[5.1]
5 AC POWER ANALYSIS
5.1 Instantaneous and Average Power
The instantaneous power is the
power at any instant of time. It is the rate at
which an element absorbs energy.
Let the voltage and current at the terminals
of the circuit shown in Fig. 5.1 be
Figure 5.1
v(t) = V m cos(ωt + θ v )
i(t) = I m cos(ωt + θ i )
where
(5.1)
(5.2)
V m , θ v : are the peak amplitude and phase angle of voltage, respectively.
I m , θ i : are the peak amplitude and phase angle of current, respectively.
The instantaneous power absorbed by the circuit is
p(t) = v(t) i(t) = V m I m cos(ωt + θ v ) cos(ωt + θ i )
1
1
p(t) = V m I m cos(θ v − θ i ) + V m I m cos(2ωt + θ v + θ i )
2
2
(5.3)
(5.4)
When p(t) is positive, power is
absorbed by the circuit. When p(t) is
negative, power is absorbed by the
source; that is, power is transferred
from the circuit to the source. This is
possible because of the storage
elements (capacitors and inductors)
in the circuit.
The average power is the average of the instantaneous power over one period.
P=
Dr. Ahmed Abdul-Kadhim
1
V m I m cos(θ v − θ i )
2
(5.5)
[5.2]
Note that
Therefore,
1
1
P = Re[VI ∗ ] = Vm I m cos(θ v − θ i )
2
2
When θ v = θ i , (purely resistive circuit or resistive load R)
(5.6)
(5.7)
When θ v − θ i = ±90◦, (purely reactive circuit)
A resistive load (R) absorbs power at all times, while a reactive load (L or C)
absorbs zero average power.
Example 5.1:
Determine the power generated by each source and the average power absorbed by
each passive element in the circuit of Fig. 5.2(a).
⇒
(b)
(a)
Figure 5.2
Solution:
We apply mesh analysis as shown in Fig. 5.2(b). For mesh 1,
I1 = 4 A
For mesh 2,
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[5.3]
The average power absorbed by the voltage source is given by:
The voltage across the current source is
The average power supplied by the current source is
For the resistor,
The voltage across the capacitor is
The average power absorbed by the capacitor is
The current through the inductor is I 1 − I 2 = 2 − j10.39 =
The voltage across it is j10(I 1 − I 2 ) =
Hence, the average power absorbed by the inductor is
Notice that the inductor and the capacitor absorb zero average power and that the
total power supplied by the current source equals the power absorbed by the
resistor and the voltage source, or
P 1 + P 2 + P 3 + P 4 + P 5 = −367.8 + 160 + 0 + 0 + 207.8 = 0
indicating that power is conserved.
5.2 Maximum Average Power Transfer
Consider the circuit in Fig.5.3. The Thevenin
impedance Z Th and the load impedance ZL are
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Figure 5.3
[5.4]
The current through the load is
The average power delivered to the load is
(5.8)
From Eq. 5.8, we obtain
Setting ∂P/∂X L to zero gives
X L = −X Th
and setting ∂P/∂R L to zero results in
(5.9)
(5.10)
Therefore, for maximum average power transfer, the load impedance Z L must be
equal to the complex conjugate of the Thevenin impedance Z Th .
(5.11)
Setting R L = R Th and X L = −X Th in Eq. 5.8 gives us the maximum average power as
(5.12)
For purely resistive load (X L =0), Eq. 5.10 become
(5.13)
This means that for maximum average power transfer to a purely resistive load, the
load impedance (or resistance) is equal to the magnitude of the Thevenin
impedance.
Example 5.2:
Determine the load impedance Z L that
maximizes the average power drawn from the
circuit of Fig. 5.4. What is the maximum
average power?
Figure 5.4
Dr. Ahmed Abdul-Kadhim