RL Circuits

RLRL Circuits CircuitsChapter 12

Thomas L. Floyd

David M. Buchla

DC/AC Fundamentals: A Systems DC/AC Fundamentals: A Systems ApproachApproach

DC/AC Fundamentals: A Systems ApproachThomas L. Floyd

© 2013 by Pearson Higher Education, IncUpper Saddle River, New Jersey 07458 • All Rights Reserved

When both resistance and inductance are in a series circuit, the phase angle between the applied voltage and total current is between 0 and 90, depending on the values of resistance and reactance.

Ch.12 Summary

Sinusoidal Response of RL Circuits



In a series RL circuit, the total impedance is the phasor sum of R and XL.

R is plotted along the positive x-axis. XL is plotted along the positive y-axis.Z is the diagonal.

It is convenient to reposition the phasors into an impedance triangle.

Ch.12 Summary

Series RL Circuit Impedance

R

Z

R

ZXL XL

qq

RXL1tanq



Sketch the impedance triangle and show the values for R = 1.2 k and XL = 960 .

Ch.12 Summary

Series RL Circuit Impedance

q

39kΩ 1.2kΩ 0.96tan

tan

1

1

RXL

R = 1.2 k

Z = 1.33 k XL = 0.96 k

q = 39o

kΩ 1.33)k (0.96)k (1.2 2222 LXRZ



Ohm’s law is applied to series RL circuits using quantities of Z, V, and I.

Because I is the same everywhere in a series circuit, you can multiply the impedance phasor values by the circuit current to obtain the voltage phasor values.

Ch.12 Summary

Series RL Circuit Analysis

IV Z

ZV I IZV



Assume the current in the previous example is 10 mA. Sketch the voltage phasors. (The impedance triangle from the previous example is shown for reference.)

The voltage phasor diagram can be found using Ohm’s law. Multiply each impedance phasor by 10 mA (as shown below):

Ch.12 Summary

Series RL Circuits Analysis

R = 1.2 k

Z = 1.33 k XL = 0.96 k

q = 39o

x 10 mA=

x 10 mA=

VR = 12 V

VS = 13.3 V VL = 9.6 V

q = 39o



Phasor diagrams that have reactance phasors can only be drawn for a single frequency because X is a function of frequency.

As frequency changes, the impedance triangle for an RL circuit changes (as illustrated here) because XL is directly proportional to f. This determines the frequency response of RL circuits.

Ch.12 Summary

Phase Angle Vs. Frequency

Z3

XC1

XC2

XC3

Z2

1

2

f

f

f

3

R

q3

q2

q1

Z1

Z 3

XC1

XC2

XC3

Z 2

1

2

f

f

f

3

R

Z1

q1

q2

q3

Increasing Increasing f



Ch.12 Summary

Phase Shift

(phase lead)

Vin L

R

Vout

Vin Vout

VR

V

Vout

Vin

A series RL circuit can be used to produce a specific phase lead between an input voltage and an output by taking the output across the inductor. This circuit is a basic high-pass filter, a circuit that passes high frequencies and rejects all others. This filter passes frequencies that are above a specific frequency, called the cutoff frequency.



Reversing the components in the previous circuit produces a circuit that is a basic lag network. This circuit is a low-pass filter, a circuit that passes low frequencies and rejects all others. This filter passes low frequencies up to a frequency called the cutoff frequency.

Ch.12 Summary

Phase Shift

(phase lag)f

R

L

Vin Vout

VL Vin

Vout

Vin

Vout



For parallel circuits, it is useful to review conductance, susceptance and admittance, introduced in Chapter 10.

Conductance is the reciprocal of resistance.

Admittance is the reciprocal of impedance.

Inductive susceptance is the reciprocal of inductive reactance.

Ch.12 Summary

AC Response of Parallel RL Circuits

RG 1

LL X

B 1

ZY 1



From the diagram, the phase angle is:

Ch.12 Summary


Y

VS G BL

BL

Gq

22LBGY

q

GBL1tan

In a parallel RL circuit, the admittance phasor is the sum of the conductance and capacitive susceptance phasors:



Draw the admittance phasor diagram for the circuit below.

The magnitudes of conductance, susceptance, and admittance are:

Ch.12 Summary


mS 1kΩ 111

R

G mS 629.mH) kHz)(25.3 (102

11

LC X

B

mS 1.18mS) (0.629mS) (1 2222 BGY L

Y = 1.18 mS

VS R L

BL = 0.629 mS

G =1.0 mSq

f =10 kHz 1 k 25.3 mH



Ohm’s law is applied to parallel RL circuits using quantities of Y, V, and I.

Because V is the same across all components in a parallel circuit, you can obtain the current in a given component by simply multiplying the admittance of the component by the voltage as illustrated in the following example.

Ch.12 Summary

Analysis of Parallel RL Circuits

VIY VYI

YIV



If the voltage in the previous example is 10 V, sketch the current phasor diagram. The admittance diagram from the previous example is shown for reference.

Ch.12 Summary

Analysis of Parallel RL Circuits

Y = 1.18 mS

BL = 0.629 mS

G = 1.0 mSx 10 V

=x 10 V

=

IL = 6.29 mA

IS = 11.8 mA

IR = 10 mA

The current phasor diagram can be found from Ohm’s law. Multiply each admittance phasor by 10 V.



Notice that the formula for inductive susceptance is the reciprocal of inductive reactance. Thus BL and IL are inversely proportional to f:

As frequency increases, BL and IL decrease, so the angle between IR and IS must decrease as well.

Ch.12 Summary

Phase Angle of Parallel RL Circuits

fLBL

2

1

IL IS

IR q



Series-parallel RL circuits are combinations of both series and parallel elements. These circuits can be solved by methods from series and parallel circuits.

For example, the components in the green box are in series:

The components in the yellow box are in parallel:

The total impedance can be found by converting the parallel components to an equivalent series combination, then adding the result to R1 and XL1 to get the total reactance.

Ch.12 Summary

Series-Parallel RL Circuits

22

22

222

L

L

XRXRZ

R CR2 C2

Z1

R1 L1

R2L 2

Z2

2211 LXRZ



As shown earlier, you can multiply the impedance phasors for a series RL circuit by the current to obtain the voltage phasors. The earlier example is shown below for review:

Ch.12 Summary

The Power Triangle

R = 1.2 k

Z = 1.33 k XL = 0.96 k

q = 39o

x 10 mA=

x 10 mA=

VR = 12 V

VS = 13.3 V VL = 9.6 V

q = 39o

Multiplying each value in the left-hand triangle gives you the corresponding value in the right-hand triangle.



Multiplying the voltage phasors by Irms (10 mA) gives the power triangle values (because P = V I ). Apparent power is the product of the magnitude of the current and magnitude of the voltage and is plotted along the hypotenuse of the power triangle.

x 10 mA =

Ch.12 Summary

The Power Triangle (Cont’d)

VR = 12 V

VS = 13.3 V VL = 9.6 V

q = 39o

Pa = 133 mVA Pr = 96 mVAR

Ptrue = 120 mW

q39o



The power factor was introduced in Chapter 10 and applies to RL circuits (as well as RC circuits). Recall that it is the relationship between the apparent power (in VA) and true power (in W). True power equals the product of Volt-amperes and power factor.

Power factor can be determined using:

Ch.12 Summary

Power Factor

Power factor can vary from 0 (for a purely reactive circuit) to 1 (for a purely resistive circuit).

qcosPF



Apparent power consists of two components; the true power component, which does the work, and a reactive power component, that is simply power shuttled back and forth between source and load.

Ch.12 Summary

Apparent Power

Components such as transformers, motors, and generators are rated in VA rather than watts.

Pr (VAR)Pa (VA)

Ptrue (W)

q



The response of a series RL circuit is similar to that of a series RC circuit. In the case of the low-pass response shown here, the output is taken across the resistor.

100 10 mH

10 V dc

VoutVin

10 V dc

0

10 V dc

0

Plotting the response:

100 ƒ = 1 kHz

8.46 V rms10 V rms 10 mH100

10 mHƒ = 10 kHz

1.57 V rms

10 V rms

100 10 mH

ƒ = 20 kHz

0.79 V rms

10 V rms

Ch.12 Summary

Frequency Response of RL Circuits

Vout (V)

9.98

8.46

1.570.79

0.1 1 10 20 100f (kHz)

987654321

Vout (V)

9.98

8.46

1.570.79

0.1 1 10 20 100f (kHz)

987654321



Reversing the position of the R and L components produces a high-pass response. In this case, the output is taken across the inductor.

Vin

10 V dc0

Vout

0 V dc10 V dc100

10 mH

Plotting the response:

ƒ = 100 Hz0.63 V rms10 V rms 100

10 mHƒ = 1 kHz

5.32 V rms10 V rms

100 10 mHƒ = 10 kHz

9.87 V rms10 V rms

100 10 mH

Ch.12 Summary

Frequency Response of RL Circuits

Vout (V)

f (kHz)

9.87

5.32

0.6300.01 0.1 1

10987654321

10

Vout (V)

f (kHz)

9.87

5.32

0.6300.01 0.1 1

10987654321

10



Inductive susceptance (BL)

The ability of an inductor to permit current; the reciprocal of inductive reactance, measured in siemens (S).

Ch.12 Summary

Key Terms



1. If the frequency is increased in a series RL circuit, the phase angle will

a. increase

b. decrease

c. remain unchanged

Ch.12 Summary

Quiz



2. If you multiply each of the impedance phasors in a series RL circuit by the current, the result is the

a. voltage phasors b. power phasorsc. admittance phasorsd. none of the above

Ch.12 Summary

Quiz



3. For the circuit shown, the output voltage

a. is in phase with the input voltage

b. leads the input voltage

c. lags the input voltage

d. leads the resistor voltage

VinVout

Ch.12 Summary

Quiz



4. Which of the equations below can be used to calculate the phase angle in a series RL circuit?

a.

b.

c. both of the above are correct

d. none of the above is correct

Ch.12 Summary

Quiz

q

RXL1tan

q

RLV

V1tan



5. In a series RL circuit, if the inductive reactance is equal to the resistance, the source current will lag the source voltage by

a. 0o

b. 30o

c. 45o

d. 90o

Ch.12 Summary

Quiz



6. Susceptance is the reciprocal of

a. resistance

b. reactance

c. admittance

d. impedance

Ch.12 Summary

Quiz



7. In a parallel RL circuit, the magnitude of the admittance can be expressed as

a.

b.

c.

d.

Ch.12 Summary

Quiz

LBG

Y11

1

22LBGY

LBGY

22LBGY



8. If you increase the frequency in a parallel RL circuit,

a. the total admittance will increase

b. the total current will increase

c. both a and b

d. none of the above

Ch.12 Summary

Quiz



9. The unit used for measuring true power is the

a. volt-ampere

b. watt

c. volt-ampere-reactive (VAR)

d. kilowatt-hour

Ch.12 Summary

Quiz



10. A power factor of zero implies that the

a. circuit is entirely reactive

b. reactive and true power are equal

c. circuit is entirely resistive

d. maximum power is delivered to the load

Ch.12 Summary

Quiz



1. a

2. a

3. c

4. a

5. c

6. b

7. d

8. d

9. b

10. a

Ch.12 Summary

Answers

RL Circuits

Documents

Transcript of RL Circuits