2. • In Pakistan, most power outlets deliver a sinusoidal voltage
having a frequency of 50 Hz and a “voltage” of 230 V
(elsewhere, 50/60 Hz and 120/240 V are typically encountered)
• But what is meant by “230 volts”?
• This is certainly not the instantaneous value of the voltage, for
the voltage is not a constant
• The value of 230 V is also not the amplitude which we have
been symbolizing as Vm
• If we displayed the voltage waveform on a calibrated
oscilloscope, we would find that the amplitude of this voltage at
one of our ac outlets is 230* 𝟐, or 325 volts
3
Introduction (1/2)
3. • We also cannot fit the concept of an average value to the 230 V
,
because the average value of the sine wave is zero
• We might come a little closer by trying the magnitude of the
average over a positive or negative half cycle; by using a
rectifier-type voltmeter at the outlet, we should measure 200 V
• As it turns out, however, 230 V is the effective value of this
sinusoidal voltage; it is a measure of the effectiveness of a
voltage source in delivering power to a resistive load
4
Introduction (2/2)
4. 1.1 Effective Value of a Periodic
Waveform (1/3)
• The effective value of any periodic current is
equal to the value of the direct current which,
flowing through an R ohm resistor, delivers the
same average power to the resistor as does the
periodic current.
• In other words, we allow the given periodic
current to flow through the resistor, determine
the instantaneous power i2R, and then find the
average value of i2R over a period; this is the
average power
• We then cause a direct current to flow through
this same resistor and adjust the value of the
direct current until the same value of average
power is obtained
5
5. 1.1 Effective Value of a Periodic
Waveform (2/3)
• The general mathematical expression for the effective value of
i(t) is now easily obtained
• The average power delivered to the resistor by the periodic
current is:
• where the period of i(t) is T. The power delivered by the direct
current is:
• Equating the power expression and solving for Ieff, we get;
6
6. 1.1 Effective Value of a Periodic
Waveform (3/3)
• The result is independent of the resistance R, as it must be to
provide us with a worthwhile concept
• A similar expression is obtained for the effective value of a
periodic voltage by replacing i and Ieff by v and Veff, respectively
• Notice that the effective value is obtained by first squaring the
time function, then taking the average value of the squared
function over a period, and finally taking the square root of the
average of the squared function.
• In short, the operation involved in finding an effective value is the
(square) root of the mean of the square; for this reason, the
effective value is often called the root-mean-square value, or
simply the rms value.
7
8. 1.2 Effective (RMS) Value of a Sinusoidal
Waveform (2/2)
• Thus the effective value of a sinusoidal current is a real quantity
which is independent of the phase angle and numerically equal to
1/ 2 = 0.707 times the amplitude of the current
• A current 2 cos 𝜔𝑡 + 𝜑 A, therefore, has an effective value of 1 A
and will deliver the same average power to any resistor as will a
direct current of 1A
• It should be noted carefully that the 2 factor that we obtained as the
ratio of the amplitude of the periodic current to the effective value is
applicable only when the periodic function is sinusoidal
• For a sawtooth waveform, for example, the effective value is equal to
the maximum value divided by 3
• The factor by which the maximum value must be divided to obtain
the effective value depends on the mathematical form of the given
periodic function; it may be either rational or irrational, depending on
the nature of the function
9
10. 1.3 Use of RMS Values to Compute
Average Power (1/3)
• The use of the effective value also simplifies slightly the
expression for the average power delivered by a sinusoidal
current or voltage by avoiding use of the factor ½
• For example, the average power delivered to an R ohm resistor
by a sinusoidal current is:
since ,
the average power is:
The other power expressions in terms of effective values are:
11
11. 1.3 Use of RMS Values to Compute
Average Power (2/3)
• Although we have succeeded in eliminating the factor 1/2 from
our average-power relationships, we must now take care to
determine whether a sinusoidal quantity is expressed in terms of
its amplitude or its effective value
• In practice, the effective value is usually used in the fields of
power transmission or distribution and of rotating machinery;
in the areas of electronics and communications, the amplitude
is more often used
• We will assume that the amplitude is specified unless the term
“rms” is explicitly used, or we are otherwise instructed
12
12. 1.3 Use of RMS Values to Compute
Average Power (3/3)
• In the sinusoidal steady state, phasor voltages and currents may
be given either as effective values or as amplitudes; the two
expressions differ only by a factor of 2
• The voltage 50/30◦ V is expressed in terms of an amplitude
• As an rms voltage, we should describe the same voltage as
35.4/30◦ Vrms
13
13. 1.4 Effective Value with Multiple-
Frequency Circuits (1/2)
• In order to determine the effective value of a periodic or non-
periodic waveform which is composed of the sum of a number
of sinusoids of different frequencies, we may use the
appropriate average-power relationship, rewritten in terms of
the effective values of the several components:
• From this we see that the effective value of a current which is
composed of any number of sinusoidal currents of different
frequencies can be expressed as
14
14. 1.4 Effective Value with Multiple-
Frequency Circuits (2/2)
• These results indicate that if a sinusoidal current of 5 A rms at 60
Hz flows through a 2 Ω resistor, an average power of 52(2) = 50 W
is absorbed by the resistor; if a second current, say 3 Arms at 120
Hz, for example, is also present, the absorbed power is 32(2) + 50 =
68 W
• Using Eq. for 𝑰𝒆𝒇𝒇 instead, we find that the effective value of the
sum of the 60 and 120 Hz currents is 5.831A
• Thus, P = 5.8312(2) = 68 W as before
• However, if the second current is also at 60 Hz, the effective value
of the sum of the two 60 Hz currents may have any value between 2
and 8A
• Thus, the absorbed power may have any value between 8 W and
128 W, depending on the relative phase of the two current
components
15
15. 2. APPARENT POWER AND POWER
FACTOR (1/5)
16
• Historically, the introduction of the concepts of apparent power and
power factor can be traced to the electric power industry, where large
amounts of electric energy must be transferred from one point to
another
• The efficiency with which this transfer is effected is related directly
to the cost of the electric energy, which is eventually paid by the
consumer
• Customers who provide loads which result in a relatively poor
transmission efficiency must pay a greater price for each
kilowatthour (kWh) of electric energy they actually receive and use
• In a similar way, customers who require a costlier investment in
transmission and distribution equipment by the power company will
also pay more for each kilowatthour unless the company is
benevolent and enjoys losing money
16. 2. APPARENT POWER AND POWER
FACTOR (2/5)
17
• Let us first define apparent power and power factor and then
show briefly how these terms are related to practical economic
situations
• We assume that the sinusoidal voltage:
is applied to a network and the resultant sinusoidal current is
• The phase angle by which the voltage leads the current is therefore
(θ − φ)
17. 2. APPARENT POWER AND POWER
FACTOR (2/5)
18
• The average power delivered to the network, assuming a passive
sign convention at its input terminals, may be expressed either in
terms of the maximum values:
or in terms of effective values:
• If our applied voltage and current responses had been dc quantities,
the average power delivered to the network would have been given
simply by the product of the voltage and the current
• Applying this dc technique to the sinusoidal problem, we should
obtain a value for the absorbed power which is “apparently’’given
by the familiar product Veff Ieff
18. 2. APPARENT POWER AND POWER
FACTOR (3/5)
• However, this product of the effective values of the voltage and
current is not the average power; we define it as the apparent power
• Dimensionally, apparent power must be measured in the same units
as real power, since cos(θ − φ) is dimensionless; but in order to
avoid confusion, the term volt-amperes, or VA, is applied to the
apparent power
• Since cos(θ − φ) cannot have a magnitude greater than unity, the
magnitude of the real power can never be greater than the magnitude
of the apparent power
• The ratio of the real or average power to the apparent power is called
the power factor, symbolized by PF
• Hence,
19
19. 2. APPARENT POWER AND POWER
FACTOR (4/5)
• For a purely resistive load, the voltage and current are in phase,
(θ − φ) is zero, and the PF is unity.
• In other words, the apparent power and the average power are
equal
• Unity PF, however, may also be achieved for loads that contain
both inductance and capacitance if the element values and the
operating frequency are carefully selected to provide an input
impedance having a zero phase angle
• A purely reactive load, that is, one containing no resistance, will
cause a phase difference between the voltage and current of
either plus or minus 90°, and the PF is therefore zero
20
20. 2. APPARENT POWER AND POWER
FACTOR (5/5)
• A PF of 0.5, for example, indicates a load having an input
impedance with a phase angle of either 60° or −60◦; the former
describes an inductive load, since the voltage leads the current
by 60°, while the latter refers to a capacitive load
• The ambiguity in the exact nature of the load is resolved by
referring to a leading PF or a lagging PF, the terms leading or
lagging referring to the phase of the current with respect to the
voltage
• Thus, an inductive load will have a lagging PF because the
current lags the voltage and a capacitive load a leading PF
because current lead the voltage
21
21. Example 11.8
The effective voltage 60 Vrms appears across the
combined load of 2 – j + 1 + j5 = 3 + j4 Ω
3+𝑗
4
𝑜
Hence 𝑰 = 60∠0
= 12∠ −53.13𝑜 Arms
The average power supplied to the top load is;
2
𝑷1 = 𝑰𝒆𝒇𝒇
𝟐
𝑅1 = 12 2 = 288 W, and
22
2
The average power supplied to the right load is; 𝑷2 = 𝑰2
𝑅2 = 12 1 =144 W
The total average power absorbed by the loads is; P = 288 + 144 = 432 W
𝑽𝑒𝑓𝑓𝑰𝑒𝑓𝑓 =
The apparent power supplied by the source is; 60 12 = 720 V
A
𝑷
𝑽𝑒𝑓𝑓𝑰𝑒𝑓𝑓 720
The power factor of the combined loads is: 𝐏𝐅 = = 432
= 0.6 lagging
since the current lags the source voltage by -53.13o
The average power supplied by the source is: P= (60)(12)cos(0 + 53.13o) = 432 W
22. 3. COMPLEX POWER (1/5)
23
• As we saw in Chap. 10, “complex” numbers do not actually
“complicate” analysis
• By allowing us to carry two pieces of information together through a
series of calculations via the “real” and “imaginary” components, they
often greatly simplify what might otherwise be tedious calculations
• This is particularly true with power, since we have resistive as well as
inductive and capacitive elements in a general load
• Now, we will define complex power to allow us to calculate the various
contributions to the total power in a clean, efficient fashion
• The magnitude of the complex power is simply the apparent power
• The real part is the average power and—as we are about to see—the
imaginary part is a new quantity, termed the reactive power, which
describes the rate of energy transfer into and out of reactive load
components (e.g., inductors and capacitors)
27. 4. POWER TRIANGLE (1/3)
28
• A commonly employed graphical
representation of complex power is
known as the power triangle, and is
illustrated in Fig. 11.15
• The diagram shows that only two of
the three power quantities are
required, as the third may be obtained
by trigonometric relationships
• If the power triangle lies in the first quadrant (θ −φ > 0); the
power factor is lagging (corresponding to an inductive load);
and if the power triangle lies in the fourth quadrant (θ −φ < 0),
the power factor is leading (corresponding to a capacitive load).
• A great deal of qualitative information concerning our load is
therefore available at a glance
28. 4. POWER TRIANGLE (2/3)
29
• Another interpretation
of reactive power may
be seen by constructing
a phasor diagram
containing Veff and Ieff
as shown in Fig. 11.16.
• If the phasor current is resolved into two components, one in phase
with the voltage, having a magnitude Ieff cos(θ − φ), and one 90° out
of phase with the voltage, with magnitude equal to Ieff sin |θ − φ|,
then it is clear that the real power is given by the product of the
magnitude of the voltage phasor and the component of the phasor
current which is in phase with the voltage
29. 3. POWER MEASUREMENT
30
r
• Strictly speaking, a wattmeter measures average real powe
P drawn by a load, and a varmeter reads the average
reactive power Q drawn by a load
• However, it is common to find both features in the same
meter, which is often also capable of measuring apparent
power and power factor (Fig. 11.17)
• It is easy to show that the complex power delivered to
several interconnected loads is the sum of the complex
powers delivered to each of the individual loads, no matter
how the loads are interconnected
• For example, consider the two loads shown connected in
parallel in Fig. 11.20. If rms values are assumed, the
complex power drawn by the combined load is
30. 5. COMPLEX POWER CALCULATION
31
A fast food restaurant has two loads connected as shown in figure. Load-1 is
drawing 24 kW at PF = 0.8 leading and Load-2 is drawing 20 kVAR at PF = 0.7
lagging. Find the totalAverage, Reactive and Complex Power
Solution: P1 = 24 kW ;
PF1 = 0.8 leading, hence the load is capacitive
1
|𝑆 | =
𝑃1
𝑃𝐹1 0.8 1
= 24
= 30 kVA ; θ = cos-1 (0.8) = -36.86o
Q1= 30sin(-36.86o) = -18 kVAR ; S1 = P1 - jQ1 = 24 - j18 = 30 /-36.86o kVA
Q2 = 20 kVAR ; PF2 = 0.7 lagging, hence the load is inductive
θ2 = cos-1 (0.7) = 45.57o ; sin(θ2) = sin(45.57o) = 0.714
2
|𝑆 | =
𝑄2 20
sin(θ2) 0.714 2 2 2
= = 28 kVA ; P = |S |xPF = 28x0.7= 19.6 kW
S2 = P2 + jQ2 = 19.6 + j20 = 28 /45.57o kVA
S = S1 + S2 = 24 - j18 + 19.6 + j20 = 43.6 + j2 = 43.65 /2.626o kVA
PT = 43.6 W ; QT = 2 kVAR ; ST = 43.65 /2.626o kVA
31. 5. POWER FACTOR CORRECTION (1/5)
32
• When electric power is being supplied to large industrial consumers by a
power company, the company will frequently include a PF clause in its
rate schedules
• Under this clause, an additional charge is made to the consumer
whenever the PF drops below a certain specified value, usually about
0.85 lagging
• Very little industrial power is consumed at leading PFs, because of the
nature of typical industrial loads
• There are several reasons that force the power company to make this
additional charge for low PFs
• In the first place, it is evident that larger current-carrying capacity must
be built into its generators in order to provide the larger currents that go
with lower-PF operation at constant power and constant voltage
• Another reason is found in the increased losses in its transmission and
distribution system
32. 5. POWER FACTOR CORRECTION (2/5)
33
• In an effort to recoup losses and encourage its customers to operate at
high PF, a certain utility charges a penalty of $0.22/kVAR for each kVAR
above a benchmark value computed as 0.62 times the average power
demand:
S = P + jQ = P + j0.62P = P(1 + j0.62) = P(1.177 /31.8o)
• This benchmark targets a PF of 0.85 lagging, as cos 31.8° = 0.85 and Q is
positive; this is represented graphically in Fig. 11.18
• Customers with a PF smaller than the
benchmark value are subject to financial
penalties
• The reactive power requirement is
commonly adjusted through the installation
of compensation capacitors placed in
parallel with the load (typically at the
substation outside the customer’s facility)
33. 5. POWER FACTOR CORRECTION (3/5)
34
• The value of the required capacitance can
be shown to be:
𝐶 = 𝑃(𝑡𝑎𝑛𝜃𝑜𝑙𝑑−𝑡𝑎𝑛𝜃𝑛𝑒𝑤)
𝜔𝑉2
𝑟𝑚𝑠
where ω is the frequency, θold is the
present PF angle and and θnew is the
target PF angle
• For convenience, however, compensation
capacitor banks are manufactured in
specific increments rated in units of
kVAR capacity
• An example of such an installation is
shown in Fig. 11.19
34. 5. POWER FACTOR CORRECTION (4/5)
Example:
35
• Aparticular industrial machine plant has a monthly peak demand of 5000 kW
and a monthly reactive requirement of 6000 kVAR
• Using the rate schedule above, what is the annual cost to this utility customer
associated with PF penalties?
• If compensation is available through the utility company at a cost of $2390 per
1000 kVAR increment and $3130 per 2000 kVAR increment, what is the most
cost-effective solution for the customer?
• The PF of the installation is the angle of the complex power S, which in this
case is 5000 + j6000 kVA
• Thus, the angle is tan (6000/5000) = 50.19o and the PF is 0.64 lagging
• The benchmark reactive power value, computed as 0.62
times the peak demand, is 0.62(5000) = 3100 kVAR
• So, the plant is drawing 6000 − 3100 = 2900 kVAR
more reactive power than the utility company is willing
to allow without penalty
35. 5. POWER FACTOR CORRECTION (5/5)
Example:
36
• This represents an annual assessment of 12(2900)(0.22) = $7656 in addition to
regular electricity costs
• If the customer chooses to have a single 1000 kVAR increment installed (at a cost
of $2390), the excess reactive power draw is reduced to:
2900 − 1000 = 1900 kVAR,
• So that the annual penalty is now 12(1900)(0.22) = $5016
• The total cost this year is then $5016 + $2390 = $7406, for a savings of $250
• If the customer chooses to have a single 2000 kVAR increment installed (at a cost
of $3130), the excess reactive power draw is reduced to:
2900 − 2000 = 900 kVAR,
• So that the annual penalty is now: 12(900)(0.22) = $2376
• The total cost this year is then $2376 + $3130 = $5506, for a first-year savings of
$2150
• If, however, the customer goes overboard and installs 3000 kVAR of
compensation capacitors so that no penalty is assessed, it will actually cost $14
more in the first year than if only 2000 kVAR were installed
38. 39
• In case we want to raise the PF to 1, then
the reactive component of the complex
power should be reduced to zero
• Hence, S2 = - j37.5 kVA and the phase
angle:
θnew = cos-1 (1) = 0o ,
θold = cos-1 (0.8) = 36.86o
• The value of capacitor for corrective load
Z2 may now be found using the expression;
𝐶 = 𝑃(𝑡𝑎𝑛𝜃𝑜𝑙𝑑−𝑡𝑎𝑛𝜃𝑛𝑒𝑤)
𝜔𝑉2
𝑟𝑚𝑠
50 × 103 × (tan(36.86𝑜) − tan(0𝑜))
𝐶 =
120 × 𝜋 × (230)2
3
𝐶 = 50×10 ×(0.75)
19942830.165
𝐶 = 1.88 mF
= 0.00188 F or
39. 4. Summary (1/2)
For convenience, key points of the chapter are summarized below:
• The instantaneous power absorbed by an element is given by the
expression p(t) = v(t) i (t)
• The average power delivered to an impedance by a sinusoidal source is
0.5 Vm*Im cos(θ − φ), where θ = the voltage phase angle and φ = the
phase angle of the current.
• Only the resistive component of a load draws nonzero average power.
The average power delivered to the reactive component of a load is
zero.
• Maximum average power transfer occurs when the condition ZL = Zth
∗
is satisfied.
• When multiple sources are present, each operating at a different
frequency, the individual contributions to average power may be
summed. This is not true for sources operating at the same frequency.
08/10/2013 40
40. 4. Summary (2/2)
• The effective or rms value of a sinusoidal waveform is obtained by
dividing its amplitude by √2.
• The power factor (PF) of a load is the ratio of its average dissipated
power to the apparent power.
• A purely resistive load will have a unity power factor. A purely
reactive load will have a zero power factor.
eff eff
• Complex power is defined as S = P + jQ, or S = V I ∗. It is measured
in units of volt-amperes (VA). (Self Study)
• Reactive power Q is the imaginary component of the complex power,
and is a measure of the energy flow rate into or out of the reactive
components of a load. Its unit is the volt-ampere-reactive (VAR).
• Capacitors are commonly used to improve the PF of industrial loads
to minimize the reactive power required from the utility company.
08/10/2013 41
