University of California, San Diego
Spring 2016
J. Connelly
ECE 45: Power in Time-Dependent Circuits
Instantaneous vs. Average Power:
In a DC circuit, all voltages and currents in the circuit are constants, so the power has no dependence
on time. However, in general, the voltages and currents in a circuit change with time, so the power
supplied/dissipated at a component varies with time.
If v(t) is the voltage across a component and i(t) is the current through a component, then the instantaneous power is p(t) = v(t)i(t).
We are often concerned with the average power a component supplies/dissipates in an interval of time.
The average power over a time interval [a, b] is:
1
b−a
Z
b
p(t) dt.
a
Example:
The average power can depend on the interval for which it is defined. Suppose the instantaneous power
of a component is given by
t 0≤t<2
p(t) =
0 else
What is the average power in [0, 2], P1 ? What is the average power in [0, 4], P2 ?
Z 2
1
P1 =
t dt = 1W
2−0 0
Z 4
1
1
P2 =
t dt = W.
4−0 0
2
Power in steady-state RLC circuits:
In circuits with sinusoidal voltages and currents, it is natural to look at the average power in a period.
Since the circuit is in steady-state, the average power will be the same in any period.
Let v(t) and i(t) be the voltage and current of a component where each is sinusoidal with frequency ω
v(t) = |V | cos(ωt + ∠V )
i(t) = |I| cos(ωt + ∠I)
p(t) = v(t)i(t) = |V ||I| cos(ωt + ∠V ) cos(ωt + ∠I)
|V ||I|
=
(cos(2ωt + ∠V + ∠I) + cos(∠V − ∠I)) .
2
Please report any typos/errors to j2connelly@uscd.edu
The average power in a period T = 2π
is then given by
ω
Z
1
PT =
p(t) dt
T T
Z
1 T |V ||I|
=
(cos(2ωt + ∠V + ∠I) + cos(∠V − ∠I)) dt
T 0
2
|V ||I|
cos(∠V − ∠I)
=
2
1 = Re |V ||I| ej(∠V −∠I)
2
1
= Re {V I ∗ }
2
where V is the phasor of v(t) and I ∗ is the complex conjugate of the phasor of i(t).
Maximum Power in DC Circuit:
Rth
Vth
RL
Suppose we want to select the resistance of the load RL that maximizes the power delivered to the load
in a Thevenin equivalent circuit. Then
RL
RL + Rth
Vth
VL
=
IL =
RL
RL + Rth
RL
PL = Vth2
(RL + Rth )2
VL = Vth
We can find PL,max by setting the derivative of PL with respect to RL equal to 0.
2
dPL
2 (RL + Rth ) − 2RL (RL + Rth )
= Vth
0 =:
dRL
(RL + Rth )4
⇒ (RL + Rth )2 = 2(RL2 + RL Rth )
2
⇒ RL2 = Rth
⇒ RL = Rth .
Then we have
PL,max = PL |RL =Rth =
Vin2
.
4Rth
Maximum Power in Steady State RLC Circuit:
Zth
Vth
ZL
Suppose we want to select the impedance of the load ZL = RL + jXth that maximizes the power
delivered to the load in a Thevenin equivalent circuit. Let Vth be the phasor of the Thevenin voltage
vth = |Vth | cos(ωt + ∠Vth ) and let Zth = Rth + jXth be the Thevenin equivalent impedance. Then
ZL
ZL + Zth
Vth
VL
=
ZL
ZL + Zth
1
Re {VL I ∗ }
2 L
ZL
1
Vth∗
Re Vth
2
ZL + Zth (ZL + Zth )∗
1
2 RL + jXL
Re |Vth |
2
|ZL + Zth |2
RL
|Vth |2
2 |ZL + Zth |2
|Vth |2
RL
2 (RL + Rth )2 + (XL + Xth )2
VL = Vth
IL =
PL =
=
=
=
=
We wish to maximize PL , and the term (XL + Xth ) only decreases PL , so set XL = −Xth . Then
RL
|Vth |2
2 (RL + Rth )2
is maximized by choosing RL = Rth . Thus the maximum power to the load occurs when ZL = Zth ∗
and
|Vth |2
8Rth .
Example:
Let R1 = 1Ω, R2 = 3Ω, C = 1F , L = 1H and vin (t) = cos(2t), and suppose a load consisting
of a resistor in series with a capacitor is placed across the output terminals. Find the values of the
resistor and the capacitor that maximize the average power deliver to the load. What is PL,max ?
R1
+
v(t)
i
−
R2
L
C
Load
From the Discussion 2 Notes, we have
Vth
jw
= H(ω) =
Vin
(jw + 2)2
Z1
Vi +
Vc
Thus when vin (t) = cos(2t),
Z2
ZC
−
+
Z L Vth
−
Z1
Z2
ZC
ZL
vth (t) = |H(2)| cos(2t+∠H(2)) =
1
cos(2t).
8
To find Zth , set Vi = 0 and solve for the effective impedance at ω = 2. We have
Zth = (Z1 //ZC + Z2 )//ZL
R1 /jωC
1
Z1 //ZC =
=
R1 + jωC
2j + 1
6j + 4
Z2 + Z1 //ZC =
2j + 1
3
Zth = 1 + j
2
The value of ZLoad that maximizes power to
∗
the load is ZLoad = Zth
= 1 − j3/2.
We also have ZLoad = RLoad + jωC1Load .
Thus RL = 1 and CL = 1/3F .