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3.2 AC Circuit Analysis

The topological analysis of basic ac electric circuits containing impedances and ideal ac supplies are presented in the following subsections. As will be demonstrated, using phasors greatly simplifies the analysis, and the VIs provide a flexible self-learning tool allowing users to create different circuit scenarios. Although Thevenin and Norton equivalent circuits as discussed for dc circuits in Chapter 2 can also be applied to ac circuits, they are not covered here.

3.2.1 Equivalent Impedances and Circuits

An ac circuit may contain a number of series and/or parallel branches. As will be studied in the following paragraphs, however, it is possible to divide any complex ac circuit into subcircuits that include simple circuit combinations. Therefore, five ac circuit combinations are identified and studied here: equivalent impedance circuit, voltage divider circuit, current divider circuit, series/parallel (combination) circuit, and circuit with dual ac supply.

The equivalent impedance of any number of impedances in series or in parallel (Fig. 3-2) is the sum of the individual impedances or the sum of the admittances that is equal to 1/Z, respectively.

Equation 3.11


Equation 3.12


03fig02.gifFigure 3-2. Series and parallel impedance circuits.

Since the current phasors and voltage phasors are related by complex impedances, Kirchhoff's Voltage and Current Laws can be applied to ac circuits containing sinusoidal sources operating in steady-state.

The voltage divider circuit in Fig. 3-3a indicates that if two impedances are connected in series and share the same current (meaning no other element is connected to the node where Z1 and Z2 join), the voltages across each of the elements are proportional to their impedances. The current and the voltages across each impedance can be given as

Equation 3.13


Equation 3.14


Equation 3.15


03fig03.gifFigure 3-3. Circuits with impedances: (a) voltage divider, (b) current divider, (c) series/parallel circuit, and (d) circuit with dual supplies.

Let's consider the current divider circuit in Fig. 3-3b now. Since the voltages across the impedances are equal to the supply voltage,

Equation 3.16


then the currents in each impedance branch can be calculated easily.

Equation 3.17


Hence the supply current I is simply the sum of all three currents.

Equation 3.18


The series/parallel circuit of Fig. 3-3c can be analyzed easily if the equivalent impedance is calculated first.

Equation 3.19


All remaining unknown parameters in the circuit can be derived as

Equation 3.20


Equation 3.21


The circuit given in Fig. 3-3d is selected since it represents a commonly used equivalent circuit in electrical engineering: single-phase transformer, and asynchronous motor. Note that the circuit given in Fig. 3-3c can be obtained easily by eliminating the supply voltage Vs2 in Fig. 3-3d.

In the analysis of this dual supply circuit, the mesh analysis can be used as described in Chapter 2. Hence the supply voltages are given as

Equation 3.22


Equation 3.23


The currents and voltages of the impedance elements can be determined after solving equations 3.22 and 3.23, in which the known parameters are Vs1, Vs2, Z1, Z2, and Z3, and the unknowns are I1 and I3. To solve for I1 and I3, the back substitution method is used in the VI provided in the following section. Then the remaining unknowns, I2, V1, V2, and V3, in the circuit are calculated as

Equation 3.24


Equation 3.25

03equ25.gif Virtual Instrument Panel

The virtual instruments provided in this section contain six circuit options. The desired circuits can be selected from the available folders. When a VI is selected, the corresponding circuit and its associated controls are displayed with their default values. Then students can run the VI and vary the complex quantities of the circuit and observe the calculated values all in the complex forms.

As demonstrated, the parameters in the circuits are given in rectangular form, which can easily be related to the practical values. In addition, the rectangular form allows the user to omit either a real or an imaginary component of the complex numbers. See Fig. 3-4 for a sample front panel.

03fig04.jpgFigure 3-4. A sample front panel and brief user guide. Self-Study Questions

Open and run the custom-written VI named Impedance Circuits.vi in the Chapter 3 folder, and investigate the following questions.


In the impedance circuits shown in Fig. 3-5, calculate the equivalent impedances.

03fig05.gifFigure 3-5. Example circuits for question 1.


Answers: Zeq(series) = 5∘−36.9° Ω, Zeq(parallel) = 3 + j1 Ω


Consider Vs = 10∘0°, Z1 = +j100 Ω, Z2 = 100 Ω, and Z3 = −j100 Ω for the circuit given in Figure 3-3b. Determine the currents in each impedance branch, and verify the results.


For the identical settings in question 2, vary the frequency of the voltage and observe its effect on the complex values of the calculated values.


Consider the voltage divider circuit, and set the supply voltage to Vs = 10∘90°. Set the impedances to Z1 = 6 and Z2 = j8, and determine the voltage across Z2.

Note: Remember to convert the complex voltage to the format that is acceptable by the custom-written VI, Impedance Circuits.vi.


Answer: 8.00∘126.87°


Compute the currents in each element for the circuit given in Fig. 3-6. Assume that the supply voltage is equal to 10∘−90°. Then verify the results analytically.

03fig06.gifFigure 3-6. Example circuit for question 5.


Answers: IL = 0.114∘−135°, IR = 0.1∘−180°, IC = 0.1∘−90°


Consider the ac circuit in Fig. 3-7, and calculate the value of the capacitor if ω = 2 rad/s and vs(t) = 90 sin 2t.

03fig07.gifFigure 3-7. Example circuit for question 6.

Hint: You can ignore one of the impedance branches shown in Fig. 3-3b or 3-3c. Remember that capacitive reactance is equal to 1/ωC.


Answer: 6 mF

3.2.2 A Reverse Study

A reverse study is presented here, where an unknown impedance is determined utilizing ac voltage and current waveforms that are set by the user.

Fig. 3-8 illustrates the block diagram of the VI model, where the ideal sine wave current is and voltage vs waveforms are defined by the user, and the corresponding equivalent impedance seen from the two terminals A–B is determined.

03fig08.gifFigure 3-8. The circuit simulated to identify the equivalent impedance of an unknown load Z.

The complex value of the equivalent impedance is displayed on the front panel, which is used to interpret the nature of the load (R, L, C, R + L, or R + C). The phasor diagram is also provided in the VI to illustrate the voltage and impedance phasors. Virtual Instrument Panel

Fig. 3-9 illustrates the front panel and the explanations diagram of the VI under investigation. The VI provided here (Single Phase AC Definitions.vi) uses the voltage and current as the base of the per-unit system, mainly for display purposes. The equivalent impedance is calculated and displayed on the front panel (as a per-unit value and a real value).

03fig09.jpgFigure 3-9. The front panel and the brief user guide of Single Phase AC Definitions.vi. Self-Study Questions

Open and run the custom-written VI named Single Phase AC Definitions.vi in the Chapter 3 folder, and investigate the following questions.


Vary the values of the phase angle, the current amplitude, and the voltage amplitude by using the knobs provided. Observe the relative positions of the voltage and current waveforms on the graph (in terms of phase angle).


Set the amplitudes of the voltage and the current equal to the base values, and analyze the estimated per-unit values of the voltage and current and the rms values.


Set the voltage and current waveforms to Vs = 50∘0° and IS = 15.8∘−18.45°, and verify that the equivalent impedance estimated in the front panel (Zeq= 3 + j1 Ω) is correct.


Without referring to the phasor diagram on the graph, plot the phasor diagram for the voltage and the current waveform that are shown on the waveform graph, and compare them with the displayed phasors. Is the load resistive, capacitive, or inductive, and why?


After making graphical observations of the voltage and current waveforms under three typical conditions (in phase, with lagging angle, and leading phase angle), observe and record the impedances, and verify the results analytically. In addition, observe that the impedances of both the pure inductor and the pure capacitor are pure imaginary numbers.


In the case of R + L or R + C load, calculate the value of the inductor L or the capacitor C. Assume that the supply frequency is 60 Hz.

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