AC POWER ANALYSISAC POWER ANALYSIS
Instantaneous & Average Power Max. Average Power Transfer RMS Value & Apparent Power Complex Power & Power Factor
Correction
Instantaneous Power
)t(i)t(v)t(p
Instantaneous power (in watts) : the power at any instant of time
)tcos(V)t(v vm Where:
)tcos(I)t(i im
)t2cos()cos(2
IV)t(p iviv
mm
Instantaneous Power
The instantaneous power p(t) entering a circuit
Average Power
)cos(2
IVP iv
mm
Average Power (in watts) : the average of the instantaneous power over one period
iv cos2
IVP mm Where:
Average Power cosIVP effeff
2
IIor I m
effrms 2
or meffrms
VVV
Example 1
V)45t377cos(120)t(v 0 A)10t377cos(10)t(i 0
Find the instantaneous power and average power absorbed by the passive linear network.
Given that
Reference : Alexander, Sadiku Chapter 11 - page 461
Exercise 1
Calculate the instantaneous power and average power if
V)20t10cos(80)t(v 0 A)60t10sin(15)t(i 0
W)10t20cos(6007.385)t(p 0
W7.385P
Circuit Elements
In purely resistive circuit, v and i are in phase. θv = θi. Therefore θ = 0.
RIR
VIV
2
IV0cos
2
IVP 2
eff
2eff
effeffmm0mm
(a) Resistors:
The average power is only dissipated in a purely resistive circuit. For a purely inductive and capacitive, the average power is zero.
Circuit Elements
In purely inductive circuit, v leads by 90o, therefore θ = 90o
0)90cos(2
omm IVP
(b) Inductors:
(c) Capacitors:
In purely capacitive circuit, I leads by 90o, therefore θ = - 90o
0)90cos(2
omm IVP
Example 2
Find the average power supplies by the source and the average power absorbed by the resistor.
Reference : Alexander, Sadiku Chapter 11 - page 462
Exercise 2
Calculate the average power absorbed by the resistor and inductor. Find the average power supplies by the voltage source.
W6.9P
W0P
W6.9P
L
R
Maximum Average Power Transfer
Finding the maximum average power transfer:
a) circuit with a load b) the Thevenin equivalent
Maximum Average Power Transfer
For maximum average power transfer, the load impedance ZL must be equal to the complex conjugate of the Thevenin impedance ZTh
LLL
ThThTh
jXRZ
jXRZ
In rectangular form, Thevenin impedance and Load impedance:
Th*
ThThLLL ZjXRjXRZ
Th*
L ZZ
Maximum Average Power Transfer
Th
2
Thmax R8
VP
In a situation in which the load is purely real or purely resistive load (XL=0), the load impedance (or resistance RL) is equal to the magnitude of the Thevenin impedance.
Th2Th
2ThL ZXRR
Example 3
Determine the load impedance ZL that maximizes the average power drawn from the circuit of figure below. Calculate the maximum average power.
Reference : Alexander, Sadiku Chapter 11 - page 466
Exercise 3
Find the load impedance ZL that absorbs the maximum average power for the circuit of figure below. Calculate the maximum average power.
7317.0j415.3ZZ Th*
LW429.1Pmax
Exercise 4
In Figure below, the resistor RL is adjusted until it absorbs the maximum average power. Calculate RL and the maximum average power absorbed by it.
W883.9Pmax 30R L
Effective or RMS Value
The effective value of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current.
Finding the effective current:
a) ac circuit b) dc circuit
Effective or RMS Value
dtxT
1X
T
0
2rms
The effective value of a periodic signals is its root mean square (rms) value.
For any perodic function x(t) in general, the rms value is given by:
Effective or RMS Value
dttcosIT
1I
T
0
22mrms
2
Idt)t2cos1(
2
1
T
II m
T
0
2m
rms
2
VV m
rms
Effective or RMS Value
The average power absorbed by resistor R can be written as:
RIR
VP 2
rms
2rms
)cos(IVP ivrmsrms
The average power can be written as:
Example 4
i(t)
t0
10
-10
2 4 6 8 10
Determine the rms value of the current waveform in figure below. If the current is passed through a 2Ω resistor, find the average power absorbed by the resistor.
A165.8Irms
W3.133P
Reference : Alexander, Sadiku Chapter 11 - page 469
Exercise 5
Find the rms value of the full wave rectified sine wave in figure below. Calculate the average power dissipated in a 6Ω resistor.
V657.5Vrms W334.5P
v(t)
t0
8
2 3
Apparent Power
The apparent power (in VA) is the product of the rms value of voltage and current.
)cos(IVP ivrmsrms
)cos(SP iv
rmsrmsIVS
S is known as the apparent power.
Power Factor
The power factor is the cosine of the phase difference between voltage and current. It is also the cosine of the angle of the load impedance.
)cos(pf iv
iv
Power Factor :
is Power Factor Angle pf is lagging if the current lags voltage (inductive load)
pf is leading if the current leads voltage (capacitive load)For purely resistive circuit, pf=1. With inductors
and capacitors in the circuit, pf may reduced to less than 1.
where
Example 5
A series connected load draw
a current
when the applied voltage is
Find the apparent power and the power factor of the load. Determine the element values that form the series connected load.
A)10t100cos(4)t(i 0
V)20t100cos(120)t(v 0
V240S )leading(866.0pf uF2.212C
Reference : Alexander, Sadiku Chapter 11 - page 472
Exercise 6
Calculate the power factor of the circuit below as seen by the source. What is the average power supplies by the source?
)lagging(936.0pf W118P
Complex Power
Complex power (in VA) is the product of the rms voltage phasor and the complex conjugate of the rms current phasor. As a complex quantity, its real part is real power P and its imaginary part is reactive power Q.
ivrmsrms* IVVI
2
1jQPS
*
2rms2
rms Z
VZIS
Complex power :
VA
*
rmsrmsIV
Complex Power
22
rmsrms QPIVSS Apparent power :
Real power :
)cos(S)SRe(P iv WReactive power :
)sin(S)SIm(Q iv VAR
VA
Q = 0 for resistive loads (unity power factor)
Q < 0 for capacitive loads (leading power factor)
Q > 0 for inductive loads (lagging power factor)
Complex Power
Q
P
SX
R
IZI
Power triangle
Impedance triangle
S
Pcospf
Complex Power
Power Triangle
Example 6
The voltage across a load is
and the current through the element in the direction
of the voltage drop is
Find
a)the complex and apparent powers
b)the real and reactive powers
c) the power factor and the load impedance
V)10tcos(60)t(v 0
A)50tcos(5.1)t(i 0
Reference : Alexander, Sadiku Chapter 11 - page 475
Example 7
A load Z draws 12kVA at a power factor of 0.856
lagging from a 120 Vrms sinusoidal source.
Calculate:
a)the average and reactive powers delivered to the load
b)the peak current
c) the load impedanceReference : Alexander, Sadiku Chapter 11 - page 476
Exercise 7
A sinusoidal source supplies 10kVAR reactive
power to load
Determine:
a)the power factor
b)the apparent power delivered to the load
c) the peak voltage
075250Z
Power Factor CorrectionThe process of increasing the power factor without altering the voltage or current to the original load is known as power factor correction.
Most loads are inductive. A load power factor is improved (to make closer to unity, pf=1) by installing a capacitor in parallel with the load.
a) Original inductive load b) inductive load with improved
power factor
Power Factor Correction
Phasor diagram showing the effect of adding a capacitor in parallel with the inductive load
Power Factor Correction
Power triangle illustrating power factor correction
Power Factor Correction
2
rms
21
2
rms
C
V
)tan(tanP
V
QC
)tan(tanPQQQ 2121C
2
rms
C
2
rmsC CV
X
VQ
Value of required shunt capacitance :
Example 8
When connected to a 120 V (rms), 60Hz power line, a load absorbs 4kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.
Reference : Alexander, Sadiku Chapter 11 - page 482
Exercise 8
Find the value of parallel capacitance needed to correct a load of 140kVAR at 0.85 lagging pf to unity pf. Assume that the load is supplied by a 110V (rms), 60Hz line.
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