Sinø ( Magnetizing Component)
Transformer on Load Condition
Equivalent Circuit Representation of Transformer
Transferred Equivalent Circuit.
Phasor Diagram of a Transformer
Basic principle of electric machines
Electric machines convert mechanical energy to electrical energy and vice versa.
Generator:
Function- Converts mechanical energy to electrical energy.
Principle- Generation of e.m.f. in a moving conductor in magnetic field.
Motor:
Function- Converts electrical energy to mechanical energy.
Principle- Force created due to interaction between magnetic field.
Machines require varying flux for their operation
Example: E.m.f. in a conductor is not induced if there is no change of flux associated
with it.
From Faraday’s laws of electromagnetic induction, induced e.m.f. in a conductor is
proportional to the rate of change of flux associated with it.
Categorizing machines with respect to variation of flux
Flux is only time varying- Transformer
Flux is only space varying- DC Machine
Flux is both time and space varying- Induction Motor
Construction of a DC Machine
Figure 1. Generalized construction of a DC Machine.
-1-
Functions of different parts of the DC Machine
Yoke-
-Stationary part of the machine.
-Made up of cast steel.
-Placeholder for main field poles and interpoles.
-Gives easy path for magnetic field lines to complete.
-Protects all the inner parts of machine.
Main Pole-
-Stationary part of the machine.
-Remains attached to the yoke.
-Produces the main magnetic field.
Armature-
-Rotating part of the machine.
-Made up of laminated sheets.
-Holds the armature winding.
-Produces electrical energy from mechanical energy in case of generator.
In this case, the armature is rotated by external source and the armature
winding generates e.m.f.
-Produces mechanical energy from electrical energy in case of motor. In
this case, armature winding is given a current supply from outside. The
armature rotates and produces mechanical energy.
Commutator- -Rotating part of the machine.
-Made from separated copper bars insulated from each other.
-Attached with armature shaft.
-Carries current to the brush from armature windings in case of generator.
-Carries current to the armature windings from brush in case of motor.
Brush-
-Stationary part of the machine.
-Made up carbon.
-Remains pressed with rotating commutator to establish electrical
connection between rotating part.
Interpole-
-Stationary part of the machine.
-Remains attached to the yoke.
-Produces magnetic field that opposes armature field.
-Reduces sparking at commutator surface.
-2-
Armature winding
Armature winding is the method to wind the conductors over the armature core.
Windings are connected in parallel to increase current rating. Windings are connected in
series to increase voltage rating. Two types of windings are mainly encountered:
Lap winding- Here the armature winding is such that part of each winding overlaps a part
of another winding.
For lap winding: Number of parallel paths (a) = Number of poles (P) in a machine
Wave winding- Here the armature winding is like a wave pattern.
For wave winding: No of parallel path (a) = 2
Generator e.m.f. equation
Let a P pole machine has  webers as its flux per pole
The total flux cut by an armature conductor in one revolution is P webers.
Let the machine rotates at n revolutions per second.
So, e.m.f. generated in one conductor = P n volts
Let the total number of conductors be Z
So e.m.f. generated = P n Z volts
Let the number of parallel paths be a
Z
P n volts
a
The total e.m.f. generated E a equals e.m.f. generated per parallel path
Z
Therefore, E a  P n volts
a
Again, let the armature rotates at a speed of N rpm.
N
So n =
60
Z
N
Then E a  P
volts. This is the expression of generator e.m.f.
a
60
So, e.m.f. generated per parallel path 
-3-
Concept of voltage build up in a generator
-Generators require strong field flux for generation of sufficient electrical power
-Strong field flux is created by electromagnets, i.e. sending current through field winding
of field poles.
-Current to the field is obtained by the generator itself, i.e. from e.m.f. generated in the
armature
-But to generate e.m.f. in the armature, field flux must exist
-This problem is overcome by what is called voltage build up
-A residual magnetism exists in the field poles which is small in magnitude
-This weak field flux creates small e.m.f. in armature when generator is started
-The small e.m.f. causes a relatively larger field current to flow causing a larger field flux
-This larger field flux creates a larger e.m.f. in armature
-The loop repeats and voltage gradually builds up to maximum value limited by the
construction, resistances of coils and the speed of operation
Points to remember in the above concept
-Voltage can only build up if residual field can generate armature e.m.f. of proper
polarity to increase the field
-The loading in case of shunt generator must not be above a critical value
-4-
Field connections in generator
Case 1- Separately excited field
A separately excited generator supplying a load.
Here the field coil is excited from a separate source
The field flux does not depend on the armature e.m.f. generated.
Voltage build up always takes place and not dependent on loading
Here, let the
e.m.f. generated in the armature = Eg volts
Resistance of armature winding = Ra ohms
Armature current
= Ia amps (This is the current through the circuit)
Resistance of load
Voltage across the load
= RL ohms
= V volts ( This is the voltage we get from generator)
Voltage supplied to the field
Field current
= Vf volts
= If amperes
Now, voltage drop due to armature resistance = Ia Ra volts
Then, Eg - Ia Ra = V (1)
And Ia = IL as load current is same as armature current.
Thus the equation (1) can be modified as
V= - Ra IL+ Eg
This equation is called the external or the load characteristics of the generator.
So, a curve of IL vs V will be a straight line with a negative slope.
-5-
Figure showing the load characteristics of a DC separately excited generator
Case 2- Shunt field
A shunt generator supplying a load.
Here the shunt field coil is excited from the armature
The field flux depends on the armature e.m.f. generated.
Voltage build up takes place only when load resistance is above a critical value
Here, let the
e.m.f. generated in the armature = Eg volts
Resistance of armature winding = Ra ohms
Armature current
= Ia amps (This is the current through the circuit)
Resistance of load
Load current
Voltage across the load
= RL ohms
= IL ohms
= V volts (This is the voltage we get from generator)
Resistance of the field
Field current
= Rf ohms
= If amps
Now, voltage drop due to armature resistance = Ia Ra volts
-6-
Voltage across field = If Rf volts
Voltage across load = IL RL volts
Now, Eg - Ia Ra = If Rf = V= IL RL
Ia = If + I L
(1)
(2)
From the above equations, load characteristics of the shunt generator can be found as
follows.
V = Eg – ( If + IL) Ra
=– Ra IL +( Eg - If Ra)
Now,
Eg  ,   If, therefore, Eg  If
Always remember that field current generates the field flux.
Then, V =– Ra IL +( k - Ra) If
k is a constant depending on the characteristics of generator.
Thus, the characteristic curve under load can be divided into two parts.
The first part is under light load where field current is constant. This represents a straight
line with a negative slope.
In the later part, the load increases and the field current is diverted to the load, so that If
decreases. Then, the curve deviates from the linear part as shown.
Figure showing the load characteristics of a DC shunt generator
-7-
Case 3- Series field
A series field generator supplying a load.
Here the series field coil is excited from the armature itself
The field flux depends on the armature e.m.f. generated.
Voltage build up takes place only when load is connected
Here, let the
e.m.f. generated in the armature = Eg volts
Resistance of armature winding = Ra ohms
Armature current
= Ia amps (This is the current through the circuit)
Resistance of load
Voltage across the load
= RL ohms
= V volts (This is the voltage we get from generator)
Resistance of the field = Rf ohms
Now, voltage drop due to armature resistance = Ia Ra volts
Voltage across field = Ia Rf volts
Then, from the circuit of the series generator, V =Eg - Ia Ra - Ia Rf
(1)
Now, (i) Eg  ,   Ia, therefore, Eg  Ia Always remember that field current generates
the field flux. Here in series circuit armature current flows through the field.
(ii) Ia = IL
Putting these values of (i) and (ii) in equation (1), we get
V = Ka IL –( Ra + Rf ) IL
Ka is constant depending on a particular generator.
This equation is called the load characteristics of the series generator.
Initially, Ka and Ra+Rf is constant, so V IL. This represents the first part of the curve.
-8-
In the next part,  becomes constant due to saturation, so that IL = constant. This
represents the peak part of the curve with a small flat top.
Next, IL = constant and ( Ra + Rf ) increases due to heating effect.
Thus
V = K2 – ( Ra + Rf ) K3 where Kn are constants. This represents the later drooping part of
the curve.
Figure showing the load characteristics of a DC series generator
Compound field generator
Here, both series and shunt field exists in a generator. If the series field and shunt field
opposes each other, it is called differential compound field generator. If the series and
shunt field supports each other, it is called cumulative compound field generator.
-9-
Motors
Electrical motor is a machine for converting electrical energy in to mechanical energy. Basic
principle of electrical motor is that a current carrying conductor placed within a magnetic field
experiences a force at right angles to the field and the direction of force is governed by Fleming
left hand rule.
Like generator, motors also have a field poles that are stationary and armature that is the rotating
part. The thing that is to be remembered is that, under influence of field, the armature rotates and
since the conductors in the armature cuts the field, e.m.f. is generated in the conductors like
generator. The direction of generated e.m.f. is such that it opposes the flow of electric current
through the armature. It is called back e.m.f. in a motor and is always kept in mind during all
calculations.
Case 1- Permanent magnet field motor
A permanent magnet field motor.
Here the field is created by a permanent magnet
The field flux does not depend on the supply voltage and armature characteristics
Example: Battery operated fan motor
Here, let the
Back e.m.f. generated in the armature = Eb volts
Resistance of armature winding = Ra ohms
Armature current
= Ia amps (This is the current drawn by the motor)
Voltage applied at motor terminal = V volts
Now, voltage drop due to armature resistance = Ia Ra volts
V  Eb
 Ia
Ra
or, Eb = V- IaRa
Then,
Case 2- Series motor
A series motor operating from a DC supply.
Here the field coil is in series with the armature
The field flux depends on the armature current
The motor is always started at loaded condition because at no load the speed of the motor
becomes dangerously high
Example- Traction motor in train
Let
Back e.m.f. generated in the armature = Eb volts
Resistance of armature winding = Ra ohms
Armature current
= Ia amps
Voltage applied
Resistance of the field
= V volts
= Rf ohms
Now, voltage drop due to armature resistance = Ia Ra volts
Voltage drop across field = Ia Rf volts
Then,
V  Eb
 Ia
R f  Ra
or, Eb = V – Ia (Rf + Ra)
Case 3- Shunt motor
A shunt motor operating from a DC supply.
Here the field coil is supplied separately from the voltage source
The field flux do not depend on the armature current
Since field flux is almost constant at constant supply voltage, speed remains more or less
constant
Example- Textile mill motors where speed needs to be constant
Here, let the
Back e.m.f. generated in the armature = Eb volts
Resistance of armature winding = Ra ohms
Armature current
= Ia amps (This is the current through the circuit)
Voltage applied
Resistance of the field
Current through the field
= V volts
= Rf ohms
= If amps
Now, voltage drop due to armature resistance = Ia Ra volts
Voltage drop across field = If Rf volts
From the figure, we get
V = Ia Rf
V - Eb = Ia Ra
and main current I = If + Ia
Characteristics of motor
Speed of a motor
All motors generate back e.m.f.
The expression of back e.m.f. is same that of generator i.e.
Z
N
Eb  P
volts
a
60
Z 1
P
= K  N where K is constant and is equal to
a 60
Then, Eb   N
This is the general relation between the back e.m.f. and speed of a motor
Case 1 – Series motor
In case of series motor
Eb = V – I a (Rf + Ra) (i)
Now,
Eb   N
and   Ia (because field is in series with the armature)
so, Eb  Ia N
therefore from (i) we get
K Ia N = V – Ia (Rf + Ra) (here K is proportionality constant) (ii)
Now, and are constants in case of motor. So, eqn (ii) can be modified as follows
N = K1 ×
V
- K2 (here K1 and K2 are proportionality constants)
Ia
Speed (N)
This is the relation between armature current and speed in a series motor.
Armature Current (Ia)
Armature Current-Speed characteristics of series motor.
Case 2 – Shunt motor
In case of shunt motor
Eb = V – I a Ra
Now,
Eb   N
and  = constant (characteristic feature of shunt motor)
Therefore, K N = V – Ia Ra
(here K is proportionality constant)
Speed (N)
This is the relation between armature current and speed in a shunt motor.
Armature Current (Ia)
Armature Current-Speed characteristics of shunt motor.
Torque developed in a motor
Torque developed in a motor determines how much mechanical power can be delivered
by the motor.
Let
Torque developed = T N-m
Field flux =  wb
Back e.m.f. = Eb volts
Armature current = Ia amps
Angular speed =  rad/sec
In this case, the electrical power consumed by the motor will be converted into
mechanical power.
Electrical power = Eb× I a
Mechanical power = T×
Therefore, Eb× Ia = T×
(i)
If the speed of the motor is given as N r.p.m,
2N
Then,  =
60
Now, from (i) , T =
or, T =
or, T =
Eb  I a

I
Z
N
P
× a
2N
a
60
60
ZP
×  Ia
2a
or, T = KT  Ia or T  I a
(ii)
where KT is the constant for a particular motor and is dependent on no. of conductors,
poles and parallel path.
Eqn. (ii) is called the generalized torque equation for a motor.
Case 1 – Series motor
Also, T  Ia (general expression)
And,   Ia (because field is in series with the armature)
Torque (T)
Therefore, T  Ia2
Armature Current (Ia)
The Armature Current-Torque characteristics of series motor
Note: In all cases of motor and generator, remember that field flux is produced by the
current through the field winding.
Case 2 – Shunt motor
T  I a (general expression)
and  = constant (characteristic feature of shunt motor)
Torque (T)
So, T Ia
Armature Current (Ia)
The Armature Current-Torque characteristics of shunt motor.
The torque speed characteristics
Case 1 – Series motor
Now, for series motor, we have seen that
N = K1 ×
V
- K2
Ia
and T  Ia2
therefore, Ia =
T
K3
(K3 is proportionality constant)
From these equations we have,
N = K4 
V
- K2
T
This is the relationship between torque and speed in a series motor.
Speed (N)
Torque (T)
The Torque-Speed Current characteristics of series motor
Case 2 – Shunt motor
If the load is applied to a motor, the motor immediately tends to slow down. With the
shunt motor, the field flux remains almost constant. Meanwhile, the decrease of speed
decreases the back e.m.f. in the armature. With decrease of the back e.m.f. more current
is drawn in to the armature. This continues until the increased armature current produces
sufficient torque to meet the demands of the increased load.
Now, we have seen that
K N = V – Ia Ra
and T Ia
From these equations we have,
Speed (N)
K N = V – K1×T
This is the relationship between torque and speed in a shunt motor.
Torque (T)
The Torque-Speed Current characteristics of shunt motor.
Compound motors
Like generators, if series and shunt field winding both exists in a motor, it is called a
compound motor. It can be of two types, namely, the cumulative compound motor and
the differential compound motor. In a cumulative compound motor, the series field
winding is connected in such a way that the flux produced by it helps the flux produced
by the main field winding, whereas in differential compound motor, the series field flux
opposes the main field flux. The characteristics of a compound motor are the combination
of both series and shunt motor.
The torque developed by a cumulative compound motor increases with sudden increase
in load and has a definite speed in no load condition (Note: series motor cannot be
operated at no load). These motors are useful in sudden loading applications such as
punching machine.
The speed of a differential compound motor remains more or less constant with increase
in load, but the torque decreases with load. These motors are not better than shunt motor
in many cases and are not used in practice.
Induction motor
The most common type of AC motor being used throughout the world today is the
induction motor. Induction motors are more rugged, require very little maintenance and
are less expensive than equally powered DC machines. Induction motors can be both
single and three phase. Three phase induction motors are widely used in industrial
applications.
Three phase induction motors are of two types, the squirrel cage motor which is very
popular, and the slip-ring type motor.
A squirrel cage induction motor.
To study the principle of operation, firstly the stationary body of the motor, called the
stator is considered. The stator comprises of the hollow body of stacked laminated steel
which is the core. This core is placed inside another hollow cast body for mounting.
Core
Winding
Stator core of induction motor.
In the stator core, three coils are placed at 120 apart, well insulated from each other and
also from the core to prevent any shock hazard or short circuit.
Production of the rotating magnetic field
Let the coils of the core be connected according to the following diagram.
Winding of the stator.
Delta connected stator.
Since this type of connection diagram looks like the letter delta, it is called delta type of
connection.
A, b and C are the terminals where three phases of the AC supply are given. Let us
consider the following sequence of three phase AC where the peaks of each phase are
120 apart from each other.
Rotating magnetic field in stator.
Since the coils are also 120 apart, application of this three phase AC will produce a
resultant magnetic field that will be different at different time instants. Considering the
direction of field, it will be continuously rotating in space.
Construction of the rotor
Now let us consider the construction of the rotating part, the rotor. Since the squirrel cage
type is used widely, it will be discussed here.
Rotor of induction motor.
The rotor consists of laminated steel cylinder with slots in which aluminium or copper
conductors are placed. Two heavy aluminium or copper end rings are placed at the two
ends of the conductors and are welded with the conductors making a complete short
circuit. This forms a squirrel cage structure and hence the name. No insulations are given
anywhere in the core as current will flow through the lower resistance aluminium/copper
path than steel. The rotor is placed inside the stator with air gap as low as possible. Shaft
from this rotor comes out through the motor housing through bearings for driving the
machines.
Principle of rotation
Let us first assume that the rotor is stationary at the beginning. As soon as the stator is
excited by three phase AC, the rotating magnetic field is created. This rotating field cuts
the rotor conductor. To facilitate this magnetic path, the rotor core is given with a
minimum air gap possible between the stator and the rotor. When the flux cuts the
conductors in this way, by electromagnetic induction, a heavy current flows through the
rotor conductors that are already short circuited by this squirrel cage construction. This
current in turn creates a counter magnetic field in the rotor core. The interaction between
the stator and the rotor fields produces a torque and the rotor starts rotating in the
direction of rotating magnetic field.
Concept of synchronous speed
The speed of rotating magnetic field depends on the supply frequency and number of
poles. The synchronous speed Ns is given by
Ns 
120 f
where f is the supply frequency and P is the number of poles in the motor.
P
Concept of slip
Whenever the supply in the stator is switched on, the rotating magnetic field starts
rotating at synchronous speed. Since the rotor is a mechanical device, it cannot
immediately gain this speed. The rotor starts rotating slow at first and then gains speed.
The difference between the speed of the rotor and the synchronous speed in the stator is
measured by the term slip. Slip is abbreviated as S and usually expressed in percentage.
Ns  Nr
 100 % where Ns is the synchronous speed of the stator and Nr is that of
Ns
rotor.
S
Frequency of induced current in the rotor
Since there is a difference of speed in the rotor and the stator, the rotor frequency is given
by
N  Nr
fr  P  s
120
Rotor can never rotate at synchronous speed
The current is induced in the rotor conductors as the rotating magnetic field cuts the
conductors. This current produces the rotor magnetic field required for rotation. In case
of induction motor, the rotor conductors also rotate in the direction of the rotating stator
field. If the rotor attains same speed same as that of stator field, the stator field will have
no relative speed with respect to that of rotor conductor. Hence the stator flux cannot cut
the rotor conductor and no rotor current will be produced. At that condition, no magnetic
field will be produced by the rotor to interact with the stator field. So no torque will be
produced in the rotor and its speed will fall. As soon as speed falls, again the stator flux
will cut the conductor and rotor field will be again produced. This field will prevent rotor
from becoming standstill, but according to the explained fact, rotor will never attain
synchronous speed. The net result is that, the rotor will always rotate, but with a slip.
Note: For all cases, slip is within 6% for small motors and can be as low as 2% for large
motors.
Problem:
A three phase induction motor of 415V, 50Hz runs at 960 rpm. Calculate its slip.