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RLC Circuit Analysis
CHAPTER 11
,~ RLC Circuit Analysis In this chapter, series and parallel circuits with resistors, capacitors, and inductors are analyzed. Many of the same techniques used in the solution of resistivecapacitive and resistive-inductive circuits are used in this analysis. Phasor diagrams provide descriptions of the circuits that lead to Pythagorean theorem solutions of certain circuit values.
BASIC AC CIRCUITS
369
RLC CIRCUIT ANALYSIS
II
Objectives
I
I
At the end of this chapter you should be able to: 1. Draw phasor diagrams showing the phase relationships of various circuit values in series and parallel RLC circuits. 2. Identify the various circuit values in series and parallel RLC circuits that can be determined by Pythagorean theorem analysis. 3. Identify the positive and negative phase angles in series and parallel RLC circuits. 4. Calculate total reactance, total reactive current, total reactive voltage, and total reactive power in RLC circuits. 5. Given schematic diagrams and typical circuit values for the circuits below, calculate current, voltage, impedance, and power values.
370
BASIC AC CIRCUITS
RLC CIRCUIT ANALYSIS
II Phase D i f f e r e n c e s 9 Current as a R e f e r e n c e 9 D e v e l o p m e n t o f a Circuit Phasor Diagram
11
INTRODUCTION
In previous chapters, series and parallel RL and RC circuits with ac voltage sources have been discussed, and you learned several useful techniques for calculating values in RL and RC circuits. In this chapter, discussion will concern the analysis of more complicated circuits consisting of series and parallel combinations of resistance, inductance, and capacitance. These circuits are called RLC circuits. In this chapter, you will apply the techniques you have already learned to determine RLC circuit values.
%%%
EA
'
( Figure 11.I Typical Series RLC Circuit
I I I
SERIES R L C CIRCUITS SUMMARIZED
The first circuit combination of resistance, capacitance and inductance to be analyzed is one that is connected in series. It is shown in Figure 11.1, and is called a series RLC circuit.
~..,B
-=-
a
i
I
I I I
Phase D i f f e r e n c e s
As in any series resistive-reactive circuit, the simple sum of the voltage drops in the circuit does not equal the applied voltage. EA ~: ER + EL + Ec
(11-1)
Recall that this occurs because of the different phase relationships between the voltage and current for each component.
Figure 11.2 Current Is the Same Throughout a Series Circuit I
r~
Current as a R e f e r e n c e
Recall that in a series circuit as shown in Figure 11.1, the current is the same t h r o u g h o u t the circuit. Therefore, as shown in Figure 11.2, it will be used as the reference quantity when discussing the phase relationships of the voltages in the circuit.
I I I I I I I
v
I
v
ER
D e v e l o p m e n t o f a Circuit Phasor Diagram
Mso recall that as shown in Figure 11.3 the voltage across the resistor, ER, is in phase with the current passing through it. This is the circuit current so the voltage across the resistor is in phase with the circuit current. BASIC AC CIRCUITS
Figure 11.3 VoltageAcross the Resistor, ER
371
RLC
CIRCUIT ANALYSIS
I1 Voltage Phasor Comparisons I1 Calculations of Ex The voltage across a capacitor, Ec, lags the current through it by 90 degrees, as in any capacitive circuit, and as shown in Figure 11.4. The current again is the circuit current so that Ec lags the circuit current by 90 degrees. Finally, the voltage across the inductor, EL, leads the current through it, the circuit current, by 90 degrees as shown in Figure 11.5. Plotting all of these on the same diagram, the phase relationships of the voltages and current in this series RLC circuit can be compared as shown in Figure 11.6.
Voltage Phasor Comparisons As you can see in Figure 11.6, the voltage across the inductor, EL, leads the voltage across the resistor, ER, by 90 degrees and leads the voltage across the capacitor, Ec by 180 degrees. Because of the 180-degree phase difference between the voltage across the inductor, EL, and the voltage across the capacitor, Ec, these two reactive voltages are opposite in phase, as shown in Figure 11.7, and one value partially cancels the effect of the other.
I I v
v
I
ER
Ec Figure 11.4 Voltageand Current Relationship of Capacitance
EL I-
v
v
I
I I
ER
Calculations of Ex Because of the partial cancellation, a net reactive voltage, Ex, exists which is equal to the difference between the two reactive voltages, EL and Ec as shown in Figure 11.8. If EL is larger than Ec, the net reactive voltage is in phase with EL. A positive Ex indicates that it is in phase with EL. If Ec is larger than EL, the net reactive voltage is in phase with Ec. A negative Ex indicates that it is in phase with Ec. The sign of Ex indicates its direction on the Y(reactive) axis.
Figure 11.5 Voltageand Current Relationship of Inductance
EL v
I
v
ER
Ec Figure 11.6 Phase Relationships in Example RLC Series Circuits 372
BASICAC CIRCUITS
RLC CIRCUIT ANALYSIS
1 Calculation of EA II Calculation of Circuit Voltage Values
111
Calculation of EA If EL and Ec are known, it is possible to calculate the applied voltage, EA. For example, assume that EL is larger than Ec. The net reactive voltage, Ex, will, therefore, be in phase with EL as shown in Figure 11.8. EA is then calculated by extending the phasor EA from the origin as shown in Figure 11.9. Note that if the phasor Ex is shifted to the right so that it extends between the tip of ER and the tip of EA, the right triangle for vector addition becomes apparent. By the Pythagorean theorem, EA equals the square root of ER squared plus Ex squared: EA = 4 E R 2 + Ex 2
E ~0 ~ L
n n
n ~" m ~ ~ r
Ec Figure 11.7 EL and Ec Partially Cancel the Effects of the Other
(11-2)
Since Ex is the net difference between EL and
EL
Ec,
Ex = E L - Ec
..~
(11-3)
Ex
Substituting this expression for Ex into equation 11-2, the applied voltage is equal to the square root of ER squared plus EL minus Ec quantity squared as shown in equation
11-4. EA -- 4 E R 2 +
(EL -- E c ) 2
ic
(11-4)
This equation described the relationship between the voltages present in the series RLC circuit.
Figure 11.8 Phase of Ex when EL Is Greater than Ec
Calculation of Circuit Voltage Values Since, by Ohm's law, E = IR
EL
(11-5)
Ex
the voltage across the resistor is the current through it times the resistance. .
ER = IR
(11-6)
.
.
.
Ex
ER
Because IR -- IL =
Ic = IT
in a series RLC circuit, ER can be expressed: ER = BASIC AC CIRCUITS
ITR
(11- 7)
Figure 11.9 Phase Relationship of Ex, ER, and EA 373
PxLC CIRCUIT ANALYSIS
II Calculations of Total Reactance
Also, recall that the voltage across an inductor, EL, is equal to the current through the inductance times the inductive reactance. Therefore EL = ILXL
4 XL
(11-8)
and substituting
.
.
.
.
IL = IT, EL = ITXL
(11-9)
The voltage across the capacitor equals the capacitive current times its capacitive reactance: Ec = I c ~
(11-10)
Ic = IT, Ec = ITXc
(11-11)
Xc Figure 11.10 Substituting IR and Ix Values
and substituting
These equivalent IR and IX quantities can be substituted for the voltages of Figure 11.6 they represent on the voltage phasor diagram as shown in Figure 11.10. Then by factoring out the common term of total current, the impedance phasor diagram is formed as shown in Figure 11.11. Calculations of Total Reactance
In a series RLC circuit, the inductive reactance and capacitive reactance are 180 degrees out of phase as shown in Figure 11.11. Because of this, the two reactive quantities are opposite in phase and one partially cancels the effect of the other. The result is a total reactance, called XT, which is equal to the difference between the inductive reactance and the capacitive reactance values: XT =XL--Xc
XL ~~,~~176~R ....,
. . , .
Xc Figure 11.11 XL and Xc Are 180 Degrees Out of Phase If in the voltage phasor diagram, EL is larger than Ec to make Ex positive, then to correspond, XL would be larger than Xc to make XT positive on the impedance phasor diagram.
(11--12)
If XL is larger than Xc, the net total reactance is in phase with XL. This would be indicated by a positive quantity for XT. On the other hand, if Xc is larger than XL, the net total reactance is in phase with Xc. This would be indicated by a negative XT. The sign of XT indicates its direction on the Y(reactive) axis. 374'
BASIC AC CIRCUITS
R L C CIRCUIT ANALYSIS
II Calculation o f Z T II Calculation o f IT II Calculation o f the Phase Angle
9 I
Calculation of Zr Recall that the total opposition to the flow of ac current in a resistive-reactive circuit is called impedance, Z, as shown in Figure 11.12. The total impedance of a series RLC circuit is equal to the vector sum of the total resistance and net total reactance.
XL , XT
Z
ZT is vector sum o f RT
(11-13)
plus X T
Note in Figure 11.12 that if phasor XT is shifted to the right so that it extends between the tip of Z and the tip of R, the right triangle for vectorial addition appears. By the Pythagorean theorem, the total impedance of the circuit is: ZT = ~ R 2 + XT 2
Xc Figure 11.12 Impedance Phasor Diagram
r
(11--14)
~x
Since XT is the net difference between XL and Xc, per equation 11-12, substituting this expression for XT into equation 11-14 gives equation 11-15 for the total impedance. ZT = ~/R 2 + (Xg -- Xc) 2
Calculation of
L
|x
ER
(11-15)
IT
Recall that the total current in a series RLC circuit is simply equal to the applied voltage divided by the total impedance of the circuit (Ohm's law for ac circuits). EA
I T --'~
ZT
(11--17)
Once the total current is known, the voltage drops across each component may be determined by Ohm's law as stated in equations 11-7, 11-9, and 11-11. Calculation o f the Phase Angle
Recall that the phase angle is defined as the phase difference between the total applied voltage and the total current being drawn from that voltage supply. Returning to the voltage phasor diagram for the series RLC BASIC AC CIRCUITS
k_
Figure 11.13 VoltagePhasor Diagram circuit, shown in Figure 11.13, remember that the total current in the circuit is used as a reference for determining phase relationships in the circuit. This total current is in phase with the voltage across the resistor. The circuit phase angle between the total applied voltage and the total current drawn from the voltage supply is represented by the angle theta. The tangent of any angle, theta, of a right triangle is: tan 0 = opposite side adjacent side
(11-17)
375
RLC CIRCUIT ANALYSIS
I1 Alternative Methods of Calculating the Phase Angle II Calculation of Series RLC Circuit Power Values For the specific angle of Figure 11.13, tan 0 = Ex ER
EL (11-18)
The arctangent of this ratio equals the value of the phase angle. Ex 0 = arctan ~ ER
(11--20)
The arctangent of this ratio also equals the value of the phase angle:
R
(11-21)
Calculation of Series RLC Circuit Power Values
Power calculations in series and parallel RLC circuits are performed in a similar manner to power calculations in series and parallel RL and RC circuits. The primary difference is that both inductive and capacitive reactive power are involved.
376
lc
a b Figure 11.14 Phase Angle Related to Voltage and Impedance Phasor Diagrams
Since the impedance phasor diagram is proportional to the voltage phasor diagram by a factor of the total current, the phase angle is also equal to the phase difference between the total impedance of the circuit and the resistance as shown in Figure 11.14. For the impedance phasor diagram, the tangent of the phase angle is, using equation 11-17 again and substituting for opposite and adjacent,
XT 0 = arctan~
Ec
(11-19)
Alternative Methods of Calculating the Phase Angle
tan 0 = XT R
XL
As you know, power is voltage times current or P = EI. The real power, PR, in watts dissipated by the resistor, is equal to ER times IR. IR is equal to IT, the total circuit current of the series circuit. Similarly, the reactive power of the inductor, PL in VAR, equals EL times IT and the reactive power of the capacitor, Pc in VAR, equals Ec times IT. In series RLC circuits, the power phasor diagram is proportional to the voltage phasor diagram by a factor of the total current as shown in Figure 11.15. The inductive power phasor and the capacitive power phasor are out of phase by 180 degrees. Thus, the resultant reactive power, which is designated Px, is equal to the difference between the inductive reactive power and the capacitive reactive power as shown in Figure 11.16. If PL is larger than Pc, the net reactive power is in phase with PL. If Pc is larger than PL, the net reactive power is in phase with Pc. The sign of Px will determine its ultimate vectorial direction.
BASICAC CIRCUITS
RLC CIRCUIT ANALYSIS
/I
i Calculation o f Series RLC Circuit Power Values
Recall that the total power in a resistivereactive circuit is the apparent power measured in volts-amperes, and is the vector sum of the resistive or real power and the net reactive power. Note in Figure 11.17 that if the phasor Px is shifted to the right so that it extends between the tip of the total apparent power and the tip of the real power, the right triangle for vector addition appears. By the Pythagorean theorem: PA - 4PR2 + Px2
(I 1-22)
Since Px is equal to the difference between PL and Pc,
Px = P , - Pc
/
"~ERIT Ec IT
~
~ PR
Pc
a
b
Figure 11.15 Voltageand PowerPhasor Diagrams
PL
Px Pc
PR
(11-24)
which is the relationship between the different types of power in a series RLC circuit.
Figure 11.16 Calculation of Total Reactive Power
PL (VAR)
The total apparent power can also be expressed in P = EI form as PA = EAIT
PL /
(11-23)
Substituting this expression for Px into equation 11-22, the total apparent power is: PA = 4 P a 2 +(PL -- Pc) 2
A
ELIT
(11--25)
where EA is the applied voltage and IT the circuit current. The apparent power calculated by using one of these two methods should be identical to the apparent power calculated by using the other method.
rx
(VAR)PC1
PA(V-A)
PR(WATTS)
Figure 11.17 Total Apparent PowerPhasor Diagram
BASICAC CIRCUITS
377
II Resistance and Reactance Phasor Diagram
RLC CIRCUIT ANALYSIS
1 Calculation of Z T 1 Calculation of IT II Calculation of Circuit Voltage Values
ANALYSIS OF A SERIES R L C CIRCUIT
Figure 11.18 is a series RLC circuit with an applied voltage of 50 volts, 20 ohms resistance, 45 ohms inductive reactance, and 30 ohms capacitive reactance. Impedance, voltage, current, and power calculations will now be made for this circuit using the techniques described.
R=20~ EA -
XLffi
-;;
45Q
50V I(
Resistance and Reactance Phasor Diagram
At this point, it is useful to sketch the resistance and reactance phasor diagram to help visualize the relationships between the resistive and reactive quantities of the circuit. Figure 11.19 shows such a diagram. Note that XL is larger than Xc; thus, the net total reactance is in phase with XL and is, using equation 11-12,
X c = 30Q
Figure 11.18 Series RLC Circuit Example
XL 45f~
XT=XL--Xc = 45f2 - 3 0 ~ = 15f~
Calculation of ZT The total impedance, ZT, is equal to
....
R 20f2
ZT = a/R2 + XT 2 = 4202 + 152
Xc
30~
= 4400 + 225 = 4625
Figure 11.19 Resistance and Rectance Phasor Diagram
=25f~
Calculation of Circuit Voltage Values Calculation of IT By using Ohm's law for ac circuits the total current, IT, c a n be calculated. IT =
EA ZT 50
25 =2A
Using Ohm's law again, the voltage drops across each of the components in the circuit can be calculated. The voltage across the resistor, ER, is calculated: ER -- I T R
= (2 A) (20 a ) =40V
378
BASIC AC CIRCUITS
RLC CIRCUIT ANALYSIS
9 Calculation of ET 9 Calculation of Phase Angle l Alternate Method of Calculating Phase Angle
11
The voltage across the inductor, EL, is calculated: EL = ITXL =
ER= 4 0 V
(2 A) (45 ~ ) EA
=90 V
The voltage across the capacitor, Ec, is calculated:
-"
";;"
EL -
90V
50V
,,(
Ec = ITXc =
'
Ec - 60 V
(2 A) (30 f~)
=60V
Figure 11.20 Calculated VoltageDrops in Example
Calculation of Er
At first, as shown in Figure 11.20, this result appears impossible since there seems to be more voltage in the circuit than the applied voltage, EA. However, you must keep in mind that the voltage across the inductor and the voltage across the capacitor are 180 degrees out of phase and one partially cancels the other. Therefore, the applied voltage, EA, is calculated using equation 11-4. EA = ~/ER2 + (EL
- -
Ec) 2
= ~402 + (90 - 60) 2 =
4402
+
302
= 41,600 + 900 = 42,500 =50V
This calculation confirms the original applied voltage value given in the example. And this method gives you a valuable check of the accuracy of the calculation of the individual voltage drops that were calculated for each component in the circuit.
Series RLC Circuit Calculation of Phase Angle
Using the voltage phasor diagram of Figure 11.13, the phase angle of the example series RLC circuit can be calculated:
0 ar tan( )
=ar tan( 0V/ V0 = arctan(0.75) =37 ~ The angle whose tangent is 0.75 is about 37 degrees. Alternate Method of Calculating Phase Angle
The same result may be obtained using the impedance phasor diagram. In this case the arctangent is XT divided by R. 0= arctan(-~) = arctan(15 ~ ] k.20 n ) = arctan(0.75) =37 ~
BASIC AC CIRCUITS
379
II II II II
Positive and Negative Phase Angles Calculation of Component Power Values Calculation of Total Reactive Power Calculation of Total Apparent Power
RLC CIRCUIT ANALYSIS
Positive and Negative Phase Angles
Calculation of Total Apparent Power
The phase angle calculated for the example series RLC circuit is considered to be a positive phase angle since the phase angle for the circuit is measured counter-clockwise. Thus, the phase angle for the circuit is stated as being a positive 37 degrees.
By equation 11-22, the apparent power, PA, is equal to the square root of PR squared plus Px squared:
The phase angle in a series RLC circuit can either be positive or negative. If XL is greater than Xc, the angle is positive. On the other hand, if Xc is greater than XL, the angle is negative.
Calculation of Component Power Values In the example series RLC circuit, the voltage drops in the circuit are: 40 volts across the resistor, 90 volts across the inductor and 60 volts across the capacitor. The applied voltage is 50 volts and a total current of 2 amperes is flowing. Multiplying the voltage times the current for each component yields these individual power values: Pg = ERIT = (40 V) (2 A) = 80 W PL = ELIT = (90 V) (2 A) = 180 VAR Pc = EclT = (60 V) (2 A) = 120 VAR
Calculation of Total Reactive Power The net reactive power, Px, is equal to the difference between PL and Pc, or 180 VAR minus 120 VAR which equals 60 VAR. Since PL is larger than Pc, Px is positive and is in phase with PL.
PA =
4PR 2 + Px2
= ~/802 + 602
= 46, 400 + 3, 600 = 410, 000 = 100 VA The total apparent power of the example series RLC circuit is 100 volt-amperes. This should be the same as the value obtained by multiplying the applied voltage, EA, by the total current, IT. PA =EAIT
= (50 V)(2 A) = (100 v a )
It is, and proves the accuracy of the first calculation. PARALLEL RLC CIRCUITS SUMMARIZED
Now that all calculations for a series RLC circuit have been made, the method of solution of a parallel RLC circuit will be discussed. The analysis of this circuit will be similar to the analysis of either a parallel RL or a parallel RC circuit. In a parallel RLC circuit, such as in Figure 11.21, the sum of the branch currents is not equal to the total current as it would be in either a purely resistive, a purely inductive, or purely capacitive circuit. That is, IT ;e IR + IL + Ic
(11-26)
This occurs because of the different phase relationships between the voltage and current for each component. Recall that this was also true for RL and RC parallel circuits.
380
BASIC AC CIRCUITS
RLC
9 Voltage as a R e f e r e n c e 9 D e v e l o p m e n t o f a Circuit P h a s o r D i a g r a m 9 Calculations of Branch Currents
CIRCUIT
11
ANALYSIS
Voltage as a R e f e r e n c e
Since the voltage across all components of a parallel circuit is the same as the applied voltage, EA, it will be used as the reference quantity in discussing the phase relationships of the currents in the circuit. Therefore,
Ir
IL
A
llc R
EA = ER = EL = E c .
L TC
D e v e l o p m e n t o f a Circuit P h a s o r D i a g r a m
Recall that the current through a resistor, IR, is in phase with the voltage across it, ER. The capacitive current, Ic, leads the voltage across the capacitor, EA, by 90 degrees; and the current through an inductor, IL, lags the voltage across it, EA, by 90 degrees. Thus, the phase relationship of the applied voltage and currents in a parallel RLC circuit are as shown in Figure 11.22.
Figure 11.21 Parallel RLC Circuit Example
Ic
Calculations of Branch Currents
The individual branch currents in the example RLC circuit can be calculated as they are in either a purely resistive, purely capacitive, or purely inductive circuit. Simply divide the voltage across the branch by the opposition to current in the branch. In the resistive branch the opposition to the flow of current is measured in ohms of resistance. The resistive current is determined by dividing the applied voltage by the value of the resistor:
EA
IR= ~ R
(11-27)
In the inductive branch, the opposition to the flow of current is measured in ohms of inductive reactance. The inductive current is determined by dividing the applied voltage by the reactance of the inductor:
EA
IL = - -
XL
BASIC AC CIRCUITS
v
la
v
Ir
IL Figure 11.22 Phase Relationships of Ea and I in a Parallel RLC Circuit
In the capacitive branch, the opposition to the flow of current is measured in ohms of capacitive reactance. The capacitive current is determined by dividing the applied voltage by the reactance of the capacitor: EA
Ic = ~ Xc
(11-29)
(11-28)
381
RLC CIRCUIT
ANALYSIS
II Current Phasor Comparisons II Calculation of Ix Current Phasor Comparisons In the parallel RLC circuit, since EA = ER = EL - Ec, and since IR is in phase with ER and thus all the voltages EL, Ec, and EA, IR becomes the reference vector for the current phasor diagram of Figure 11.23. Comparing the current phase relationships using the current phasor diagram of Figure 11.23, you can see that the capacitive branch current, Ic, leads the resistive branch current, IR, by 90 degrees; the inductive branch current, IL, lags IR by 90 degrees. Because of the 180-degree difference between the capacitive branch current and the inductive branch current, these two reactive current values are opposite in phase, and one partially cancels the effect of the other.
[r IX ll
ll
ll
Ix = ~]IR 2 + Ix 2
(11-30)
Ic
382
IT
,x
IL Figure 11.24 Current Phasor Diagram with Iv Phasor
Ix = I c - L
(11-31)
Ix is equal to Ic minus IL. Substituting this expression for Ix into equation 11-30, the total current is equal to
IT 4IR 9 + (Ic - IL) 2 =
Sine Ix is the net difference between Ic and IL, as shown in equation 11-31,
I
Figure 11.23 Current Phasor Diagram for a Parallel RLC Circuit
Calculation of Ix
Applying the Pythagorean theorem,
IR
It.
The difference between Ic and IL is Ix, which is the net reactive current. If Ic is larger than IL, the net reactive current is in phase with Ic and the value of Ix is positive. If, on the other hand, IL is larger than Ic, the net reactive current is in phase with IL and the value of Ix is negative. The sign of Ix indicates its direction on the Y(reactive) axis. By adding the total current vector, Ix, to the current phasor diagram, and shifting vector Ix to the right so that it extends between the tip of IR and the tip of Ix, a right triangle is produced. This is shown in Figure 11.24. The right triangle can then be used to add the phasors IR and Ix vectorially.
,180~ [A
(11-32)
which is the relationship between the branch currents in a parallel RLC circuit.
BASIC AC CIRCUITS
RLC CIRCUIT ANALYSIS
9 Calculation of Zr
, I
9 Calculation of Phase Angle 9 Calculation of Power I
Calculation of Zx Once the total current is known, the total impedance, ZT, of the circuit is easily determined using Ohm's law for ac circuits: EA
zT = ~
IC
Ix
IT
(11-33)
Ix
IL
Calculation of Phase Angle
Recall that the phase angle is the n u m b e r of degrees of phase difference between the applied voltage and the total current. Also recall that the applied voltage is in phase with the resistive current. Therefore, the phase angle is measured from the horizontal vector IR (which is also EA) tO the vector IT; it is identified by the angle, theta, on the current phasor diagram of Figure 11.25. The tangent of the phase angle, theta, is equal to the ratio of the net reactive current divided by the resistive current as shown in equation 11-34: tan 0 = I--Ex IR
(11-34)
O= arctan(~RX )
(11-35)
i: A I R
Figure 11.25 PhaseAngle on the Current Phasor Diagram
Pc
Px
Px
PLI
ex
PR
Figure 11.26 TotalApparent PowerPhasor Diagram
Therefore,
the arctangent of the net reactive current divided by the resistive current equals the value of the phase angle.
The total apparent power is found by using the Pythagorean theorem solution of resistive and net reactive power shown in the power phasor diagram of Figure 11.26 defined by these equations:
Calculation of Power
PA = ~/pg2 + px 2
(11-39)
Power calculations in parallel RLC circuits are the same as power calculations in series RLC circuits. The individual power values are calculated by multiplying current through a component times voltage across a component.
Px = P c - PL
(11-40)
PA = ~/pg2 + (Pc - PL)2
(11-41)
Therefore, PR = ERIR = EAIR PL = ELIL = EAIL Pc = EcIc = EAIc BASICAC CIRCUITS
(11--36) (11--37) (11-38) 383
RLC CIRCUIT ANALYSIS II Calculation of Branch Currents II Current Phasor Diagrams
ANALYSIS OF A P A R A L L E L R L C CIRCUIT
Figure 11.27 shows a typical parallel RLC circuit with 180 VAC, 60 kilohms resistance, 30 kilohms inductive reactance, and 18 kilohms capacitive reactance. Impedance, voltage, current, and power measurements will be performed for this circuit using the techniques just described.
R=60kQ
EA 18
'lXc = 8kQ
Calculation of Branch Currents
XL=3OkD
The resistive branch current, IR, is calculated by dividing the voltage across the resistor, 180 volts, by the value of the resistor, 60 kilohms.
Figure 11.27 Typical Parallel RLC Circuit
EA
IR = ~
R 180 V
60 lffl
Ic *I IT Ix . . . . . . . ~,lx i
=3mA The inductive branch current is determined in a similar manner by dividing the voltage across the inductor by the inductive reactance.
I
.....
IR
EA IL ~ m
XL 180 V
IL
30 kD
=6mA The capacitive branch current is calculated by dividing the voltage across the capacitor by the capacitive reactance. EA Ic - - m
Xc 180 V 18k~
=10m~
Figure 11.28 Current Phasor Diagram for Parallel RLC Circuit Example
Current Phasor Diagrams
The current phasor diagram can now be drawn to show the relationships between the resistive, inductive, and capacitive branch currents as shown in Figure 11.28. Note that Ic is larger than IL; therefore, the net reactive current is in phase with Io The net reactive current, Ix, equals Ix = Ic - IL = 1 0 m A - 6mA
=4mA 384
BASIC AC CIRCUITS
RLC CIRCUIT ANALYSIS
I Calculation o f IT I Calculation o f Z 1 Calculations o f the Phase Angle
Calculation of IT As you can see in Figure 11.28, by shifting the phasor Ix to the right so that it extends from the tip of the vector representing IR to the tip of the vector for IT, a right triangle is formed. The right triangle can then be used to determine the value of IT. Using the Pythagorean theorem, the total current, IT, is now equal to the vector sum of the values of Ix and IR.
I c 10 mA 4mA
Ix
=49+16
~
I T 5mA v
IR 3mA
IT = ~/IR2 + IX2
= 432 + 42
11
IL6mA Figure 11.29 Current Phasor Diagram for Calculating Theta
=4-~g =5mA The total current in the circuit is 5 milliamperes.
[c
Ix~~IT,,.
Ic
Calculation o f Z The total impedance of this circuit can now be calculated by dividing the applied voltage by the total current. Z=~
IR
IL
Ix
I "
EA
IL
IT =
180 V ~
5mA =36k~
a
b
Figure 11.30 Current Phasor Diagramsfor Positive and Negative Thetas
Calculations o f the Phase Angle
The phase angle, theta is the angle between the resistive current and the total current as shown on the current phasor diagram in Figure 11.29. The value of the phase angle is equal to the arctangent of the ratio of net reactive current to the resistive current. Therefore, 0 = arctan(Ix ~ k, I R )
-artan/mama/
Thus, the phase angle of this parallel RLC circuit is approximately 53 degrees. The phase angle is described as a positive phase angle because the total current phasor is rotated counter-clockwise from the reference. However, the phase angle of parallel RLC circuit may be either positive or negative depending upon the relationships of IL and Ic. If Ic is greater than IL, the phase angle theta will be positive as shown in Figure 11.30a. If IL is greater than Ic, theta will be negative as shown in Figure 11.30b.
= arctan(1.33) =53 ~ BASIC AC CIRCUITS
385
RLC CIRCUIT ANALYSIS
II Calculation of Component Power Values II Calculation of Total Reactive Power II Calculation of Total Power
l
Calculation of Component Power Values
l
II
l
II
As you know power values depend on the various voltages and currents in the circuit. In the parallel RLC circuit example shown in Figure 11.31, currents are: I~ equals 3 milliamperes, IL equals 6 milliamperes, and Ic equals 10 milliamperes. The applied voltage is 180 volts; the total current is 5 milliamperes.
Figure 11.31 Branch Currents in Parallel RLC Circuit Example
The real power dissipated by the resistor, P~, is calculated:
Calculation of Total Power
Pp. =EAIR = (180 V)(3 mA) = 540 mW The inductive reactive power, PL is calculated: PL =EAIL
180V
A
Calculation of Total Reactive Power
Ic 10mA
= ,1540 m W ~ + 720 mVAR 2
= 4291,600 + 518, 400 = 900 mVA
= 1,800 mVAR
C
PA = 4PR 2 + Px2
= 4810,000
= (180 V) (10 mA)
L
The apparent total power, PA, is calculated using equation 11-39:
= 1,080 mVAR
Pc =EAIc
' I
It. IR 3mA 6mA
= (180 V) (6 mA)
The capacitive reactive power, Pc, is calculated:
,R
This value should be the same as the apparent power value obtained by multiplying the applied voltage by the total current:
PA =EAIT = (180 V)(5 mA) = 900 mVA
The net reactive power, Px is calculated: Px = P c - PL = 1,800 mVAR - 1,080 mVAR = 720 mVAR The power phasor diagram will look like the one in Figure 11.26. Since Pc is larger than PL, Px is positive and is in phase with Pc.
386
BASICAC CIRCUITS
RLC
II
Summary I
I
I
II
I
CIRCUIT ANALYSIS
9 I
SUMMARY
In this chapter, the concepts and techniques you learned in analyzing RL and RC circuits were applied to determine circuit values in RLC circuits. You were shown how to determine current, voltage, reactance, and the phase angle for any RLC circuit. Also, a new concept was i n t r o d u c e d - that when values are in opposite phase, one value partially cancels the effect of the other. This difference is called the net effect. The net voltage, Ex, net total reactance, XT, and net reactive current Ix, were calculated. The methods you learned in this chapter should enable you to determine equivalent circuit values in any series or parallel RLC circuit.
BASIC AC CIRCUITS
387
RLC CIRCUIT ANALYSIS ll Worked-Out
Examples
1. Draw a phasor diagram for this circuit showing current and impedance phasors.
XL : 10~
Xc = 6N Solution:
X L > Xc;
therefore, XT is plotted in phase with
XL
XT
R I(all) Xc
2. Draw voltage phasor diagrams for the circuit in Example 1. Calculate the value and sign of the phase angle. Solution:
tan
XT
= ~
=
4
R 6 arctan 0.667 = 33.7 ~
EXT
Erl
EXT
0 = ~
ER
=0.667
E.
The angle is rotated counter-clockwise; therefore, the sign of the angle is positive.
388
BASICAC CIRCUITS
RLC CIRCUIT ANALYSIS
II W o r k e d - O u t E x a m p l e s
/I
3. Draw p h a s o r d i a g r a m s showing P y t h a g o r e a n t h e o r e m r e l a t i o n s h i p s in parallel RLC circuits. Write e q u a t i o n s for e a c h solution.
Solution:
Ix
PT
Px
Pa
In
IT = N / i a 2 + I x ~ = X/I'R' + ( I L -
PT =
Ic)'
N/PR 2 + Px 2
= N/PR' + (Pt.-
Pr
4. Given the circuit a n d typical circuit values shown, calculate the circuit values specified.
R = lOft
f-
.1,o
.....T L : 0.25mH
a.
XL =
h.
ER =
b.
Xc=
i.
PL--
C. XT--
j.
Pc =
d.
ZT=
k.
PR =
e.
IT =
1.
Px -
f.
EL =
m.
PA--
g.
E C
n.
Phase angle =
--
Solution: a.
X L = 27rfL = 6.28 • 5 x 103 x 0.25 x 10 -~ = 7.85 f l
b.
Xc .
C.
XT = XL -- X c = 7.85 f2 - 15.9 f~ = 8 . 0 5
d.
ZT =
e.
IT
f.
EL = I T X L = 0.781 A x 7.85 f2 = 6.13 VAC
g.
Ec = I T X c = 0.781 A x 15.9 f2 = 12.4 VAC
1
2xfC
.
1
. . 6.28•215215
4R 2 +
XT 2 =
EA
10 VAC
= ~
ZT
=
BASIC AC CIRCUITS
12.8f~
1
-6
6 . 2 8 x 1 0 -2
= 15.9 f l
4102 +8.052 = 4i~65 • i02 = 12.8 f l = 0 . 7 8 1 Arms
389
RLC
CIRCUIT ANALYSIS
II W o r k e d - O u t E x a m p l e s I
I
h.
ER = ITR = 0.781A x 10 f~ = 7.81VAC
i.
PL = ITEL = 0 . 7 8 1 A x 6.13 V = 4.79 V A R ~
j.
Pc = ITEc = 0.781 A x 12.4 V = 9.68 VAR~s
k.
PR = ITER = 0.781 A x 7.81V - 6.1 W ~
1.
Px = PL - Pc - 4.79 V A R - 9.68 VAR = 4.89 V A R ~
m. PA = a/PR2 + Px 2 = ~/6.1W 2 +4.89 VAR 2 = 64~6~.11 = 7.82 VAn.~ n.
phase a n g l e = a r c t a n ( - - ~ - ) = a r c t a n 0 . 8 0 5 = - 3 8 . 8 ~ kKJ
5. Given this circuit a n d typical circuit values shown, calculate the circuit values specified.
--
A
I= ~
a.
X L ~-~
h"
PR
b.
Xc=
i.
PL =
C.
IR =
j.
Pc =
k.
Px = PA =
lOOHz
~k~l
~5;
O.eS~l~T d" IL =
%~k~ -
"
e.
Ic =
1.
f.
Ix
=
m. Phase angle =
g,
IT
--
Solution: a.
X L = 2~zfL = 6.28 x 100 x 15 = 9.42 kD.
b.
X c = ~ 2~fC
C.
IR .
1
6.28 x 100 x 0.5 x 10 .6
Ea 50 VAC . . . . R
390
3.14 x 10 -4
10 mAtins
5 kD.
d.
IL .
EA 50 VAC . . . . XL 9.42 k.Q
5.31 m A r ~
e.
Ic
EA . Xc
15.7 rearms
f.
Ix = I L - I c = 5 . 3 1 m A - 1 5 . 7 m A = 1 0 . 4 m A
.
= 3.18 k ~
.
50 VAC . 3.18 k ~
BASIC AC CIRCUITS
RLC
CIRCUIT ANALYSIS
II Worked-Out Examples I
I
I
11 I
I
I
4IR 2 + Ix 2 = 4 i 0 m A 2 + 10.4 m A 2 = 14.4 mA,m~
g.
IT
=
h.
PR
=IREa=lOmA+5OV=O.5Wrms
i.
PL = ILEA = 5.31 m A + 50 V = 0.27 V A R ~
j.
Pc
k.
Px = PL - P c = 0 . 2 7 V A R - 0 . 7 9
1.
I
= IcEA = 15.7 m A + 5 0 V = 0 . 7 9 V A R ~ s V A R = 0 . 5 2 VAR~ms
PA = 4PR 2 + Px 2 = 4 0 . 5 W 2 + 0.52 V A R 2 = 040.52 = 0.72 VArm, o r Pa
= ITEA = 14.4 m A x 50 V = 0.72 VA,n,
m. t a n 0 = I___~_x= ~ 1 0 " 4m A = 1.04 IR 10mA a r c t a n l . 0 4 = 46.1 ~
BASIC AC CIRCUITS
391
RLC CIRCUIT ANALYSIS
I Practice P r o b l e m s
1. Draw a phasor diagram for this circuit showing voltage and current phasors.
/•
IR = 15mA
..
8Cm~
. r
2. Draw power phasor diagrams for the circuit in problem 1. Calculate the value and sign of the phase angle. 3. Draw phasor diagrams showing Pythagoren theorem relationships in series RLC circuits. Write equations for each solution. 4. Given the circuit and typical circuit values below, calculate the circuit values specified. t=SmH
a. X L =
g. E c =
b. Xx=
h. PR=
c. Z T - -- -
i.
P L - ---
d. IT =
j.
Pc =
e. ER =
k. PA =
f. EL
1. Phase angle
t-2.SRH
R=
E, =
1501"1
2or
xc:2oon
=
=
5. Given this circuit and typical circuit values shown, calculate the circuit values specified.
A
2'kzL I
392
A
a. X c =
g. PR =
b. IR =
h. PL =
c. IL =
i. Pc =
d. Ic =
j.
e. I x =
k. PA =
f. IT =
1. Phase angle =
.
Px =
BASIC AC CIRCUITS
RLC CIRCUIT ANALYSIS
11
a
I 1. Draw a phasor diagram for this circuit showing power and c u r r e n t phasors.
a. X c = b. XT=
Xc = lkfl
d. IT =
g. h. i. j. k. 1.
C. ZT-"
I
X, = 1.5kfl
R=
e.
ER =
f.
EL =
Ec= PR = PL =
Pc= PA = Phase angle
8. Given this circuit and typical values shown, calculate the circuit values specified.
lkfl
2. Draw a phasor diagram for this circuit showing power and voltage phasors.
E,0; ~ ~ , /
3. Calculate the value and sign of the phase angle in the circuit of Problem 1. 4. Calculate the value and sign of the phase angle in the circuit of Problem 2. 5. Calculate the value of ZT in Problem 1. 6. Calculate the value of IT in Problem 2.
a. b. c. d. e. f.
XL= IR = IL = Ic = IT = PR =
1Okfl
.io.!.
XC--'-- i
g. h. i. j. k.
PL = Pc = Px = PA = Phase angle =
9. In a series RCL circuit, which c o m p o n e n t will have the most voltage across it? 10. In a parallel RCL circuit, which c o m p o n e n t will have the most c u r r e n t t h r o u g h it?
7. Given this circuit and typical circuit values shown, calculate the circuit values specified. XL : 40fl
R = SOft BASIC AC C I R C U I T S
393
,~ RLC Circuit Analysis In this chapter, series and parallel circuits with resistors, capacitors, and inductors are analyzed. Many of the same techniques used in the solution of resistivecapacitive and resistive-inductive circuits are used in this analysis. Phasor diagrams provide descriptions of the circuits that lead to Pythagorean theorem solutions of certain circuit values.
BASIC AC CIRCUITS
369
RLC CIRCUIT ANALYSIS
II
Objectives
I
I
At the end of this chapter you should be able to: 1. Draw phasor diagrams showing the phase relationships of various circuit values in series and parallel RLC circuits. 2. Identify the various circuit values in series and parallel RLC circuits that can be determined by Pythagorean theorem analysis. 3. Identify the positive and negative phase angles in series and parallel RLC circuits. 4. Calculate total reactance, total reactive current, total reactive voltage, and total reactive power in RLC circuits. 5. Given schematic diagrams and typical circuit values for the circuits below, calculate current, voltage, impedance, and power values.
370
BASIC AC CIRCUITS
RLC CIRCUIT ANALYSIS
II Phase D i f f e r e n c e s 9 Current as a R e f e r e n c e 9 D e v e l o p m e n t o f a Circuit Phasor Diagram
11
INTRODUCTION
In previous chapters, series and parallel RL and RC circuits with ac voltage sources have been discussed, and you learned several useful techniques for calculating values in RL and RC circuits. In this chapter, discussion will concern the analysis of more complicated circuits consisting of series and parallel combinations of resistance, inductance, and capacitance. These circuits are called RLC circuits. In this chapter, you will apply the techniques you have already learned to determine RLC circuit values.
%%%
EA
'
( Figure 11.I Typical Series RLC Circuit
I I I
SERIES R L C CIRCUITS SUMMARIZED
The first circuit combination of resistance, capacitance and inductance to be analyzed is one that is connected in series. It is shown in Figure 11.1, and is called a series RLC circuit.
~..,B
-=-
a
i
I
I I I
Phase D i f f e r e n c e s
As in any series resistive-reactive circuit, the simple sum of the voltage drops in the circuit does not equal the applied voltage. EA ~: ER + EL + Ec
(11-1)
Recall that this occurs because of the different phase relationships between the voltage and current for each component.
Figure 11.2 Current Is the Same Throughout a Series Circuit I
r~
Current as a R e f e r e n c e
Recall that in a series circuit as shown in Figure 11.1, the current is the same t h r o u g h o u t the circuit. Therefore, as shown in Figure 11.2, it will be used as the reference quantity when discussing the phase relationships of the voltages in the circuit.
I I I I I I I
v
I
v
ER
D e v e l o p m e n t o f a Circuit Phasor Diagram
Mso recall that as shown in Figure 11.3 the voltage across the resistor, ER, is in phase with the current passing through it. This is the circuit current so the voltage across the resistor is in phase with the circuit current. BASIC AC CIRCUITS
Figure 11.3 VoltageAcross the Resistor, ER
371
RLC
CIRCUIT ANALYSIS
I1 Voltage Phasor Comparisons I1 Calculations of Ex The voltage across a capacitor, Ec, lags the current through it by 90 degrees, as in any capacitive circuit, and as shown in Figure 11.4. The current again is the circuit current so that Ec lags the circuit current by 90 degrees. Finally, the voltage across the inductor, EL, leads the current through it, the circuit current, by 90 degrees as shown in Figure 11.5. Plotting all of these on the same diagram, the phase relationships of the voltages and current in this series RLC circuit can be compared as shown in Figure 11.6.
Voltage Phasor Comparisons As you can see in Figure 11.6, the voltage across the inductor, EL, leads the voltage across the resistor, ER, by 90 degrees and leads the voltage across the capacitor, Ec by 180 degrees. Because of the 180-degree phase difference between the voltage across the inductor, EL, and the voltage across the capacitor, Ec, these two reactive voltages are opposite in phase, as shown in Figure 11.7, and one value partially cancels the effect of the other.
I I v
v
I
ER
Ec Figure 11.4 Voltageand Current Relationship of Capacitance
EL I-
v
v
I
I I
ER
Calculations of Ex Because of the partial cancellation, a net reactive voltage, Ex, exists which is equal to the difference between the two reactive voltages, EL and Ec as shown in Figure 11.8. If EL is larger than Ec, the net reactive voltage is in phase with EL. A positive Ex indicates that it is in phase with EL. If Ec is larger than EL, the net reactive voltage is in phase with Ec. A negative Ex indicates that it is in phase with Ec. The sign of Ex indicates its direction on the Y(reactive) axis.
Figure 11.5 Voltageand Current Relationship of Inductance
EL v
I
v
ER
Ec Figure 11.6 Phase Relationships in Example RLC Series Circuits 372
BASICAC CIRCUITS
RLC CIRCUIT ANALYSIS
1 Calculation of EA II Calculation of Circuit Voltage Values
111
Calculation of EA If EL and Ec are known, it is possible to calculate the applied voltage, EA. For example, assume that EL is larger than Ec. The net reactive voltage, Ex, will, therefore, be in phase with EL as shown in Figure 11.8. EA is then calculated by extending the phasor EA from the origin as shown in Figure 11.9. Note that if the phasor Ex is shifted to the right so that it extends between the tip of ER and the tip of EA, the right triangle for vector addition becomes apparent. By the Pythagorean theorem, EA equals the square root of ER squared plus Ex squared: EA = 4 E R 2 + Ex 2
E ~0 ~ L
n n
n ~" m ~ ~ r
Ec Figure 11.7 EL and Ec Partially Cancel the Effects of the Other
(11-2)
Since Ex is the net difference between EL and
EL
Ec,
Ex = E L - Ec
..~
(11-3)
Ex
Substituting this expression for Ex into equation 11-2, the applied voltage is equal to the square root of ER squared plus EL minus Ec quantity squared as shown in equation
11-4. EA -- 4 E R 2 +
(EL -- E c ) 2
ic
(11-4)
This equation described the relationship between the voltages present in the series RLC circuit.
Figure 11.8 Phase of Ex when EL Is Greater than Ec
Calculation of Circuit Voltage Values Since, by Ohm's law, E = IR
EL
(11-5)
Ex
the voltage across the resistor is the current through it times the resistance. .
ER = IR
(11-6)
.
.
.
Ex
ER
Because IR -- IL =
Ic = IT
in a series RLC circuit, ER can be expressed: ER = BASIC AC CIRCUITS
ITR
(11- 7)
Figure 11.9 Phase Relationship of Ex, ER, and EA 373
PxLC CIRCUIT ANALYSIS
II Calculations of Total Reactance
Also, recall that the voltage across an inductor, EL, is equal to the current through the inductance times the inductive reactance. Therefore EL = ILXL
4 XL
(11-8)
and substituting
.
.
.
.
IL = IT, EL = ITXL
(11-9)
The voltage across the capacitor equals the capacitive current times its capacitive reactance: Ec = I c ~
(11-10)
Ic = IT, Ec = ITXc
(11-11)
Xc Figure 11.10 Substituting IR and Ix Values
and substituting
These equivalent IR and IX quantities can be substituted for the voltages of Figure 11.6 they represent on the voltage phasor diagram as shown in Figure 11.10. Then by factoring out the common term of total current, the impedance phasor diagram is formed as shown in Figure 11.11. Calculations of Total Reactance
In a series RLC circuit, the inductive reactance and capacitive reactance are 180 degrees out of phase as shown in Figure 11.11. Because of this, the two reactive quantities are opposite in phase and one partially cancels the effect of the other. The result is a total reactance, called XT, which is equal to the difference between the inductive reactance and the capacitive reactance values: XT =XL--Xc
XL ~~,~~176~R ....,
. . , .
Xc Figure 11.11 XL and Xc Are 180 Degrees Out of Phase If in the voltage phasor diagram, EL is larger than Ec to make Ex positive, then to correspond, XL would be larger than Xc to make XT positive on the impedance phasor diagram.
(11--12)
If XL is larger than Xc, the net total reactance is in phase with XL. This would be indicated by a positive quantity for XT. On the other hand, if Xc is larger than XL, the net total reactance is in phase with Xc. This would be indicated by a negative XT. The sign of XT indicates its direction on the Y(reactive) axis. 374'
BASIC AC CIRCUITS
R L C CIRCUIT ANALYSIS
II Calculation o f Z T II Calculation o f IT II Calculation o f the Phase Angle
9 I
Calculation of Zr Recall that the total opposition to the flow of ac current in a resistive-reactive circuit is called impedance, Z, as shown in Figure 11.12. The total impedance of a series RLC circuit is equal to the vector sum of the total resistance and net total reactance.
XL , XT
Z
ZT is vector sum o f RT
(11-13)
plus X T
Note in Figure 11.12 that if phasor XT is shifted to the right so that it extends between the tip of Z and the tip of R, the right triangle for vectorial addition appears. By the Pythagorean theorem, the total impedance of the circuit is: ZT = ~ R 2 + XT 2
Xc Figure 11.12 Impedance Phasor Diagram
r
(11--14)
~x
Since XT is the net difference between XL and Xc, per equation 11-12, substituting this expression for XT into equation 11-14 gives equation 11-15 for the total impedance. ZT = ~/R 2 + (Xg -- Xc) 2
Calculation of
L
|x
ER
(11-15)
IT
Recall that the total current in a series RLC circuit is simply equal to the applied voltage divided by the total impedance of the circuit (Ohm's law for ac circuits). EA
I T --'~
ZT
(11--17)
Once the total current is known, the voltage drops across each component may be determined by Ohm's law as stated in equations 11-7, 11-9, and 11-11. Calculation o f the Phase Angle
Recall that the phase angle is defined as the phase difference between the total applied voltage and the total current being drawn from that voltage supply. Returning to the voltage phasor diagram for the series RLC BASIC AC CIRCUITS
k_
Figure 11.13 VoltagePhasor Diagram circuit, shown in Figure 11.13, remember that the total current in the circuit is used as a reference for determining phase relationships in the circuit. This total current is in phase with the voltage across the resistor. The circuit phase angle between the total applied voltage and the total current drawn from the voltage supply is represented by the angle theta. The tangent of any angle, theta, of a right triangle is: tan 0 = opposite side adjacent side
(11-17)
375
RLC CIRCUIT ANALYSIS
I1 Alternative Methods of Calculating the Phase Angle II Calculation of Series RLC Circuit Power Values For the specific angle of Figure 11.13, tan 0 = Ex ER
EL (11-18)
The arctangent of this ratio equals the value of the phase angle. Ex 0 = arctan ~ ER
(11--20)
The arctangent of this ratio also equals the value of the phase angle:
R
(11-21)
Calculation of Series RLC Circuit Power Values
Power calculations in series and parallel RLC circuits are performed in a similar manner to power calculations in series and parallel RL and RC circuits. The primary difference is that both inductive and capacitive reactive power are involved.
376
lc
a b Figure 11.14 Phase Angle Related to Voltage and Impedance Phasor Diagrams
Since the impedance phasor diagram is proportional to the voltage phasor diagram by a factor of the total current, the phase angle is also equal to the phase difference between the total impedance of the circuit and the resistance as shown in Figure 11.14. For the impedance phasor diagram, the tangent of the phase angle is, using equation 11-17 again and substituting for opposite and adjacent,
XT 0 = arctan~
Ec
(11-19)
Alternative Methods of Calculating the Phase Angle
tan 0 = XT R
XL
As you know, power is voltage times current or P = EI. The real power, PR, in watts dissipated by the resistor, is equal to ER times IR. IR is equal to IT, the total circuit current of the series circuit. Similarly, the reactive power of the inductor, PL in VAR, equals EL times IT and the reactive power of the capacitor, Pc in VAR, equals Ec times IT. In series RLC circuits, the power phasor diagram is proportional to the voltage phasor diagram by a factor of the total current as shown in Figure 11.15. The inductive power phasor and the capacitive power phasor are out of phase by 180 degrees. Thus, the resultant reactive power, which is designated Px, is equal to the difference between the inductive reactive power and the capacitive reactive power as shown in Figure 11.16. If PL is larger than Pc, the net reactive power is in phase with PL. If Pc is larger than PL, the net reactive power is in phase with Pc. The sign of Px will determine its ultimate vectorial direction.
BASICAC CIRCUITS
RLC CIRCUIT ANALYSIS
/I
i Calculation o f Series RLC Circuit Power Values
Recall that the total power in a resistivereactive circuit is the apparent power measured in volts-amperes, and is the vector sum of the resistive or real power and the net reactive power. Note in Figure 11.17 that if the phasor Px is shifted to the right so that it extends between the tip of the total apparent power and the tip of the real power, the right triangle for vector addition appears. By the Pythagorean theorem: PA - 4PR2 + Px2
(I 1-22)
Since Px is equal to the difference between PL and Pc,
Px = P , - Pc
/
"~ERIT Ec IT
~
~ PR
Pc
a
b
Figure 11.15 Voltageand PowerPhasor Diagrams
PL
Px Pc
PR
(11-24)
which is the relationship between the different types of power in a series RLC circuit.
Figure 11.16 Calculation of Total Reactive Power
PL (VAR)
The total apparent power can also be expressed in P = EI form as PA = EAIT
PL /
(11-23)
Substituting this expression for Px into equation 11-22, the total apparent power is: PA = 4 P a 2 +(PL -- Pc) 2
A
ELIT
(11--25)
where EA is the applied voltage and IT the circuit current. The apparent power calculated by using one of these two methods should be identical to the apparent power calculated by using the other method.
rx
(VAR)PC1
PA(V-A)
PR(WATTS)
Figure 11.17 Total Apparent PowerPhasor Diagram
BASICAC CIRCUITS
377
II Resistance and Reactance Phasor Diagram
RLC CIRCUIT ANALYSIS
1 Calculation of Z T 1 Calculation of IT II Calculation of Circuit Voltage Values
ANALYSIS OF A SERIES R L C CIRCUIT
Figure 11.18 is a series RLC circuit with an applied voltage of 50 volts, 20 ohms resistance, 45 ohms inductive reactance, and 30 ohms capacitive reactance. Impedance, voltage, current, and power calculations will now be made for this circuit using the techniques described.
R=20~ EA -
XLffi
-;;
45Q
50V I(
Resistance and Reactance Phasor Diagram
At this point, it is useful to sketch the resistance and reactance phasor diagram to help visualize the relationships between the resistive and reactive quantities of the circuit. Figure 11.19 shows such a diagram. Note that XL is larger than Xc; thus, the net total reactance is in phase with XL and is, using equation 11-12,
X c = 30Q
Figure 11.18 Series RLC Circuit Example
XL 45f~
XT=XL--Xc = 45f2 - 3 0 ~ = 15f~
Calculation of ZT The total impedance, ZT, is equal to
....
R 20f2
ZT = a/R2 + XT 2 = 4202 + 152
Xc
30~
= 4400 + 225 = 4625
Figure 11.19 Resistance and Rectance Phasor Diagram
=25f~
Calculation of Circuit Voltage Values Calculation of IT By using Ohm's law for ac circuits the total current, IT, c a n be calculated. IT =
EA ZT 50
25 =2A
Using Ohm's law again, the voltage drops across each of the components in the circuit can be calculated. The voltage across the resistor, ER, is calculated: ER -- I T R
= (2 A) (20 a ) =40V
378
BASIC AC CIRCUITS
RLC CIRCUIT ANALYSIS
9 Calculation of ET 9 Calculation of Phase Angle l Alternate Method of Calculating Phase Angle
11
The voltage across the inductor, EL, is calculated: EL = ITXL =
ER= 4 0 V
(2 A) (45 ~ ) EA
=90 V
The voltage across the capacitor, Ec, is calculated:
-"
";;"
EL -
90V
50V
,,(
Ec = ITXc =
'
Ec - 60 V
(2 A) (30 f~)
=60V
Figure 11.20 Calculated VoltageDrops in Example
Calculation of Er
At first, as shown in Figure 11.20, this result appears impossible since there seems to be more voltage in the circuit than the applied voltage, EA. However, you must keep in mind that the voltage across the inductor and the voltage across the capacitor are 180 degrees out of phase and one partially cancels the other. Therefore, the applied voltage, EA, is calculated using equation 11-4. EA = ~/ER2 + (EL
- -
Ec) 2
= ~402 + (90 - 60) 2 =
4402
+
302
= 41,600 + 900 = 42,500 =50V
This calculation confirms the original applied voltage value given in the example. And this method gives you a valuable check of the accuracy of the calculation of the individual voltage drops that were calculated for each component in the circuit.
Series RLC Circuit Calculation of Phase Angle
Using the voltage phasor diagram of Figure 11.13, the phase angle of the example series RLC circuit can be calculated:
0 ar tan( )
=ar tan( 0V/ V0 = arctan(0.75) =37 ~ The angle whose tangent is 0.75 is about 37 degrees. Alternate Method of Calculating Phase Angle
The same result may be obtained using the impedance phasor diagram. In this case the arctangent is XT divided by R. 0= arctan(-~) = arctan(15 ~ ] k.20 n ) = arctan(0.75) =37 ~
BASIC AC CIRCUITS
379
II II II II
Positive and Negative Phase Angles Calculation of Component Power Values Calculation of Total Reactive Power Calculation of Total Apparent Power
RLC CIRCUIT ANALYSIS
Positive and Negative Phase Angles
Calculation of Total Apparent Power
The phase angle calculated for the example series RLC circuit is considered to be a positive phase angle since the phase angle for the circuit is measured counter-clockwise. Thus, the phase angle for the circuit is stated as being a positive 37 degrees.
By equation 11-22, the apparent power, PA, is equal to the square root of PR squared plus Px squared:
The phase angle in a series RLC circuit can either be positive or negative. If XL is greater than Xc, the angle is positive. On the other hand, if Xc is greater than XL, the angle is negative.
Calculation of Component Power Values In the example series RLC circuit, the voltage drops in the circuit are: 40 volts across the resistor, 90 volts across the inductor and 60 volts across the capacitor. The applied voltage is 50 volts and a total current of 2 amperes is flowing. Multiplying the voltage times the current for each component yields these individual power values: Pg = ERIT = (40 V) (2 A) = 80 W PL = ELIT = (90 V) (2 A) = 180 VAR Pc = EclT = (60 V) (2 A) = 120 VAR
Calculation of Total Reactive Power The net reactive power, Px, is equal to the difference between PL and Pc, or 180 VAR minus 120 VAR which equals 60 VAR. Since PL is larger than Pc, Px is positive and is in phase with PL.
PA =
4PR 2 + Px2
= ~/802 + 602
= 46, 400 + 3, 600 = 410, 000 = 100 VA The total apparent power of the example series RLC circuit is 100 volt-amperes. This should be the same as the value obtained by multiplying the applied voltage, EA, by the total current, IT. PA =EAIT
= (50 V)(2 A) = (100 v a )
It is, and proves the accuracy of the first calculation. PARALLEL RLC CIRCUITS SUMMARIZED
Now that all calculations for a series RLC circuit have been made, the method of solution of a parallel RLC circuit will be discussed. The analysis of this circuit will be similar to the analysis of either a parallel RL or a parallel RC circuit. In a parallel RLC circuit, such as in Figure 11.21, the sum of the branch currents is not equal to the total current as it would be in either a purely resistive, a purely inductive, or purely capacitive circuit. That is, IT ;e IR + IL + Ic
(11-26)
This occurs because of the different phase relationships between the voltage and current for each component. Recall that this was also true for RL and RC parallel circuits.
380
BASIC AC CIRCUITS
RLC
9 Voltage as a R e f e r e n c e 9 D e v e l o p m e n t o f a Circuit P h a s o r D i a g r a m 9 Calculations of Branch Currents
CIRCUIT
11
ANALYSIS
Voltage as a R e f e r e n c e
Since the voltage across all components of a parallel circuit is the same as the applied voltage, EA, it will be used as the reference quantity in discussing the phase relationships of the currents in the circuit. Therefore,
Ir
IL
A
llc R
EA = ER = EL = E c .
L TC
D e v e l o p m e n t o f a Circuit P h a s o r D i a g r a m
Recall that the current through a resistor, IR, is in phase with the voltage across it, ER. The capacitive current, Ic, leads the voltage across the capacitor, EA, by 90 degrees; and the current through an inductor, IL, lags the voltage across it, EA, by 90 degrees. Thus, the phase relationship of the applied voltage and currents in a parallel RLC circuit are as shown in Figure 11.22.
Figure 11.21 Parallel RLC Circuit Example
Ic
Calculations of Branch Currents
The individual branch currents in the example RLC circuit can be calculated as they are in either a purely resistive, purely capacitive, or purely inductive circuit. Simply divide the voltage across the branch by the opposition to current in the branch. In the resistive branch the opposition to the flow of current is measured in ohms of resistance. The resistive current is determined by dividing the applied voltage by the value of the resistor:
EA
IR= ~ R
(11-27)
In the inductive branch, the opposition to the flow of current is measured in ohms of inductive reactance. The inductive current is determined by dividing the applied voltage by the reactance of the inductor:
EA
IL = - -
XL
BASIC AC CIRCUITS
v
la
v
Ir
IL Figure 11.22 Phase Relationships of Ea and I in a Parallel RLC Circuit
In the capacitive branch, the opposition to the flow of current is measured in ohms of capacitive reactance. The capacitive current is determined by dividing the applied voltage by the reactance of the capacitor: EA
Ic = ~ Xc
(11-29)
(11-28)
381
RLC CIRCUIT
ANALYSIS
II Current Phasor Comparisons II Calculation of Ix Current Phasor Comparisons In the parallel RLC circuit, since EA = ER = EL - Ec, and since IR is in phase with ER and thus all the voltages EL, Ec, and EA, IR becomes the reference vector for the current phasor diagram of Figure 11.23. Comparing the current phase relationships using the current phasor diagram of Figure 11.23, you can see that the capacitive branch current, Ic, leads the resistive branch current, IR, by 90 degrees; the inductive branch current, IL, lags IR by 90 degrees. Because of the 180-degree difference between the capacitive branch current and the inductive branch current, these two reactive current values are opposite in phase, and one partially cancels the effect of the other.
[r IX ll
ll
ll
Ix = ~]IR 2 + Ix 2
(11-30)
Ic
382
IT
,x
IL Figure 11.24 Current Phasor Diagram with Iv Phasor
Ix = I c - L
(11-31)
Ix is equal to Ic minus IL. Substituting this expression for Ix into equation 11-30, the total current is equal to
IT 4IR 9 + (Ic - IL) 2 =
Sine Ix is the net difference between Ic and IL, as shown in equation 11-31,
I
Figure 11.23 Current Phasor Diagram for a Parallel RLC Circuit
Calculation of Ix
Applying the Pythagorean theorem,
IR
It.
The difference between Ic and IL is Ix, which is the net reactive current. If Ic is larger than IL, the net reactive current is in phase with Ic and the value of Ix is positive. If, on the other hand, IL is larger than Ic, the net reactive current is in phase with IL and the value of Ix is negative. The sign of Ix indicates its direction on the Y(reactive) axis. By adding the total current vector, Ix, to the current phasor diagram, and shifting vector Ix to the right so that it extends between the tip of IR and the tip of Ix, a right triangle is produced. This is shown in Figure 11.24. The right triangle can then be used to add the phasors IR and Ix vectorially.
,180~ [A
(11-32)
which is the relationship between the branch currents in a parallel RLC circuit.
BASIC AC CIRCUITS
RLC CIRCUIT ANALYSIS
9 Calculation of Zr
, I
9 Calculation of Phase Angle 9 Calculation of Power I
Calculation of Zx Once the total current is known, the total impedance, ZT, of the circuit is easily determined using Ohm's law for ac circuits: EA
zT = ~
IC
Ix
IT
(11-33)
Ix
IL
Calculation of Phase Angle
Recall that the phase angle is the n u m b e r of degrees of phase difference between the applied voltage and the total current. Also recall that the applied voltage is in phase with the resistive current. Therefore, the phase angle is measured from the horizontal vector IR (which is also EA) tO the vector IT; it is identified by the angle, theta, on the current phasor diagram of Figure 11.25. The tangent of the phase angle, theta, is equal to the ratio of the net reactive current divided by the resistive current as shown in equation 11-34: tan 0 = I--Ex IR
(11-34)
O= arctan(~RX )
(11-35)
i: A I R
Figure 11.25 PhaseAngle on the Current Phasor Diagram
Pc
Px
Px
PLI
ex
PR
Figure 11.26 TotalApparent PowerPhasor Diagram
Therefore,
the arctangent of the net reactive current divided by the resistive current equals the value of the phase angle.
The total apparent power is found by using the Pythagorean theorem solution of resistive and net reactive power shown in the power phasor diagram of Figure 11.26 defined by these equations:
Calculation of Power
PA = ~/pg2 + px 2
(11-39)
Power calculations in parallel RLC circuits are the same as power calculations in series RLC circuits. The individual power values are calculated by multiplying current through a component times voltage across a component.
Px = P c - PL
(11-40)
PA = ~/pg2 + (Pc - PL)2
(11-41)
Therefore, PR = ERIR = EAIR PL = ELIL = EAIL Pc = EcIc = EAIc BASICAC CIRCUITS
(11--36) (11--37) (11-38) 383
RLC CIRCUIT ANALYSIS II Calculation of Branch Currents II Current Phasor Diagrams
ANALYSIS OF A P A R A L L E L R L C CIRCUIT
Figure 11.27 shows a typical parallel RLC circuit with 180 VAC, 60 kilohms resistance, 30 kilohms inductive reactance, and 18 kilohms capacitive reactance. Impedance, voltage, current, and power measurements will be performed for this circuit using the techniques just described.
R=60kQ
EA 18
'lXc = 8kQ
Calculation of Branch Currents
XL=3OkD
The resistive branch current, IR, is calculated by dividing the voltage across the resistor, 180 volts, by the value of the resistor, 60 kilohms.
Figure 11.27 Typical Parallel RLC Circuit
EA
IR = ~
R 180 V
60 lffl
Ic *I IT Ix . . . . . . . ~,lx i
=3mA The inductive branch current is determined in a similar manner by dividing the voltage across the inductor by the inductive reactance.
I
.....
IR
EA IL ~ m
XL 180 V
IL
30 kD
=6mA The capacitive branch current is calculated by dividing the voltage across the capacitor by the capacitive reactance. EA Ic - - m
Xc 180 V 18k~
=10m~
Figure 11.28 Current Phasor Diagram for Parallel RLC Circuit Example
Current Phasor Diagrams
The current phasor diagram can now be drawn to show the relationships between the resistive, inductive, and capacitive branch currents as shown in Figure 11.28. Note that Ic is larger than IL; therefore, the net reactive current is in phase with Io The net reactive current, Ix, equals Ix = Ic - IL = 1 0 m A - 6mA
=4mA 384
BASIC AC CIRCUITS
RLC CIRCUIT ANALYSIS
I Calculation o f IT I Calculation o f Z 1 Calculations o f the Phase Angle
Calculation of IT As you can see in Figure 11.28, by shifting the phasor Ix to the right so that it extends from the tip of the vector representing IR to the tip of the vector for IT, a right triangle is formed. The right triangle can then be used to determine the value of IT. Using the Pythagorean theorem, the total current, IT, is now equal to the vector sum of the values of Ix and IR.
I c 10 mA 4mA
Ix
=49+16
~
I T 5mA v
IR 3mA
IT = ~/IR2 + IX2
= 432 + 42
11
IL6mA Figure 11.29 Current Phasor Diagram for Calculating Theta
=4-~g =5mA The total current in the circuit is 5 milliamperes.
[c
Ix~~IT,,.
Ic
Calculation o f Z The total impedance of this circuit can now be calculated by dividing the applied voltage by the total current. Z=~
IR
IL
Ix
I "
EA
IL
IT =
180 V ~
5mA =36k~
a
b
Figure 11.30 Current Phasor Diagramsfor Positive and Negative Thetas
Calculations o f the Phase Angle
The phase angle, theta is the angle between the resistive current and the total current as shown on the current phasor diagram in Figure 11.29. The value of the phase angle is equal to the arctangent of the ratio of net reactive current to the resistive current. Therefore, 0 = arctan(Ix ~ k, I R )
-artan/mama/
Thus, the phase angle of this parallel RLC circuit is approximately 53 degrees. The phase angle is described as a positive phase angle because the total current phasor is rotated counter-clockwise from the reference. However, the phase angle of parallel RLC circuit may be either positive or negative depending upon the relationships of IL and Ic. If Ic is greater than IL, the phase angle theta will be positive as shown in Figure 11.30a. If IL is greater than Ic, theta will be negative as shown in Figure 11.30b.
= arctan(1.33) =53 ~ BASIC AC CIRCUITS
385
RLC CIRCUIT ANALYSIS
II Calculation of Component Power Values II Calculation of Total Reactive Power II Calculation of Total Power
l
Calculation of Component Power Values
l
II
l
II
As you know power values depend on the various voltages and currents in the circuit. In the parallel RLC circuit example shown in Figure 11.31, currents are: I~ equals 3 milliamperes, IL equals 6 milliamperes, and Ic equals 10 milliamperes. The applied voltage is 180 volts; the total current is 5 milliamperes.
Figure 11.31 Branch Currents in Parallel RLC Circuit Example
The real power dissipated by the resistor, P~, is calculated:
Calculation of Total Power
Pp. =EAIR = (180 V)(3 mA) = 540 mW The inductive reactive power, PL is calculated: PL =EAIL
180V
A
Calculation of Total Reactive Power
Ic 10mA
= ,1540 m W ~ + 720 mVAR 2
= 4291,600 + 518, 400 = 900 mVA
= 1,800 mVAR
C
PA = 4PR 2 + Px2
= 4810,000
= (180 V) (10 mA)
L
The apparent total power, PA, is calculated using equation 11-39:
= 1,080 mVAR
Pc =EAIc
' I
It. IR 3mA 6mA
= (180 V) (6 mA)
The capacitive reactive power, Pc, is calculated:
,R
This value should be the same as the apparent power value obtained by multiplying the applied voltage by the total current:
PA =EAIT = (180 V)(5 mA) = 900 mVA
The net reactive power, Px is calculated: Px = P c - PL = 1,800 mVAR - 1,080 mVAR = 720 mVAR The power phasor diagram will look like the one in Figure 11.26. Since Pc is larger than PL, Px is positive and is in phase with Pc.
386
BASICAC CIRCUITS
RLC
II
Summary I
I
I
II
I
CIRCUIT ANALYSIS
9 I
SUMMARY
In this chapter, the concepts and techniques you learned in analyzing RL and RC circuits were applied to determine circuit values in RLC circuits. You were shown how to determine current, voltage, reactance, and the phase angle for any RLC circuit. Also, a new concept was i n t r o d u c e d - that when values are in opposite phase, one value partially cancels the effect of the other. This difference is called the net effect. The net voltage, Ex, net total reactance, XT, and net reactive current Ix, were calculated. The methods you learned in this chapter should enable you to determine equivalent circuit values in any series or parallel RLC circuit.
BASIC AC CIRCUITS
387
RLC CIRCUIT ANALYSIS ll Worked-Out
Examples
1. Draw a phasor diagram for this circuit showing current and impedance phasors.
XL : 10~
Xc = 6N Solution:
X L > Xc;
therefore, XT is plotted in phase with
XL
XT
R I(all) Xc
2. Draw voltage phasor diagrams for the circuit in Example 1. Calculate the value and sign of the phase angle. Solution:
tan
XT
= ~
=
4
R 6 arctan 0.667 = 33.7 ~
EXT
Erl
EXT
0 = ~
ER
=0.667
E.
The angle is rotated counter-clockwise; therefore, the sign of the angle is positive.
388
BASICAC CIRCUITS
RLC CIRCUIT ANALYSIS
II W o r k e d - O u t E x a m p l e s
/I
3. Draw p h a s o r d i a g r a m s showing P y t h a g o r e a n t h e o r e m r e l a t i o n s h i p s in parallel RLC circuits. Write e q u a t i o n s for e a c h solution.
Solution:
Ix
PT
Px
Pa
In
IT = N / i a 2 + I x ~ = X/I'R' + ( I L -
PT =
Ic)'
N/PR 2 + Px 2
= N/PR' + (Pt.-
Pr
4. Given the circuit a n d typical circuit values shown, calculate the circuit values specified.
R = lOft
f-
.1,o
.....T L : 0.25mH
a.
XL =
h.
ER =
b.
Xc=
i.
PL--
C. XT--
j.
Pc =
d.
ZT=
k.
PR =
e.
IT =
1.
Px -
f.
EL =
m.
PA--
g.
E C
n.
Phase angle =
--
Solution: a.
X L = 27rfL = 6.28 • 5 x 103 x 0.25 x 10 -~ = 7.85 f l
b.
Xc .
C.
XT = XL -- X c = 7.85 f2 - 15.9 f~ = 8 . 0 5
d.
ZT =
e.
IT
f.
EL = I T X L = 0.781 A x 7.85 f2 = 6.13 VAC
g.
Ec = I T X c = 0.781 A x 15.9 f2 = 12.4 VAC
1
2xfC
.
1
. . 6.28•215215
4R 2 +
XT 2 =
EA
10 VAC
= ~
ZT
=
BASIC AC CIRCUITS
12.8f~
1
-6
6 . 2 8 x 1 0 -2
= 15.9 f l
4102 +8.052 = 4i~65 • i02 = 12.8 f l = 0 . 7 8 1 Arms
389
RLC
CIRCUIT ANALYSIS
II W o r k e d - O u t E x a m p l e s I
I
h.
ER = ITR = 0.781A x 10 f~ = 7.81VAC
i.
PL = ITEL = 0 . 7 8 1 A x 6.13 V = 4.79 V A R ~
j.
Pc = ITEc = 0.781 A x 12.4 V = 9.68 VAR~s
k.
PR = ITER = 0.781 A x 7.81V - 6.1 W ~
1.
Px = PL - Pc - 4.79 V A R - 9.68 VAR = 4.89 V A R ~
m. PA = a/PR2 + Px 2 = ~/6.1W 2 +4.89 VAR 2 = 64~6~.11 = 7.82 VAn.~ n.
phase a n g l e = a r c t a n ( - - ~ - ) = a r c t a n 0 . 8 0 5 = - 3 8 . 8 ~ kKJ
5. Given this circuit a n d typical circuit values shown, calculate the circuit values specified.
--
A
I= ~
a.
X L ~-~
h"
PR
b.
Xc=
i.
PL =
C.
IR =
j.
Pc =
k.
Px = PA =
lOOHz
~k~l
~5;
O.eS~l~T d" IL =
%~k~ -
"
e.
Ic =
1.
f.
Ix
=
m. Phase angle =
g,
IT
--
Solution: a.
X L = 2~zfL = 6.28 x 100 x 15 = 9.42 kD.
b.
X c = ~ 2~fC
C.
IR .
1
6.28 x 100 x 0.5 x 10 .6
Ea 50 VAC . . . . R
390
3.14 x 10 -4
10 mAtins
5 kD.
d.
IL .
EA 50 VAC . . . . XL 9.42 k.Q
5.31 m A r ~
e.
Ic
EA . Xc
15.7 rearms
f.
Ix = I L - I c = 5 . 3 1 m A - 1 5 . 7 m A = 1 0 . 4 m A
.
= 3.18 k ~
.
50 VAC . 3.18 k ~
BASIC AC CIRCUITS
RLC
CIRCUIT ANALYSIS
II Worked-Out Examples I
I
I
11 I
I
I
4IR 2 + Ix 2 = 4 i 0 m A 2 + 10.4 m A 2 = 14.4 mA,m~
g.
IT
=
h.
PR
=IREa=lOmA+5OV=O.5Wrms
i.
PL = ILEA = 5.31 m A + 50 V = 0.27 V A R ~
j.
Pc
k.
Px = PL - P c = 0 . 2 7 V A R - 0 . 7 9
1.
I
= IcEA = 15.7 m A + 5 0 V = 0 . 7 9 V A R ~ s V A R = 0 . 5 2 VAR~ms
PA = 4PR 2 + Px 2 = 4 0 . 5 W 2 + 0.52 V A R 2 = 040.52 = 0.72 VArm, o r Pa
= ITEA = 14.4 m A x 50 V = 0.72 VA,n,
m. t a n 0 = I___~_x= ~ 1 0 " 4m A = 1.04 IR 10mA a r c t a n l . 0 4 = 46.1 ~
BASIC AC CIRCUITS
391
RLC CIRCUIT ANALYSIS
I Practice P r o b l e m s
1. Draw a phasor diagram for this circuit showing voltage and current phasors.
/•
IR = 15mA
..
8Cm~
. r
2. Draw power phasor diagrams for the circuit in problem 1. Calculate the value and sign of the phase angle. 3. Draw phasor diagrams showing Pythagoren theorem relationships in series RLC circuits. Write equations for each solution. 4. Given the circuit and typical circuit values below, calculate the circuit values specified. t=SmH
a. X L =
g. E c =
b. Xx=
h. PR=
c. Z T - -- -
i.
P L - ---
d. IT =
j.
Pc =
e. ER =
k. PA =
f. EL
1. Phase angle
t-2.SRH
R=
E, =
1501"1
2or
xc:2oon
=
=
5. Given this circuit and typical circuit values shown, calculate the circuit values specified.
A
2'kzL I
392
A
a. X c =
g. PR =
b. IR =
h. PL =
c. IL =
i. Pc =
d. Ic =
j.
e. I x =
k. PA =
f. IT =
1. Phase angle =
.
Px =
BASIC AC CIRCUITS
RLC CIRCUIT ANALYSIS
11
a
I 1. Draw a phasor diagram for this circuit showing power and c u r r e n t phasors.
a. X c = b. XT=
Xc = lkfl
d. IT =
g. h. i. j. k. 1.
C. ZT-"
I
X, = 1.5kfl
R=
e.
ER =
f.
EL =
Ec= PR = PL =
Pc= PA = Phase angle
8. Given this circuit and typical values shown, calculate the circuit values specified.
lkfl
2. Draw a phasor diagram for this circuit showing power and voltage phasors.
E,0; ~ ~ , /
3. Calculate the value and sign of the phase angle in the circuit of Problem 1. 4. Calculate the value and sign of the phase angle in the circuit of Problem 2. 5. Calculate the value of ZT in Problem 1. 6. Calculate the value of IT in Problem 2.
a. b. c. d. e. f.
XL= IR = IL = Ic = IT = PR =
1Okfl
.io.!.
XC--'-- i
g. h. i. j. k.
PL = Pc = Px = PA = Phase angle =
9. In a series RCL circuit, which c o m p o n e n t will have the most voltage across it? 10. In a parallel RCL circuit, which c o m p o n e n t will have the most c u r r e n t t h r o u g h it?
7. Given this circuit and typical circuit values shown, calculate the circuit values specified. XL : 40fl
R = SOft BASIC AC C I R C U I T S
393