4. Circuit Theorems with Phasors
PROPORTIONALITY
The proportionality property states that phasor output responses are proportional
to the input phasor. Mathematically, proportionality means that
Y = KX
Eq.(8-28)
where X is the input phasor, Y is the output phasor, and K is the proportionality
constant. In phasor circuit analysis, the proportionality constant is generally a
complex number.
To apply the unit output method in the phasor domain, we assume that the output
is a unit phasor Y= 1
0 . By successive application of KCL, KVL, and the
element impedances, we solve for the input phasor required to produce the unit
output. Because the circuit is linear, the proportionality constant relating input
and output is
Once we have the constant K, we can find the output for any input or the input
required to produce any specified output.
The next example illustrates the unit output method for phasor circuits.
EXAMPLE 8-13
Use the unit output method to find the input impedance, current I1, output voltage
VC, and current I3 of the circuit in Fig. 8-20 for Vs=
SOLUTION:
The following steps implement the unit output method for the circuit in Fig. 8-20:
Fig. 8-20
1.
Assume a unit output voltage
2.
3.
4.
By Ohm's law,
By KVL,
By Ohm's law,
5.
6.
By KCL,
By KCL,
Given VS and I1, the input impedance is
.
.
.
The proportionality factor between the input VS and output voltage Vc is
Given K and ZIN, we can now calculate the required responses for an input
SUPERPOSITION
The superposition principle applies to phasor responses only if all of the
independent sources driving the circuit have the same frequency. That is , when
the input sources have the same frequency, we can find the phasor response
due to each source acting alone and obtain the total response by adding the
individual phasors. If the sources have different frequencies, then superposition
can still be used but its application is different. With different frequency sources,
each source must be treated in a separate steady-state analysis because the
element impedances change with frequency. The phasor response for each
source must be changed into waveforms and then superposition applied in the
time domain. In other words, the superposition principle always applies in the
time domain. It also applies in the phasor domain when all independent sources
have the same frequency. The following examples illustrate both cases.
EXAMPLE 8-14
Fig. 8-21
Use superposition to find the steady - state voltage vR (t) in Fig. 8 - 21 for R=20
, L1 = 2mH, L2 = 6mH, C = 20
F, V s1= 100cos 5000t V , and Vs2=120cos (5000t
+30 )V.
SOLUTION:
In this example, the two sources operate at the same frequency. Fig. 8-22(a)
shows the phasor domain circuit with source no.2 turned off and replaced by a
short circuit. The three elements in parallel in Fig. 8-22(a) produce an equivalent
impedance of
Fig. 8-22
By voltage division, the phasor response VR1 is
Fig. 8-22(b) shows the phasor-domain circuit with source no.1 turned off and
source no. 2 on. The three elements in parallel in Fig. 8-22(b) produce an
equivalent impedance of
By voltage division, the response VR2 is
Since the sources have the same frequency, the total response can be found by
adding the individual phasor responses VR1 and VR2:
The time-domain function corresponding to the phasor sum is
The overall response can also be obtained by adding the time-domain functions
corresponding to the individual phasor responses VR1 and VR2:
You are encouraged to show that the two expressions for V R(t) are equivalent
using the additive property of sinusoids.
EXAMPLE 8-15
Fig. 8-23
Use superposition to find the steady-state current i(t) in Fig. 8-23 for R=10k
,L=200mH, vS1=24cos20000t V, and vS2=8cos(60000t+30 ).
SOLUTION:
In this example the two sources operate at different frequencies. With source no.
2 off, the input phasor is Vs1 = 24
0 V at a frequency of
=20 krad/s . At this
frequency the equivalent impedance of the inductor and resistor is
The phasor current due to source no.1 is
With source no.1 off and no.2 on, the input phasor V S2 = 8
of
30
at a frequency
= 60 krad/s. At this frequency the equivalent impedance of the inductor and
resistor is
The phasor current due to source no.2 is
The two input sources operate at different frequencies, so that phasors
responses I1 and I2 cannot be added to obtain the overall response. In this case
the overall response is obtained by adding the corresponding time-domain
functions.
THEVENIN AND NORTON EQUIVALENT CIRCUITS
Fig. 8-24 Thevenin and Norton equivalent circuits in the phasor analysis
In the phasor domain, a two-terminal circuit containing linear elements and
sources can be replaced by the Thevenin or Norton equivalent circuits shown in
Fig. 8-24. The general concept of Thevenin's and Norton's theorems and their
restrictions are the same as in the resistive circuit studied in Chapter 3. The
important difference here is that the signals VT, IN, V, and I are phasors, and
VT=1/YN and ZL are complex numbers representing the source and load
impedances.
Finding the Thevenin or Norton equivalent of a phasor circuit involves the same
process as for resistance circuits, except that now we must manipulate complex
numbers. The thevenin and Norton circuits are equivalent to each other, so their
circuit parameters are related as follows:
Eq.(8-29)
Algebraically, the results in Eq.(8-29) are identical to the corresponding
equations for resistance circuits. The important difference is that these equations
involve phasors and impedances rather than waveforms and resistances. These
equations point out that we can determine a Thevenin or Norton equivalent by
finding any two of the following quantities: (1) the open-circuit voltage VOC, (2) the
short-circuit current ISC, and (when there are no dependent sources) (3) the
impedance ZT looking back into the source circuit with all independent sources
turned off.
The relationships in Eq. (8-29) define source transformations that allow us to
convert a voltage source in series with an impedance into a current source in
parallel with the same impedance, or vice versa. Phasor-domain source
transformations simplify circuits and are useful in formulating general nodevoltage or mesh-current equations, discussed in the next section.
The next two examples illustrate applications of source transformation and
Thevenin equivalent circuits.
EXAMPLE 8-17
Both sources in Fig. 8-25(a) operate at a frequency of
steady-state voltage vR(t) using source transformations.
=5000 rad/s. Find the
SOLUTION:
Example 8-14 solves this problem using superposition. In this example we use
source transformations. We observe that the voltage sources in Fig. 8-25(a) are
connected in series with an impedance and can be converted into the following
equivalent current sources:
Fig. 8-25
Fig. 8-25 (b) shows the circuit after these two source transformations. The two
current sources are connected in parallel and can be replaced by a single
equivalent current source:
The four passive elements are connected in parallel and can be replaced by an
equivalent impedance:
The voltage across this equivalent impedance equals V R,since one of the parallel
elements is the resistor R. Therefore, the unknown phasor voltage is
The value of VR is the same as found in Example 8-14 using superposition. The
corresponding time-domain function is
EXAMPLE 8-18
Use Thevenin's theorem to find the current Ix in the bridge circuit shown in Fig. 826.
Fig. 8-26
SOLUTION:
Example 8-12 solves this problem using a
-to-Y transformation. In this
example, we determine Ix by find the Thevenin equivalent circuit seen by the
impedance j200. The Thevenin equivalent will be found by determining the opencircuit voltage and the lookback impedance.
Disconnecting the impedance j200 form the circuit in Fig. 8-26 produces the
circuit shown in Fig. 8-27(a). The voltage between nodes A and B is the
Thevenin voltage since removing the impedance j200 leaves an open circuit. The
voltages at nodes A and B can each be found by voltage division. Since the
open-circuit voltage is the difference between these node voltages, we have
Fig. 8-27
Turning off the voltage source in Fig. 8-27(a) and replacing it by a short circuit
produces the situation shown in Fig. 8-27(b). The lookback impedance seen at
the interface is a series connection of two pairs of elements connected in parallel.
The equivalent impedance of the series/parallel combination is
Given the Thevenin equivalent circuit, we treat the impedance j200 as a load
connected at the interface and calculate the resulting load current Ix as
The value of Ix found here is the same as the answer obtained in Example 8-12
using a
-to Y transformation.