EE 204
Lecture 24
Instantaneous, Average and Apparent Powers &
Power Factor
Average Value of a Periodic Function of Time:
The average value of the periodic function f (t ) is given by:
to +T
f avg =
∫
f (t )dt
to
T
=
net area under one period
period
Where to = an arbitrary time instant and T = period of f (t )
Figure 1
Average Value of the Sinusoidal Function:
The general sinusoidal function f (t ) = A cos(ωt + θ ) is periodic
∴ f (t ) has an average value
The positive and negative areas under one period T cancel each other perfectly
The net area under one period of f (t ) is zero
∴ f avg =
net area under one period 0
= =0
T
period
∴ the average value of a general sinusoidal function of time is zero
f(t)
t
Figure 2
Instantaneous Power in Sinusoidal Circuits:
The current and voltage in any element of a sinusoidal circuit can be expressed as:
i(t ) = I cos(ωt + θi ) [ A] & v(t ) = V cos(ωt + θ v ) [V ]
The instantaneous power absorbed by the element is:
p (t ) = i (t )v(t ) = IV cos(ωt + θi ) cos(ωt + θ v ) [W ]
[Note: if i (t ) enters from the (-) side of v(t ) ⇒
p (t ) = −i (t )v(t ) ]
p (t ) = IV cos(ωt + θi ) cos(ωt + θ v )
is called the instantaneous power, because it changes from one time instant to another
+
i(t)
v(t)
Arbitrary
Element
Figure 3
Average Power in Sinusoidal Circuits:
The instantaneous power p(t ) = IV cos(ωt + θ v ) cos(ωt + θi ) is actually a periodic function of time.
Using the trigonometric identity:
cos( A) cos( B ) =
1
1
cos( A − B ) + cos( A + B)
2
2
We can express p (t ) = IV cos(ωt + θ v ) cos(ωt + θi ) as:
p (t ) =
1
1
IV cos(ωt + θ v − ωt − θi ) + IV cos(ωt + θ v + ωt + θ i )
2
2
p (t ) =
1
1
IV cos(θ v − θi ) + IV cos(2ωt + θ v + θi )
2
2
p (t ) =
1
1
IV cos(θ ) + IV cos(2ωt + θ v + θi )
2
2
Where θ ≡ θ v − θi (the phase difference between voltage and current)
The average value of p(t ) can then be obtained as follows:
1
1
P = pavg = avg[ IV cos(θ ) + IV cos(2ωt + θ v + θi )]
2
2
1
1
P = avg[ IV cos(θ )] + avg[ IV cos(2ωt + θ v + θi )]
2
2
1
avg[ IV cos(2ωt + θ v + θi )] = 0 (why?)
2
1
1
1
avg[ IV cos(θ )] = IV cos(θ ) (because IV cos(θ ) is a constant, independent of time)
2
2
2
∴ P=
1
IV cos(θ ) = average power absorbed by an element in a sinusoidal circuit
2
To calculate P =
1
IV cos(θ ) we can use either:
2
1- i (t ) = I cos(ωt + θi )
2- I = I θi
&
& v(t ) = V cos(ωt + θ v )
V = V θv
From which we can obtain I , V and θ = θ v − θi needed to calculate P =
+
i(t)
v(t)
Arbitrary
Element
+
I
V
Arbitrary
Element
Figure 4
Average Power Absorbed by a Resistor, Inductor and Capacitor:
The voltage and current in the resistor are in phase ⇒ θ = θ v − θi = 0
The average power absorbed by R ⇒ PR =
1
1
IV cos(0) = IV
2
2
Other forms for PR are:
PR =
1
1
1
IV = I ( IR) = I 2 R
2
2
2
PR =
1
1V
1V2
IV =
V=
2
2R
2 R
V
I
R
Figure 5
For the inductor ⇒ θ = 90o
⇒
PL =
1
IV cos(90o ) = 0
2
For the capacitor ⇒ θ = −90o
⇒
Pc =
1
IV cos(−90o ) = 0
2
1
IV cos(θ )
2
+
I
V
jwL = wL 90o
+
I
1
=
jwC
V
1
wC
−90o
Figure 6
Summary of Average Power Expressions:
For an arbitrary load ⇒
For the resistor ⇒ PR =
P=
1
IV cos(θ )
2
1
1
1V2
IV = I 2 R =
2
2
2 R
For the inductor or the capacitor ⇒ PL = Pc = 0
This means that the average power is absorbed only by the resistive part of the load. The reactive
part of the load does not absorb any average power.
Example 1:
Calculate the average power absorbed by the load using two methods
30
j10
V S = 100 70o
[V]
-j25
Figure 7
Solution:
Method 1:
30 Ω & j10 Ω & − j 25 Ω are in series
∴ Z eq = 30 + j10 + (− j 25) = 30 − j15 = 33.541 −26.565o Ω
The current through the load:
I =
Vs
100 70o
=
= 2.981 96.565o A
o
Z eq 33.541 −26.565
For the load impedance Z eq , the voltage and current are:
Vs = 100 70o V
I = 2.981 96.565o A
θ v = 70o & θi = 96.565o ⇒ θ = θ v − θi = 70 − 96.565 = −26.565o
∴ Pload =
1
1
IV cos θ = (2.981)(100) cos(−26.565o ) = 133.3W
2
2
Method 2:
PR =
1 2
1
I R = (2.981) 2 × 30 = 133.3 W
2
2
PL = Pc = 0
Pload = PR + PL + Pc = 133.3 + 0 + 0 = 133.3 W
Which is the same as the previous answer
In this example the average power delivered by the source must be 133.3 W
I
VS
V S =100 70o
IL
V S =100 70o
VS
Z eq = 33.54 −26.57o Ω
Figure 8
The Apparent Power and the Power Factor:
The average power absorbed by a load is given by:
P=
1
IV cos θ
2
The term
1
IV is called the Apparent Power, with the unit of [ Volt × Ampere ] or [ VA ]
2
The Power Factor (p.f.) is defined as:
Power Factor ≡
Average Power 0.5IV cos θ
=
= cos θ
Apparent Power
0.5 IV
∴ p.f. = cos θ
The angle θ in the above equation is called the power factor angle
The power factor angle = the phase difference between V and I [ because θ = θ v − θi ]
V
I
Figure 9
Leading and Lagging Power Factor:
For an arbitrary load:
1) When I leads V ⇒ we have a leading power factor
2) When I lags V ⇒ we have a lagging power factor
V
I
Figure 9
In other words, a leading power factor implies:
1) I leads V
2) θi > θ v [because I leads V ]
3) θ < 0 [the power factor angle is negative, because θ = θ v − θi and θi > θ v ]
4) the impedance is capacitive [for a capacitive impedance, θ < 0 ]
R
I
V
jX C
XC < 0
Figure 10
A lagging power factor implies:
1) I lags V
2) θi < θ v [because I lags V ]
3) θ > 0 [the power factor angle is positive, because θ = θ v − θi and θi < θ v ]
4) the impedance is inductive [for an inductive impedance, θ > 0 ]
V
R
jX L
XL > 0
I
Figure 11
When I and V are in phase ⇒ θ v = θi
⇒ θ = θ v − θi = 0 ⇒ p.f. = cos(0) = 1
In this case we a have a unity power factor
Thus, a unity power factor implies:
1) I and V are in phase
2) θ = 0
3) the impedance is purely resistive
V
I
R
Figure 12
Relation between the Power Factor Angle and the Impedance Angle:
For any load impedance Z :
the power factor angle θ = θ v − θi
Since Z =
V V θv V
=
= θ v − θi = Z θ
I
I θi
I
The impedance angle θ = θ v − θi
∴ p.f. angle = impedance angle = θ
V
I
Z
Figure 13
Example 2:
Using the circuit of the previous example, calculate:
a) The apparent power absorbed by the load
b) The p.f. of the load
c) Does the load have a leading or a lagging power factor?
30
j10
V S = 100 70o
[V]
-j25
Figure 7
Solution:
a) From the previous example:
The current through the load is I = 2.981 96.565o A
The voltage across the load is Vs = 100 70o V
Apparent power absorbed by the load =
b) θ v = 70o & θi = 96.565o
c) Since θi > θ v
⇒
1
1
IV = IVs = 0.5 × 2.981× 100 = 149.05 VA
2
2
⇒ θ = θ v − θi = 70 − 96.565 = −26.565o
I leads Vs ⇒ leading p.f.
Alternatively, since θ = −26.565o < 0 ⇒ leading p.f.
I = 2.981 96.565o A
+
30
j10
V S = 100 70o
[V]
-j25
Figure 14
VS
Example 3:
The load shown absorbs electric power from the source. For this load, calculate:
a) the average power
b) the apparent power
c) the power factor
d) the power factor angle
e) the load impedance angle
1
F
50
Figure 15
Solution:
The circuit is shown in the phasor domain.
a) Z eq = j 20Ω + (30Ω − j10Ω) = j 20 +
Z eq = j 20 +
30 × (− j10)
30 − j10
300 −90o
= j 20 + 9.487 −71.565o
o
31.623 −18.435
Z eq = j 20 + (3 − j 9) = 3 + j11 = 11.402 74.745o Ω
I
V s = 15 0o V
Figure 16
I =
Vs
15 0o
=
= 1.316 −74.745o A
Z eq 11.402 74.745o
The voltage and current in the load are:
Vs = 15 0o V
& I = 1.316 −74.745o A
∴ θ = θ v − θi = 0 − (−74.745o ) = 74.745o
The average power is then given by:
P = 0.5IV cos θ = 0.5 × 1.316 × 15 × cos(74.745o ) = 2.597 W
b) The apparent power = 0.5IV = 0.5 × 1.316 × 15 = 9.87 VA
c) p.f. = cos θ = cos(74.745o ) = 0.263 (lagging p.f. , because θ = 74.745o > 0 )
d) The power factor angle is θ = 74.745o
e) The power factor angle = impedance angle = 74.745o
I
Vs =15 0o
VS
Z eq = 11.402 74.745o Ω
Figure 17