1
16.202:AC Power Analysis
• Instantaneous Power p(t) = v(t) × i(t) Watts
– Rate at which energy is absorbed by an element
– v(t) : Voltage across the element
– i(t) : Current through the element
– Recall Convention for Power Absorbed:
+
i(t)
v(t)
p(t)=v(t)i(t)
-
Figure 1: Power Absorbed
c
Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
2
• Consider:
v(t) = Vmcos(ωt + θv )
i(t) = Imcos(ωt + θi)
p(t) = VmImcos(ωt + θv )cos(ωt + θi)
1
= VmIm [cos(θv − θi) + cos(2ωt + θv + θi)]
2
= A + Bcos(2ωt + θv + θi)
• Note that p(t) is a sinusoid with frequency 2ω
• DC offset A =
Vm Im
2 cos(θv
− θi )
• p(t) > 0 :Power Absorbed ; p(t) < 0: Power Delivered
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
3
Average Power P
• Average Power : Constant : P =
• where T = 1/f =
• Therefore P =
1 T
T 0
p(t)dt
2π
ω
1
2 Vm Im cos(θp )
• θp = θv − θi or = θi − θv
• In terms of phasors:
v(t) : V = Vm θv
i(t) : I = Im θi
VI∗ = VmIm (θv − θi) = VmIm[cos(θv − θi) + jsin(θv − θi)
1
P = Re[VI∗]
2
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
4
• Resistor : θv = θi
P =
Vm Im
2
=
Vm2
2R
=
2R
Im
2
: Always Positive.
• Inductor or Capacitor :
θv − θi = ± 90o cos(±90) = 0
Therefore P = 0:
– Purely Reactive Elements; No Average Power Absorption
I = 10 30
P =?
20 −22
Figure 2: Example 1: Average Power
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
5
Maximum Average Power Transfer Theorem
I
Linear
Circuit
Load : ZL
+
- Vth
Zth
ZL
Figure 3: Thevenin equivalent
• Represent the circuit to the left of the load by its Thev.equiv.
• Load ZL represents any element that is absorbing the power
generated by the circuit
• Find the load ZL that will generate the maximum average
power transfer
c
Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
6
• Approach:
Write the expression for average power associated with ZL:P (ZL)
ZL = RL + jXL
∂P
Set ∂R
= 0: Solve for RL
L
Set
∂P
∂XL
= 0: Solve for XL
Vth
I =
ZL + ZT h
= Im θi
Im =
|VT h|
[(RL + RT h)2 + (XL + XT h)2]
2
RL
Im
|VT h|2RL/2
P =
=
2
(RL + RT h)2 + (XL + XT h)2
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
7
∂P
= 0 → XL = − XT h
∂XL
∂P
= 0 → RL = RT h
∂RL
• Therefore: Zl = RT h − XT h = ZT∗ h will generate the maximum average power transfer.
• Pmax =
IL2 RL
2
2
T h|
= |V
8R
Th
• If load is constrained to be purely real: ZL = RL and
XL = 0, solution above yields ZL = |ZT h| for maximum
average power transfer.
c
Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell