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LESSON 28
AC POWER
&
POWER FACTOR
Lesson Objectives
• Compute and define apparent, reactive, and
average power for capacitors, inductors, and
resistors.
• Compute and draw the power triangle for RC, RL,
and RLC circuits.
• Define and compute the power factor for RC, RL,
and RLC circuits.
• Summarize the basic steps to compute AC power
in all or part of a circuit.
AVERAGE POWER
AVERAGE POWER
• For DC power:
P  IV
• For AC avg power:
p  i(t )  v(t ) 
Vm I m
2
cos( )
AVERAGE POWER
p  i(t )  v(t ) 
where;
Vm I m
2
cos( )
  v i
• Avg power is independent of whether v leads i,
or i leads v.
AVERAGE POWER
• It can be further shown that
P  Veff I eff cos 
• since,
Vm
Im
Veff 
; I eff 
2
2
Average Power in RESISTOR
• Since ||=0o and cos (0o) =1
PR  Veff I eff cos0
2
eff
V
Vm I m
2
P
 Veff I eff 
 I eff R
2
R
Average Power in L and C
• PAV in a capacitor and inductor is 0, since;
|C|= |L|= 90o and cos (90o) =0.
PL / C  Veff I eff cos90  0
POWER FACTOR
POWER FACTOR
• The factor that has the significant control over the
delivered power level is the cos (), where
   
• No matter what level Ivand V are,
i if:
cos ()=0, >> the power delivered is zero. cos ()=1,
>> the power delivered is max.
POWER FACTOR
• Power Factor equation:
P
Fp  cos   
Veff I eff
• where,
  v i
Power Factor Leading or
Lagging?
•
•
•
•
Inductive circuits have lagging power factors.
Capacitive circuits have leading power factors.
Power factors follow the current.
Remember ELI and ICE
Ex.
• Find power factor if,
i  2 sin t  20 ; v  50 sin t  40
Fp  cos(  v  i ) 


cos 40   20   0.5leading
Leading because current is leading and ICE.
Equivalent Circuits
• Any circuit impedance with combinations of resistors, capacitors, and
inductors may be written in rectangular format as a phasor impedance.
ZT = R K jX
• In other words, any circuit may be modeled with just two components:
– R (real component)
– X (reactive component)
R1
VS
aZT = R1+(-jXC1)+R2||R3
C1
R2
R3
aZT = R123 – jXC1
VS
R123
• This will work for any circuit (RC/RL/RLC)
C1
True and Reactive Power
• Since we now have two types of components that oppose current in a
circuit, we must distinguish and define two types of power.
• True power – power dissipated in a resistor
2
V
P  VR I R  I R2 R  R
R
• Reactive power – power repeatedly stored and returned to a circuit in
either a capacitor or an inductor.
2
V
Q  VI  I L2 X  L
XL
or
2
V
Q  VI  I C2 X  C
XC
Power Triangle and Apparent Power
• The impedance triangle with R, X, and Z may be shown to
be similar to the power triangle with P, Q, and S,
respectively as components.
• Apparent power – A useful quantity combining the vector
sum of P and Q.
Z
XL

R
I2Z
I2XL
PSa

I2R

PtP
IMPEDANCE
• S = VI = I2 Z = V2/Z
POWER PHASORS
POWER TRIANGLE
P = S cos T
Q = S sin T
P
Qr
Names for Power units
• Since all three types of power use the same units
of volts, amps, and ohms, it might be helpful to
define names for each type of power
– P will have units of Watts (W)
– Q will have units of VAR’s (VAR) – (Volt-AmpsReactive)
– S will have units of VA’s (VA) – (Volt-Amps)
• Notice the units are the same. Only the names are
different.
POWER IN AC CIRCUITS
• The product of instantaneous voltage v(t) and
instantaneous current i(t) will give us instantaneous
power. p(t)=v(t) ·i(t)
• Power follows a sine curve and fluctuates at twice the
freq of v(t) or i(t). Which means we can’t use phasors
to calculate the frequency
• But we just did this in previous slides…
Calculating Power using Phasors
• Recall the power triangle is just a scaled up
version of the impedance triangle. Z is the same.
• Recall S is the vector sum of P and Q.
• In order to compare the three types of power at
any time, t, we will only use RMS values of V and
I.
Procedure
1. Convert VM and IM to VRMS and IRMS.
2. Use only magnitude |VRMS| and |IRMS|. Throw
away the phase shift angles for voltage and
current. Keep T.
3. S = VRMS IRMS
P = S cos T
Q= S sin |T|
S<T = P + jR
P = I2RMS R = V2RMS/R
Q = I2RMS X = V2RMS/X
S = P + jQ = S<T
Ex:
• Find Reactive Power, Q, and Real Power, P
I  553A _ 44
Z  6.5k  j 6.28k
Z  9.04k44
• Q = 1.92 mVAR
• P = 1.99 mW
R=
6.5 k
E(rms)
50

L
100
mH
Ex: Find P, Q and S; Draw the Power Triangle
(Note: E = VS)
ZT = 330-74.7S
IT = 45.4674.7mARMS
R1
VS
C1
R2
R3
S = I VS = I2 ZT
S = (45.46)2 x 330 = 682 mVA
Q = S sin T
Q = 682 mVA sin(74.7) = 658 mVAR
P = S cos T
P = 682 mVA cos(74.7) = 180 mW
P
T
Q
S
Power Triangle
POWER FACTOR
• True Power is the ‘usable’ power,
• The power factor tells us what fraction of the
Apparent Power is ‘useful’
P  S cos  T
P
 cos  T
S
• Power Factor (PF) = cos 2z
– Range: 0  PF  1
POWER FACTOR CORRECTION
• Recall that PF= P/S = cos zt
• The maximum PF of a circuit occurs when cos zt =
1
– zt =0 °
–  Zt has only a real component
• For a parallel RL circuit, the addition of a capacitor
in parallel can produce zt = 0 °
– PF increases
– Useful power is maximized.
Called “Power Factor Correction”
Complex Example
• Determine Is, S, P, Q, and PF. (Note: E = VS)
X L1
R3
5.03k
5.6k
VS
R1
R2
X L2
18V 0
1.2k
3.3k
4.02k


Z T  R1 X L1  R2 R3  X L2

Z T  682  j 232
Z T  72018.8
PSa  VRMS I RMS
IS 
VS
18V  0

Z T 72018.8
I S  25mA18.8
18V  25mA

 225mVA
2
Complex Example (cont)
Sa cos Z
Pt  P
Pt  225mVA cos18.8 OR
Pt  213mW
2
Pt  I RMS
Z TREAL
Pt  25mA.707  682
2
Pt  213mW
QPr  P
Sa sin  Z
PF  cos Z
QPr  (225mVA) sin 18.8
PF  cos( 18.8)
PF  .947
QPr  72.5mVAR
Review Quiz
• Name the three types of power.
• Q has units of … ?
P has units of … ?
S has units of … ?
• What are the first two steps in calculating AC
power?
• Power factor is … ?
• T/F: Power factor can never be greater than one
or less than zero.