6. Energy and Power
In the sinusoidal steady state, ac power is transferred from sources to various
loads. To study the transfer process, we must calculate the power delivered in
the sinusoidal steady state to any specified load. It turns out that there is an
upper bound on the available load power; hence, we need to understand how to
adjust the load to extract the maximum power from the rest of the circuit. In this
section the load is assumed to be made up of passive resistance, inductance,
and capacitance. To reach our objectives, we must first study the power and
energy delivered to these passive elements in the sinusoidal steady state.
In the sinusoidal steady state the current through a resistor can be pressed as
iR(t)=IAcos(wt). The instantaneous power delivered to the resistor is
Eq.8-42
where the identity cos2(x)=1/2[1+cos(2x)] is used to obtain the last line in Eq. (842). The energy delivered for t 0 is found to be
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Fig. 8-61: Resistor power and energy in the sinusoidal steady state
Figure 8-61 shows the time variation of pR(t) and wR(t). Note that the power is a
periodic function with twice the frequency of the current, that both pr(t) and wR(t)
are always positive, and that wR(t) increases without bound. These observations
remind us that a resistor is a passive element that dissipates energy.
In the sinusoidal steady state an inductor operates with a current iL(t)=IAcos(wt).
The corresponding energy stored in the element is
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where the identity cos2(x)=1/2[1+cos(2x)] is again used to produce the last line.
The instantaneous power delivered to the inductor is
Eq. 8-43
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Fig. 8-62: Inductor power and energy in the sinusoidal steady state.
Figure 8-62 shows the time variation of pL(t) and wL(t). Observe that both PL(t)
and wL(t) are periodic functions at twice the frequency of the ac current, that P L(t)
is alternately positive and negative, and that wL(t) is never negative. Since wL(t)
0 , the inductor does not deliver net energy to the rest of the circuit. Unlike the
resistor's energy in Figure 8-61, the energy in the inductor is bounded by
1/2LIA2 wL(t), which means that the inductor does not dissipate energy. Finally,
since PL(t) alternates signs, we see that the inductor stores energy during a
positive half cycle and then returns the energy undiminished during the next
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negative half cycle. Thus, in the sinusoidal steady state there is a lossless
interchange of energy between an inductor and the rest of the circuit.
In the sinusoidal steady state the voltage across a capacitor is vc(t)=VAcos(wt).
The energy stored in the element is
The instantaneous power delivered to the capacitor is
Eq. 8-44
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Fig. 8-63: Capacitor power and energy in the sinusoidal steady state.
Figure 8-63 shows the time variation of pC(t) and wC(t). Observe that these
relationships are the duals of those found for the inductor. Thus, in the sinusoidal
steady state the element power is sinusoidal and there is a lossless interchange
of energy between the capacitor and the rest of the circuit.
AVERAGE POWER
We are now in a position to calculate the average power delivered to various
loads. The instantaneous power delivered to any of the three passive elements is
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a periodic function that can be described by an average value. The average
power in the sinusoidal steady state is defined as
The power variation of the inductor in Eq. (8-43) and capacitor in Eq. (8-44) have
the same sinusoidal form. The average value of any sinusoid is zero since the
areas under alternate cycles cancel. Hence, the average power delivered to an
inductor or capacitor is zero:
Inductor: PL=0
Capacitor: PC=0
The resistor power in Eq. (8-42) has both a sinusoidal ac component and a
constant de component 1/2LIA2. The average value of the ac component is zero,
but the dc component yields
Resistor:
To calculate the average power delivered to an arbitrary load ZL=RL+jXL, we use
phasor circuit analysis to find the phasor current IL through ZL. The average
power delivered to the load is dissipated in RL, since the reactance XL represents
the net inductance or capacitance of the load. Hence the average power to the
load is
Eq. 8-45
Caution: When a circuit contains two or more sources, superposition applies only
to the total load current and not to the total load power. You cannot find the total
power to the load by summing the power delivered by each source acting alone.
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The following example illustrates a power transfer calculation.
EXAMPLE 8-31
Find the average power delivered to the load to the right of the interface in Figure
8-64.
Fig. 8-64
SOLUTION:
The equivalent impedance to the right of the interface is
The current delivered to the load is
Hence the average power delivered across the interface is
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Note: All of this power goes into the 100
resistor since the inductor and
capacitor do not absorb average power.
Fig. 8-65
MAXIMUM POWER
To address the maximum power transfer problem, we model the source/load
interface as shown in Figure 8-66. The source circuit is represented by a
Thevenin equivalent circuit with source voltage VT and source impedance
ZT=RT+jXT. The load circuit is represented by an equivalent impedance
ZL=RL+jXL. In the maximum power transfer problem the source parameters VT,
RT and XT are given, and the objective is to adjust the load impedance RL and XL
so that average power to the load is a maximum.
The average power to the load is expressed in terms of the phasor current and
load resistance:
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Then, using series equivalence, we express the magnitude of the interface
current as
Combining the last two equations, yields the average power delivered across the
interface:
Eq. 8-46
The quantities |VT|, RT , and XT in Eq. (8-46) are fixed. Our problem is to select
RL and XL to maximize P.
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Fig. 8-66: A source-load interface in the sinusoidal steady state.
Clearly, for every value of RL the denominator in Eq.(8-46) is minimized and P
maximized when XL=-XT. This choice of XL is possible because a reactance can
be positive or negative. When the source Thevenin equivalent has an inductive
reactance (XT>0), we modify the load to have a capacitive reactance of the same
magnitude, and vice versa. This step reduces the net reactance of the series
connection in Figure 8-66 to zero, creating a condition in which the net
impedance seen by the Thevenin voltage source is purely resistive.
When the source and load reactances cancel out, the expression for average
power in Eq.(8-46) reduces to
Eq. 8-47
This equation has the same form encountered in Chapter 3 in dealing with
maximum power transfer in resistive circuits. From the derivation in Sect 3-5, we
know P is maximized when RL=RT. In summary, to obtain maximum power
transfer in the sinusoidal steady state, we select the load resistance and
reactance so that
Eq. 8-48
These conditions can be compactly expressed in the following way:
Eq. 8-49
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The condition form maximum power transfer is called
a conjugate match, since the load impedance is the
conjugate of the source impedance. When the
conjugate-match conditions are inserted into Eq. (846), we find that the maximum average power
available from the source circuit is
Eq. 8-50
where |VT| is the peak amplitude of the Thevenin equivalent voltage.
It is important to remember that conjugate matching applies when the source is
fixed and the load is adjustable. These conditions arise frequently in powerlimited communication systems. However, as we will see in Chapter 14,
conjugate matching does not apply to electrical power systems because the
power transfer constraints are different.
EXAMPLE 8-32
(a) Calculate the average power delivered to the load in the circuit shown in
Figure 8-67 for Vs(t)=5cos106t, R=200 ohm, and RL=200 ohm.
(b) Calculate the maximum average power available at the interface and specify
the load required to draw the maximum power.
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SOLUTION:
(a) To find the power delivered to the 200 ohm load resistor, we use a Thevenin
equivalent circuit. By voltage division, the open-circuit voltage at the interface is
By inspection, the short-circuit current at the interface is
Give VT and IN, we calculate the Thevenin source impedance.
Using the Thevenin equivalent shown in Figure 8-67(b), we find that the current
through the 200-ohm resistor is
and the average power delivered to the load resistor is
(b) Using Eq.(8-50), the maximum average power available at the interface is
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The 200-ohm load resistor in part (a) draws about half of the maximum available
power. To extract maximum power, the load impedance must be
This impedance can be obtained using a 40-ohm resistor in series with a
reactance of +80-ohm. The required reactance is inductive (positive) and can be
produced by an inductance of
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