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BASIC NETWORK ANALYSIS 1.19

FIGURE 1.25 A decaying sinusoidal wave.

Related Calculations. This concept of a dc component superimposed on a sinusoidal accomponent is illustrated in Fig. 1.25. This figure shows the decay of a dc componentbecause of a short circuit and also shows how the asymmetrical short-circuit current grad-ually becomes symmetrical when the dc component decays to zero.

STEADY-STATE AC ANALYSIS OF A SERIES

RLC CIRCUIT

Calculate the current in the circuit ofFig. 1.26a.

Calculation Procedure

1. CalculateZ

Angular frequency 2f (2)(3.1416)(60) 377 rad/s. ButXL L; therefore,XL (377)(0.5) 188.5 . Also,XC 1/C 1/[(377)(26.5) 10

6] 100 . ThenZ R j(XL XC) R jXEQ, whereXEQ XL XC net equivalent reactance.

In polar form, the impedance for the series RLC circuit is expressed as

Z Z . Z 100 j(188.5 100) 100 +j88.5

133.5 . The impedance triangle (Fig.1.26b) illustrates the results of the preceding solution.

Apply KVL to the circuit: E VR jVL jVC VR jVX where VX VL VC net reactive voltage.

2. Draw the Phasor Diagram

The phasor diagram of Fig. 1.26c shows the voltage relations with respect to the

current as a reference.

41.5 tan1(88.5/100)(100 )2 (88.5)2 tan1(XEQ/R)R 2 X2EQ


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3. Calculate I

From Ohms law for ac circuits, I 120/133.5 0.899 A. Because I is a referenceit can be expressed in polar form as I 0.899 A. The angle between the voltage andcurrent in Fig. 1.26c is the same as the angle in the impedance triangle of Fig. 1.26b.

Therefore E 120 V.

Related Calculations. In a seriesRLCcircuit the net reactive voltage may be zero (whenVL VC), inductive (when VL VC), or capacitive (when VL VC). The current in such acircuit may be in phase with, lag, or lead the applied emf. When VL VC, the condition isreferred to as series resonance. Voltages VL and VC may be higher than the applied voltageE, because the only limiting opposition to current is resistanceR. A circuit in series reso-nance has maximum current, minimum impedance, and a power factor of 100 percent.

STEADY-STATE AC ANALYSIS OF A PARALLELRLC CIRCUIT

Calculate the impedance of the parallelRLCcircuit ofFig. 1.27a.


1. Calculate the Currents in R, L,andC

In a parallel circuit, it is convenient to use the voltage as a reference; thereforeE 200 V. Because the R, L, and C parameters of this circuit are the same as in0

41.5

0

1.20 HANDBOOK OF ELECTRIC POWER CALCULATIONS

FIGURE 1.26 Series RLCac circuit: (a) circuit with component values; (b) imped-ance triangle; and (c) phasor diagram.


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Fig. 1.26a and the frequency (60 Hz) is the same, XL 188.5 and XC 100 .From Ohms law: IR E/R 200 /100 2 A. IL E/XL 200 /188.5

1.06 j1.06 A, and IC E/XC 200 /100 2 j2A.But IT IR jIL jIC; therefore, IT 2 j1.06 j2 2 j0.94 2.21 A.

2. Calculate ZEQ

Impedance is ZEQ E/IT 200 /2.21 90.5 . ZEQ, changed to rec-tangular form, is ZEQ 82.6 j39 REQ jXEQ. Figure 1.27billustrates the volt-age-current phasor diagram. The equivalent impedance diagram is given in Fig. 1.27c.

Note that ZEQ can also be calculated by

sinceZL jXL andZC jXC.

Related Calculations. The impedance diagram ofFig. 1.27chas a negative angle. Thisindicates that the circuit is anRCequivalent circuit.Figure 1.27bverifies this observation

ZEQ 1

1

R

1

jXL

1

jXC

25.225.20

25.2909009090

0000


FIGURE 1.27 Parallel RLC circuit: (a) circuit with component values; (b) phasordiagram; and (c) impedance triangle.


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because the total circuit current IT leads the applied voltage. In a parallelRLCcircuit thenet reactive current may be zero (when IL IC), inductive (when IL IC), or capacitive(when IL IC). The current in such a circuit may be in phase with, lag, or lead the ap-plied emf. When IL IC, this condition is referred to as parallel resonance. Currents IL

and IC may be much higher than the total line current, IT. A circuit in parallel resonancehas a minimum current, maximum impedance, and a power factor of 100 percent. Note inFig. 1.27bthat IT IR jIX, whereIX IC IL.

ANALYSIS OF A SERIES-PARALLEL

AC NETWORK

A series-parallel ac network is shown in Fig. 1.28. Calculate ZEQ, I1, I2, and I3.


1. Combine All Series Impedances

The solution to this problem is similar to that for the first problem in the section,except that vector algebra must be used for the reactances. Z1 300 j600 j200 300 j400 500 , Z2 500 j1200 1300 , and Z3 800

j600 1000 .

2. Combine All Parallel Impedances

Using the product-over-the-sum rule, we find ZBC Z2Z3/(Z2 Z3) (1300 )(1000 )/[(500 j1200) (800 j600)] 908 901 j90.2 .

3. Combine All Series Impedances to Obtain the Total Impedance, ZEQ

ZEQ Z1 ZBC (300 j400) (901 j90.2) 1201 j490 1290 .22.4

5.736.967.4

36.967.453.1


FIGURE 1.28 Series-parallel ac circuit to be analyzed.


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4. Calculate the Currents

I1 E/ZEQ 100 /1290 0.0775 A. From the current-dividerrule: I2 I1Z3/(Z2 Z3) (0.0775 )(1000 )/[(500 j1200) (800

j600)] 0.0541 A. I3 I1Z2/(Z2 Z3) (0.0775 )(1300 )/ [(500 j1200) (800 j600)] 0.0709 A.

Related Calculations. Any reducible ac circuit (i.e., any circuit that can be reduced toone equivalent impedance ZEQ with a single power source), no matter how complex, canbe solved in a similar manner to that just described. The dc network theorems used in pre-vious problems can be applied to ac networks except that vector algebra must be used forthe ac quantities.

ANALYSIS OF POWER IN AN AC CIRCUIT

Find the total watts, total VARS, and total volt-amperes in the ac circuit ofFig. 1.29a.Recall that watts, VARS, and volt-amperes are all dimensionally the same, that is, theproduct of voltage and current. However, we use the designators of watts (W) to representreal power (instantaneous or average), volt-amperes-reactive (VARS) to represent reactivepower, and volt-amperes (VA) to represent complex (or apparent) power.


1. Study the Power Triangle

Figure 1.30 shows power triangles for ac circuits. Power triangles are drawn followingthe standard of drawing inductive reactive power in the j direction and capacitive reac-tive power in thej direction. Two equations are obtained by applying the Pythagoreantheorem to these power triangles: S2 P2 and S2 P2 . These equations canbe applied to series, parallel, or series-parallel circuits.

The net reactive power supplied by the source to an RLC circuit is the differencebetween the positive inductive reactive power and the negative capacitive reactive power:QX QL QC, where QXis the net reactive power, in VARS.

2. Solve for the Total Real Power

Arithmetic addition can be used to find the total real power. PT P1 P2 200 500 700 W.

Q2CQ2

L

20.267.422.484.1

36.922.422.422.40


FIGURE 1.29 Calculating ac power: (a) circuit and (b) power triangle.


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3. Solve for the Total Reactive Power

QX QL QC 1200 500 700 VARS. Because the total reactive power is posi-tive, the circuit is inductive (see Fig. 1.29b).

4. Solve for the Total Volt-Amperes

S

Related Calculations. The principles used in this problem will also be applied to solvethe following two problems.

ANALYSIS OF POWER FACTOR

AND REACTIVE FACTOR

Calculate the power factor (pf) and the reactive factor (rf) for the circuit shown in Fig. 1.31.


1. Review Power-Factor Analysis

P 2T Q2

X (700)2 (700)2 989.8VA.


FIGURE 1.30 Power triangles for (a)RCand (b)RL equivalent circuits.

FIGURE 1.31 Calculating power and reactive factors of circuit.


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The power factor of an ac circuit is the numerical ratio between the true power P andthe apparent power S. It can be seen by referring to the power triangles ofFig. 1.30 thatthis ratio is equal to the cosine of the power-factor angle . The power-factor angle is thesame as the phase angle between the voltage across the circuit (or load) and the current

through the circuit (or load). pf cos P/S.

2. Review Reactive-Factor Analysis

The numerical ratio between the reactive power and the apparent power of a circuit (orload) is called the reactive factor. This ratio is equal to the sine of the power-factor angle(see Fig. 1.30). rf sin Q/S.

3. Calculate the Power and Reactive Factors

Z1 R jXL 100 j100 141.4 . I1 E/Z1 120 /141.4 0.849

A.I

1 (0.6 j0.6)A.I

2 E/XC 120 /60 2 (0 j2)A.IT I1 I2 (0.6 j0.6) (0 j2) (0.6 j1.4)A 1.523 A. S E LT

(120)(1.523) 182.8 VA. Power factor cos cos 66.8 0.394 or 39.4 percent;rf sin sin 66.8 0.92 or 92 percent.

Related Calculations. Inductive loads have a lagging power factor; capacitive loadshave a leading power factor. The value of the power factor is expressed either as a deci-mal or as a percentage. This value is always less than 1.0 or less than 100 percent. Themajority of industrial loads, such as motors and air conditioners, are inductive (laggingpower factor). Thus, power engineers often refer to capacitors or capacitive loads as

sources of reactive power.

POWER-FACTOR CORRECTION

Calculate the value of the capacitor needed to obtain a circuit power factor of 100 percent(Fig. 1.32).


1. Calculate the Motor Current

66.8

90

900

45

45045


FIGURE 1.32 Power-factor correction: (a) given circuit and (b) adding a capacitor (C) in parallel to

improve power factor.


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S P/cos 1200/0.7 1714 VA. Hence, the motor current I is: I S/E (1714 VA) (120 V) 14.29 A. The active component of this current is the component inphase with the voltage. This component, which results in true power consumption, is:I cos (14.29 A)(0.7) 10 A. Because the motor has a 70 percent power factor, thecircuit must supply 14.29 A to realize a useful current of 10 A.

2. Calculate the Value ofC

In order to obtain a circuit power factor of 100 percent, the inductive apparentpower of the motor and the capacitive apparent power of the capacitor must be equal.

QL E I where reactive factor. Hence, QL (120)(14.29)

1714 1224 VARS (inductive). QC must equal 1224 VARS for100 percent power factor. XC /QC (120)

2/1224 11.76 (capacitive). There-

fore, C 1/XC 1/(377)(11.76) 225.5 F.

Related Calculations. The amount of current required by a load determines the sizes ofthe wire used in the windings of the generator or transformer and in the conductors con-necting the motor to the generator or transformer. Because copper losses depend upon thesquare of the load current, a power company finds it more economical to supply 10 A at apower factor of 100 percent than to supply 14.29 A at a power factor of 70 percent.

A mathematical analysis of the currents in Fig. 1.32bfollows: IC QC/VC (1220VARS)/(120 V) 10.2 A (0 j10.2) A. (for motor) cos1 0.7 45.6; there-fore, IM 14.29 (10 j10.2) A. Then IT IM IC (10 j10.2) (0

j10.2) 10 A (100 percent power factor). Typically, power factor correction capaci-tors are rated in kVARS (kilo-VARS) and may be installed in switched banks to provide arange of pf correction.

MAXIMUM POWER TRANSFER

IN AN AC CIRCUIT

Calculate the load impedance in Fig. 1.33 for maximum power to the load.


045.6

V2C

0.511(0.7)21 cos21 cos2


FIGURE 1.33 Finding value ofZL for maximum power transfer.


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1. Statement of the Maximum Power Theorem

The maximum power theorem, when applied to ac circuits, states that maximumpower will be delivered to a load when the load impedance is the complex conjugate ofthe Thevenin impedance across its terminals.

2. Apply Thevenins Theorem to the Circuit

ZTh Z1Z2/(Z1 Z2) (10 )(8 )/[6 j8) j8] 13.3 , or ZTh 10.6 j8 whereR 10.6 andXL 8 . Then, ZL must be 13.3 10.6

j8 , whereRL 10.6 andXC 8 .In order to find the maximum power delivered to the load, ETh must be found using the

voltage-divider rule: ETh EZ2/(Z1 Z2) (9 )(8 )/[(6 j8) j8] 12 V.Pmax /4RL; therefore Pmax (12)

2/(4)(10.6) 3.4 W.

Related Calculations. The maximum power transfer theorem, when applied to dccircuits, states that a load will receive maximum power from a dc network when itstotal resistance is equal to the Thevenin resistance of the network as seen by theload.

ANALYSIS OF A BALANCED

WYE-WYE SYSTEM

Calculate the currents in all lines of the balanced three-phase, four-wire, wye-connectedsystem ofFig. 1.34. The system has the following parameters: VAN 120 V, VBN 120 V, VCN 120 V, and ZA ZB ZC 12 .


1. Calculate Currents

IA VAN/ZA 120 /12 10 A. IB VBN/ZB 120 /12 10 A. IC VCN/ZC 120 /12 10 A. IN IA IB IC; hence,IN 10 10 10 0 A.

Related Calculations. The neutral current in a balanced wye system is always zero.Each load current lags or leads the voltage by the particular power factor of the load. Thissystem, in which one terminal of each phase is connected to a common star point, isoften called a star-connected system.

ANALYSIS OF A BALANCED

DELTA-DELTA SYSTEM

Calculate the load currents and the line currents of the balanced delta-delta system ofFig. 1.35. The system has the following load parameters: VAC 200 V, VBA

200 V, VCB 200 V, and ZAC ZBA ZCB 4 .0120120

0

12012001200120120

0120000

01201200

E2Th

90900

37379053



10/13


FIGURE 1.35 A balanced delta-delta system.

FIGURE 1.34 A balanced three-phase, four-wire, wye-connected system:(a) circuit and (b) load-phasor diagram.


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1. Solve for the Load Currents

IAC VAC/ZAC 200 /4 50 A, IBA VBA/ZBA 200 /4

50 A, and ICB VCB/ZCB 200 /4 50 A.

2. Solve for the Line Currents

Convert the load currents to rectangular notation: IAC 50 50 j0, IBA 50 25 j43.3, and ICB 50 25 j43.3. Apply KCL at loadnodes: IA IAC IBA (50 j0) (25 j43.3) 86.6 A, IB IBA ICB (25 j43.3) (25 j43.3) 86.6 A, IC ICB IAC (25

j43.3) (50 j0) 86.6 A.

Related Calculations. In comparing a wye-connected system with a delta-connectedsystem, one can make the following observations:

1. When a load is wye-connected, each arm of the load is connected from a line to theneutral. The impedance Z is shown with a single subscript, such as ZA.

2. When a load is delta-connected, each arm of the load is connected from a line to line.The impedance Z is shown with a double subscript such as ZAC.

3. In a wye-connected system, the phase current of the source, the line current, and thephase current of the load are all equal.

4. In a delta-connected system, each line must carry components of current for twoarms of the load. One current component moves toward the source, and the other

15090

30120120

0

12001201200120000


FIGURE 1.36 Relationships between phase and line currents in a bal-

anced delta-connected system.


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current component moves away from the source. The line current to a delta-connected load is the phasor difference between the two load currents at the enter-ing node.

5. The line current in a balanced delta load has a magnitude of times the phase cur-rent in each arm of the load. The line current is 30 out of phase with the phase current(Fig. 1.36).

6. The line-line voltage in a balanced, wye-connected, three phase source has a magni-

tude of times the line-neutral voltage. The line-line voltage is 30 out of phase withthe line-neutral voltage.

RESPONSE OF AN INTEGRATOR

TO A RECTANGULAR PULSE

A single 10-V pulse with a width of 100 s is applied to theRCintegrator ofFig. 1.37.Calculate the voltage to which the capacitor charges. How long will it take the capacitorto discharge (neglect the resistance of the pulse source)?


1. Calculate the Voltage to Which the Capacitor Charges

The rate at which a capacitor charges or discharges is determined by the time constantof the circuit. The time constant of a series RCcircuit is the time interval that equals theproduct ofR and C. The symbol for time constant is (Greek letter tau): RC, where

R is in ohms, Cis in farads, and is in seconds.The time constant of this circuit is: RC (100 k)(0.001 F) 100 s.

Because the pulse width equals 200 s (2 time constants), the capacitor will charge to 86percent of its full charge, or to a voltage of 8.6 V. The expression for RCcharging is:vC(t) VF(1 e

t/RC), where VF is the final value. In this case the final value, VF 10

V, would be reached if the pulse had a width of 5 or more time constants. See theRCtimeconstant charging table (Table 1.1).

2. Calculate the Discharge Time

The capacitor discharges back through the source at the end of 200 s. The total dis-charge time for practical purposes is 5 time constants or (5)(100 s) 500 s. The ex-pression for RC discharging is: vC(t) Vt(e

t/RC), where Vi is the initial value. In thiscase, the initial value before discharging is 8.6 V. Table 1.2 shows theRCtime constantdischarge characteristics.

3

3


FIGURE 1.37 Pulse input to anRCintegrator.


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Related Calculations. Figure 1.38 illustrates the output charging and dischargingcurves.

BIBLIOGRAPHY

Hayt, William J., and Jack Kemmerly. 1993.Engineering Circuit Analysis, 5th ed. New York:McGraw-Hill.

Nilsson, James W., and Susan A. Riedel. 1999.Electric Circuits, 6th ed. Englewood Cliffs, N.J.:Prentice Hall.

Stanley, William D. 1999.Network Analysis with Applications, 3rd ed. Englewood Cliffs, N.J.:

Prentice Hall.


FIGURE 1.38 Output charging and discharging curves for theRCintegrator ofFig. 1.37.

TABLE 1.1 RCTime Constant Charging

Characteristics

% Full charge

1 63

2 86

3 95

4 98

5 99*

*For practical purposes, 5 time constants are

considered to result in 100 percent charging.

TABLE 1.2 RCTime Constant Discharging

Characteristics

% Full charge

1 37

2 14

3 5

4 2

5 1*

*For practical purposes, 5 time constants are consid-

ered to result in zero charge or 100 percent discharge.

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