Average Power
Average Power,
For Resistive Load
Average Power,
For Inductive & Capacitive Load
Average Power < Pmax
For large Inductive & Capacitive Loads
Average Power = 0
)cos(
2
IV
Piv
mm
2
IV
Pmm
max
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Average Power-Electrical Circuit Analysis-Lecture Slides, Slides of Electrical Circuit Analysis
This lecture is part of lecture series on Electrical Circuit Analysis course. It was delivered by Prof. Mursleen Sayed at Bengal Engineering and Science University. It includes: Average, Power, Resistive, Load, Inductive, Capacitive, Rms, Value, Current, Voltage, Energy
Typology: Slides
2011/2012
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Average Power
Average Power,
For Resistive Load^
Average Power,
For Inductive & Capacitive Load^
Average Power < P
max
For large Inductive & Capacitive Loads^
Average Power = 0
cos( I 2 V P^
i
v
m m^
I 2
V
P^
m m
max
The RMS Value
Average Power delivered to a resistor^
RMS value is the effective value of the sinusoidalvoltage or current^
Energy delivered by a sinusoidal source of 100 V
rms
in T
sec is the same as the energy delivered by a 100 V dcsource in T sec.
V^ R 2
dt)
t ( cos
V TR
dt
R
t (
cos V
(^1) T P
(^2) m
T t t
v
2
(^2) m
T t t
v
2
(^2) m
0 0
0 0
^
R
I
V R
P
2 rms
(^2) rms
Complex & Apparent Power
Complex sum of real and reactive power^
S = P + jQ ^
units of complex power are volt-amps (VA) ^
Power factor angle
is
θ
= tan
-1^ (Q/P)
Apparent Power is the magnitude of Complex Power^
Apparent power is measured in volt-amps (VA)
2
2
Q
P
S^
Complex & Apparent Power
Apparent power requirements of a device designedto convert electric energy to a non-electric form ismore important than the average power requirement
Average power represents the useful output of theenergy converting device
Apparent power represents the volt-ampere capacityrequired to supply the average power
Complex & Apparent Power
The power factor of inductive loads iscorrected by adding a capacitor to the deviceitself or connecting capacitors across the linefeeding the load.
Obviously WAPDA cannot get to the devicesa easily as they can get to the line feedingthe area.
Example ^
An electric load operates at 240 V
rms
^
The load absorbs an average power of 8 kW at alagging power factor of 0.
^
Complex power of the load?
^
Impedance of the load?
^
P = 8 kW
cos
^
P = |S|cos
^
|S| = P/cos
= 8/0.8 = 10 kVA
^
Q = |S|sin
= 10sin(36.87) = 6 kVAR
^
S = 8 + j6 kVA
Power Calculations
S = P + jQ
V
eff
Ieff
*^
V
m
I^ m
θ
θ sin( I 2 Vj
) θ
θ cos( I 2 V
i v m m i v m
m^
^
θ
θ sin(j ) θ
θ cos( I 2 V
i v
i v
m m^
)
(j
m m
i v
e
I 2
V
I
V
i
v
eff eff
I 2
V
i
v
m m
(j
eff eff
i v
e
I
V
) (j
eff ) (j
eff
i
v^
e
I
e
V
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Alternate forms for ComplexPower ^
V
eff
Z
Ieff
^
S =
V
eff
I eff
S = Z
Ieff
I eff
S = |
Ieff
Z = |
Ieff
(R + jX)
S = |
Ieff
R + j|
Ieff
X = P + jQ
^
P = |
Ieff
R = I
m
2 R/
^
Q = |
Ieff
X = I
(^2) m
X/
^
S =
V
eff
( V
eff
/ Z)*= |
V
eff
/Z* = P + jQ
^
P = |
V
eff
/R
^
Q = |
V
eff