Resistive-Inductive (RL) Circuits

AC Circuits I

2

Series RL Circuits

IT = IR1 = IR2 IT = IL1=IL2 IT = IL=IR

VT =VR1+VR2 VT = j(VL1 + VL2)|VT| = sqrt(VR

2 + VL2)

RT = R1+R2 XT = XL1 + XL2 |ZT| = sqrt(R2 + XL2)

Series circuits: A comparison (Magnitude)

3

Series RL Circuits

• Series Voltages

4

Series RL Circuits

• Series Voltages (Continued)– Since VL leads VR by 90°, the values of the component

voltages are usually represented using phasors– Phasor – a vector used to represent a value that constantly

changes phase. Usually denoted in a polar form r@θ.

5

Series RL circuits

• Series Impedance

900

90

I

V

I

VX LL

L

00

0

I

V

I

VR RR

22 RXZ LT

R

X L1tan

Polar form

Polar form

Rectangular form

Rectangular form

6

Example

• The component voltages shown below were measured using an ac voltmeter (HINT: rms values!!!). Calculate the source voltage for the circuit.

rmso

S

oS

LRS

VV

VVV

57.2671.6

57.266

3tan

71.64536

1

2222

7

Example

• Calculate the total impedance for the circuit shown (remember Z = R +jXL , where XL=2πfL )

oT

oZ

LT

LT

L

Z

XRZ

jjXRZ

R

mHfLX

T

49.5160.160

49.51100

66.125tan

60.16066.125100

66.125100

100

66.12540)1)(20000(22

1

2222

8

Phase Reference• In a series RL circuit the current I and voltage across the resistor VR are

considered to have 0o phase.• Sometimes however (especially in the lab) the source voltage Vs is

assumed to have 0o phase.

9

Example• Calculate the circuit impedance, and current in the circuit shown below

• I and Vs can be represented as either: (I lags Vs or Vs leads I)

mAkZ

VI

kZ

kXRZ

kjXRZ

kR

kkmHfLX

oo

o

T

s

oT

oZ

LT

LT

L

T

95.3902.195.39568.19

020

95.39568.19

95.3915

566.12tan

568.19566.1215

566.1215

15

566.124)200)(10000(22

1

2222

mAI

VV os

4002.1

020

mAI

VV os

002.1

4020

10

Example• Determine the voltage, current and impedance values for the circuit shown below

mVA

jAjXIV

mVAIRV

VAI

AZ

VI

Z

XRZ

jjXRZ

R

mHfLX

ooo

oLL

oooR

os

o

oo

o

T

s

oT

oZ

LT

LT

L

T

907.979069.41405.235

69.41405.235)(

04.2109105.235

62.771.005.235

62.775.23562.7756.424

01.0

62.7756.424

62.7715

566.12tan

56.42469.41491

69.41491

91

69.414132)3.3)(20000(22

1

2222

11

Voltage Dividers

• In a series RL circuit where I(VR )is used as the 0o phase reference, Vs and ZT always have the same phase angle!!!

• RL Voltage Dividers

T

nSn Z

ZVV

where Zn = magnitude of R or XL

Vn = voltage across the component

12

Example• Determine the voltages VL and VR in the circuit below

mVZ

jXVV

mVZ

RVV

Z

XRZ

jjXRZ

R

mHfLX

oo

o

T

LsL

oo

o

TsR

o

oZ

L

L

L

905.9762.7756.424

9069.41462.771.0

04.2162.7756.424

09162.771.0

62.7756.424

62.7715

566.12tan

56.42469.41491

69.41491

91

69.414132)3.3)(20000(22

1

2222

13

Example

• Calculate ZT, VL1 VL2 and VR for the circuit shown below.

VV

VV

VV

kX

kX

kZ

oR

oL

oL

L

L

oT

091.0

9084.6

9010.3

49.2

13.1

8.8464.3

2

1

2

1

14

Series RL Circuit Frequency Response

• Frequency Response – used to describe any changes that occur in a

circuit as a result of a change in operating frequency

– An increase in frequency causes XL to increase

– An increase in XL causes ZT and to increase

– An increase in ZT causes IT to decrease

– An increase in XL causes VL to increase

– An increase in VL causes VR to decrease

15

Power

• Power is also a complex quantity in AC circuits

XRAPP jPPP

16

Apparent, True and Reactive Power

Value Definition

Resistive power (PR)

The power dissipated by the resistance in an RL circuit. Also known as true power.

Reactive power (PX)

The value found using P = I2XL. Also known as imaginary power. The energy stored by the inductor in its electromagnetic field. PX is measured in volt-amperes-reactive (VARs) to distinguish it from true power.

Apparent power (PAPP)

The combination of resistive (true) power and reactive (imaginary) power. Measured in volt-amperes (VAs).

17

Power Factor (PF)

θ cosAPP

R

P

PPF

• The ratio of resistive power to apparent power

18

Example• Calculate PR, PX and PAPP for the following circuit - (see slide 13)

oXRAPP

oooLX

oooR

WjjPPP

WAXIP

WARIP

625.7755.23998.22046.5

90998.229069.41405.235

0046.509105.23522

22

19

Parallel RL Circuits

• Note that VS, IR and VR are all in phase

• IL is 90o out of phase with (lags) VS, IR and VR.

VS = VR1 = VR2 VS = VL1 = VL2 VT = VL= VR

IT =IR1+IR2 IT = j(IL1 + IL2) |IT|= sqrt(IR2 + IL

2)

RT = 1/[ (1/R1 )+(1/R2)] XT = 1/[(1/XL1 ) + (1/XL2 )] XT = 1/[(1/R ) + (1/jXL )]

20

Parallel RL Circuits• Branch Currents

– IL lags IR , VR and VS by 90º

21

Parallel RL Circuits

• Parallel Circuit Impedance– Impedance phase angle is positive since current phase

angle is negative

– Angles of ZT and IT have the same magnitude but opposite signs

• Calculating Parallel-Circuit Impedance (polar form)

T

ST I

VZ

22 RX

RXZ

L

LT

LITZ X

RT

1tan

22

Parallel RL Circuits

• Calculating Parallel-Circuit Impedance– rectangular form approach:

L

LT jXR

jXRZ

L

T

jXR

Z11

1

23

Parallel RL Circuits

• Parallel-Circuit Frequency Response

– The increase in frequency causes XL to increase– The increase in XL causes:

• IL to decrease• IT to increase• ZT to increase

– The decrease in IL causes IT to decrease

24

Example 1• Calculate the total current for circuit (a) in both rectangular and polar forms • Calculate total impedance of circuit (a)• Repeat for circuit (b)

oT

o

R

LI

LRT

T

LR

I

I

I

III

mAjI

mAImAI

T

04.1462.20

04.1420

5tantan

62.20425

520

5,20

11

22

kmA

V

I

VZ

mAI

VV

oo

o

T

ST

oT

oS

04.1429.004.1462.20

06

04.1462.20

06

25

Example 2 (Rectangular)• Calculate the total impedance for circuit (a) in both rectangular and polar forms • Calculate total current of circuit (a)• Repeat for circuit (b)

mAZ

VI

Z

jj

j

j

j

j

j

j

jXR

jXRZ

XR

oo

o

T

ST

oT

L

LT

L

04.1462.2004.1404.291

06

04.1404.291

59.7035.2821530000

108000000432000000

1200300

1200300

1200300

360000

1200300

360000

1200,300

26

Example 2 (Polar)• Calculate the total impedance for circuit (a) in both rectangular and polar forms • Calculate total current of circuit (a)• Repeat for circuit (b)

mAZ

VI

Z

X

R

XR

XRZ

XR

oo

o

T

ST

oT

o

LZ

L

LT

L

T

04.1462.2004.1404.291

06

04.1404.291

04.141200

300tantan

04.29193.1236

360000

1530000

360000

1200,300

11

22

27

28

Series –Parallel RL Circuits

• First collapse the parallel RL portion into a single impedance ZP

• If this single impedance is in polar form, convert it to rectangular form

• Add any resistances/ reactances in series with ZP to get ZT.

– Remember you cannot add complex numbers in their polar form.

• Always assume that IT is the zero Phase reference!! And that VS and ZT have the same phase i.e. Same assumption as for series RL circuit

29

Example• Calculate the Equivalent Impedance, the total current and the current in each branch of the

parallel RL circuit shown below

oP

o

LZ

L

LP

kZ

k

k

X

R

kkk

kk

RX

RXZ

P

8.6428.1

8.6441.1

3tantan

28.1341.1

341.1

121

2222

2

2

oP kZ 8.6428.1

oT

ST

S

ooS

osP

kZ

jZZ

jZ

jkZ

kZZ

8.4854.1

11581015470

1158545

8.64sin8.64cos28.1

8.6428.1

Now convert to rect and back to polar

os kZ 8.6428.1

oT kZ 8.4854.1

30

Example Cont’d

• Knowing that

• Remember in a series circuit, IT is the phase reference and VS and ZT always have the same phase!!!

oT kZ 8.4854.1

oo

o

T

SRCT mA

Z

VI 009.9

8.481540

8.4814

oooTR VmAIRV 0273.4009.9047011

31

Example Cont’d

• Alternatively using a current divider equation

• Does IT = IL + IR2?

oo

oP

R

oo

o

L

PL

oooTPP

mAR

VI

mAX

VI

mAkIZV

8.6488.303000

8.6464.11

2.2523.8901410

8.6464.11

8.6464.11009.98.6428.1

22

oo

ooP

TR

oo

oo

L

PTL

mAk

mAR

ZII

mAk

mAX

ZII

8.6488.303000

8.6428.1009.9

2.2523.8901410

8.6428.1009.9

22

32

33

Remember!!

• For Series RL Circuits– IT and VR are the 0o phase reference

– VS and ZT have the same phase

• For Parallel RL Circuits– VS is the 0o phase reference

– IT and ZT have the same phase magnitude but opposite signs

• For Series-Parallel RL Circuits– Use same rules as Series RL Circuits