4. Circuit Theorems with Phasors

4. Circuit Theorems with Phasors

PROPORTIONALITY

The proportionality property states that phasor output responses are proportional to the input phasor. Mathematically, proportionality means that

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= 10. 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. By Ohm's law, .
3. By KVL,
4. By Ohm's law,
5. By KCL,
6. 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 = 20F, 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 VR(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 = 240V 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 VS2 = 830 at a frequency of = 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 node-voltage 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 =5000 rad/s. Find the steady-state voltage vR(t) using source transformations.

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 VR,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. 8-26.

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 open-circuit 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.

2. Phasor Circuit Analysis

Phasor circuit analysis is a method of finding sinusoidal steady-state responses directly from the circuit without using differential equations. How do we perform phasor circuit analysis? At several points in our study we have seen that circuit analysis is based on two kinds of constraints: (1) connection constraints (Kirchhoff's laws), (2) device constraints (element equations). To analyze phasor circuits, we must see how these constraints are expressed in phasor form.

Connection Constraints in Phasor Form

In the steady state all of the voltages and currents are sinusoids with the same frequency as the driving force. Under these conditions, the application of KVL around a loop could take the form.

Therefore, if the sum of the waveforms is zero, then the corresponding phasors must also sum to zero.

Clearly the same result applies to phasor currents and KCL. In other words, we can state Kirchhoff's laws in phasor form as follows:

KVL: The algebraic sum of phasor voltages around a loop is zero.

KCL: The algebraic sum of phasor currents at a node is zero.

Device Constraints in Phasor Form

Eq.(8-12)

In the sinusoidal steady state, all of these currents and voltages are sinusoids. Given that the signals are sinusoid, how do these i-v relationships constrain the corresponding phasors?

This relationship holds only if the phasor voltage and current for a resistor are related as

Eq.(8-13)

Since L is a real constant, moving it inside the real part operation does not change things. Written this way, we see that the phasor voltage and current for an inductor are related as

Eq.(8-14)

Fig. 8-3: Phasor iv characteristics of the inductor

The resulting phasor diagram in Fig. 8-3 shows that the inductor voltage current are 90 out of phase. The voltage phasor is advanced by 90counterclockwise, which is in the direction of rotation of the complex exponential . When the voltage phasor is advanced counterclockwise(that is, ahead of the rotating current phasor), we say that the voltage phasor leads the current phasor by 90or, equivalently, the current lags the voltage by 90.

The phasor voltage and current for a capacitor are related as

Solving for VC yields

Eq.(8-15)

Fig. 8-4: Phasor iv characteristics of the capacitor.

The resulting phasor diagram in Fig. 8-7 shows that voltage and current are 90out of phase. In this case, the voltage phasor is retarded by 90clockwise, which is in a direction opposite to rotation of the complex exponential . When the voltage is retarded clockwise (that is, behind the rotating current phasor), we say that the voltage phasor lags the current phasor by 90or, equivalently, the current leads the voltage by 90.

The Impedance Concept

The IV constraints Eqs.(8-13), (8-14, and (8-15) are all of the form

V=ZI Eq.(8-16)

where Z is called the impedance of the element. Equation (8-16) is analogous to Ohm's law in resistive circuits. Impedance is the proportionality constant relating phasor voltage and phasor current in linear, two-terminal elements. The impedances of the three passive elements are

Eq.(8-17)

Since impedance relates phasor voltage to phasor current, it is a complex quantity whose units are ohms. Although impedance can be a complex number, it is not a phasor. Phasors represent sinusoidal signals, while impedances characterize circuit elements in the sinusoidal steady state. Finally, it is important to remember that the generalized two-terminal device constraint in Eq.(8-16) assumes that the passive sign convention is used to assign the reference marks to the voltage and current.

EXAMPLE 8-5

Fig. 8-5

The circuit in Fig. 8-5 is operating in the sinusoidal steady state with and . Find the impedance of the elements in the rectangular box.

SOLUTION:

The purpose of this example is to show that we can apply basic circuit analysis tools to phasors. The phasors representing the given signals are and . Using Ohm's law in phasor form, we find that . Applying KVL around loop A yields . Solving form the unknown voltage yields

V

This voltage appears across the load resistor R1; hence . Applying KCL at node A yields . Given the phasor voltage across and current through the unknown element, we find the impedance to be

3. Basic Circuit Analysis with Phasors

To perform ac circuit analysis in this way, we obviously need to develop methods of analyzing circuits in the phasor domain.

In the preceding section, we showed that KVL and KCL applying the phasor domain and that the phasor element constraints all have the form V=ZI. Therefore, familiar algebraic circuit analysis tools, such as series and parallel equivalence, voltage and current division, proportionality and superposition, and Thevenin and Norton equivalent circuits, are applicable in the phasor domain. We do not need new analysis techniques to handle circuits in the phasor domain. The only difference is that circuit responses are phasors (complex numbers) rather than dc signals (real numbers).

Fig. 8-6: Flow diagram for phasor circuit analysis.

We can think of phasor domain circuit analysis in terms of the flow diagram in Fig. 8-6. The analysis begins in the time domain with a linear circuit operating in the sinusoidal steady state and involves three major steps:

Step 1: The circuit is transformed into the phasor domain by representing the input and response sinusoids as phasor and the passive circuit elements by their impedances.

Step 2: Standard algebraic circuit techniques are applied to solve the phasor domain circuit for the desired unknown phasor responses.