Download (RL) Circuits

Resistive-Inductive (RL) Circuits
AC Circuits I
Series RL Circuits
Series circuits: A comparison (Magnitude)
IT = IR1 = IR2
IT = IL1=IL2
VT =VR1+VR2
VT = j(VL1 + VL2)
RT = R1+R2
XT = XL1 + XL2
IT = IL=IR
|VT| = sqrt(VR2 + VL2)
|ZT| = sqrt(R2 + XL2)
2
Series RL Circuits
• Series Voltages
3
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.
4
Series RL circuits
• Series Impedance
VL 90 VL
XL 
 90
I0
I
VR 0 VR
R

0
I0
I
Z T  X L2  R 2
1 
XL 
  tan 

 R 
Polar form
Polar form
Rectangular form
Rectangular form
5
Example
• The component voltages shown below were measured
using an ac voltmeter (HINT: rms values!!!). Calculate
the source voltage for the circuit.
VS  VR2  VL2  6 2  32  45  6.71
3
6
VS  6.71  26.57 o Vrms
 S  tan 1    26.57 o
6
Example
• Calculate the total impedance for the circuit
shown (remember Z = R +jXL , where XL=2πfL )
X L  2fL  2 (20000)(1mH )  40  125.66
R  100
Z T  R  jX L  100  j125.66 
Z T  R 2  X L2  100 2  125.66 2  160.60
 125.66 
o

51
.
49

T
 100 
Z T  160.60 51.49o 
 Z  tan 1 
7
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.
8
Example
• Calculate the circuit impedance, and current in the circuit shown below
X L  2fL  2 (10000)( 200mH )  4 k  12.566 k
R  15 k
Z T  R  jX L  15  12.566 k
Z T  R 2  X L2  152  12.566 2  19.568 k
 12.566 
o

39
.
95

T
 15 
Z T  19.568 39.95o k
 Z  tan 1 
Vs
20 0o
o
I


1
.
02


39
.
95
mA
o
Z T 19.568 39.95 k
• I and Vs can be represented as either: (I lags Vs or Vs leads I)
Vs  200o V
I  1.02  40 mA
Vs  2040o V
I  1.020 mA
9
Example
•
Determine the voltage, current and impedance values for the circuit shown below
X L  2fL  2 (20000)(3.3mH )  132   414.69 
R  91
Z T  R  jX L  91  j 414.69 
Z T  R 2  X L2  912  414.69 2  424.56 
 12.566 
o

77
.
62

T
 15 
Z T  424.56 77.62 o 
 Z  tan 1 
Vs
0.1 0 o
I

 235.5   77.62 o A
o
Z T 424.56 77.62 
I  235.5 0 o A
Vs  0.1 77.62 o
VR  IR  235.5 0 o A  91 0 o  21.4 0 o mV
VL  I ( jX L )  235.5 0 o A   j 414.69 
 235.5 0 o A  414.69 90 o   97.7 90 o mV
10
Voltage Dividers
• RL Voltage Dividers
Zn
Vn  VS
ZT
where
Zn
Vn
= magnitude of R or XL
= voltage across the component
• In a series RL circuit where I(VR )is used as the 0o phase
reference, Vs and ZT always have the same phase angle!!!
11
Example
•
Determine the voltages VL and VR in the circuit below
X L  2fL  2 (20000)(3.3mH )  132   414.69 
R  91
Z  R  jX L  91  j 414.69 
Z  R 2  X L2  912  414.69 2  424.56 
 12.566 
o
  77.62
 15 
Z  424.56 77.62o 
R
910
o
VR  Vs
 0.1 77.62o

21
.
4

0
mV
ZT
424.56 77.62o
 Z  tan 1 
VL  Vs
jX L
414.6990
 0.1 77.62o
 97.590o mV
o
ZT
424.56 77.62
12
Example
• Calculate ZT, VL1 VL2 and VR for the circuit
shown below.
Z T  3.6484.8 k
o
X L1  1.13 k
X L 2  2.49 k
VL1  3.1090 V
o
VL 2  6.8490 V
o
VR  0.910 V
o
13
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
14
Power
• Power is also a complex quantity in AC circuits
PAPP  PR  jPX
15
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 voltamperes (VAs).
16
Power Factor (PF)
• The ratio of resistive power to apparent power
PR
PF 
 cos θ
PAPP
17
Example
• Calculate PR, PX and PAPP for the following
circuit - (see slide 13)


PR  I R  235.5 0 A 910o   5.0460o W
2
2
o


PX  I X L  235.5 0 A 414.69 90o   22.99890o W
2
o
2
PAPP  PR  jPX  5.046  j 22.998W  23.55 77.625o 
18
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 + IL2)
RT = 1/[ (1/R1 )+(1/R2)]
XT = 1/[(1/XL1 ) + (1/XL2 )]
XT = 1/[(1/R ) + (1/jXL )]
19
Parallel RL Circuits
• Branch Currents
– IL lags IR , VR and
VS by 90º
20
Parallel RL Circuits
• Parallel Circuit Impedance
– Impedance phase angle is positive since current phase
angle is negative
VS
ZT 
IT
– Angles of ZT and IT have the same magnitude but
opposite signs
• Calculating Parallel-Circuit Impedance (polar form)
ZT 
X LR
X L2  R 2
R
 Z T   IT  tan
XL
1
21
Parallel RL Circuits
• Calculating Parallel-Circuit Impedance
– rectangular form approach:
R  jX L
ZT 
R  jX L
ZT 
1
1
1

R jX L
22
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
23
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)
I R  20mA, I L  5mA
VS  60o V
I T  20  j 5 mA
I T  20.62  14.04o mA
I T  I R  I L  425  20.62
VS
60o V
o
ZT  

0
.
29

14
.
04
k
o
I T 20.62  14.04 mA
2
2
I 
 5
 IT  tan 1  L   tan 1    14.04o
 20 
 IR 
I T  20.62  14.04o
24
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)
R  300, X L  1200
ZT 
R  jX L
j 360000
j 360000
300  j1200



R  jX L
300  j1200 300  j1200 300  j1200
432000000  j108000000
 282.35  j 70.59
1530000
Z T  291.0414.04 o 

VS
60 o
o
IT 


20
.
62


14
.
04
mA
o
ZT
291.0414.04 
25
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)
R  300, X L  1200
ZT 
R XL
R  XL
2
2

360000
360000

 291.04
1236
.
93
1530000
 R 
 300 
o
  tan 1 
 ZT  tan 
  14.04
 1200 
 XL 
Z T  291.0414.04 o 
1
VS
60 o
o
IT 


20
.
62


14
.
04
mA
o
Z
291.0414.04 
26
27
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
28
Example
• Calculate the Equivalent Impedance, the total current and the current in each
branch of the parallel RL circuit shown below
o
ZT  1.54k48.8
Z P  1.28k64.8o
ZP 
X L  R2
X L  R2
2
2

1.41k  3k
1.41k  3k
2
2
 1.28k
R 
 3k 
o
 Z P  tan  2   tan 1 
  64.8
 1.41k 
 XL 
Z P  1.28k64.8o Now convert to rect
1
Z s  1.28k64.8o
Z P  Z s  1.28k64.8o
 


Z S  1.28k cos 64.8o  j sin 64.8o

Z S  545  j1158
ZT  470  Z S  1015  j1158
ZT  1.54k48.8o and back to polar
29
Example Cont’d
• Knowing that
ZT  1.54k48.8o
VSRC
1448.8o
o
IT 


9
.
09
mA

0
ZT
154048.8o
• Remember in a series circuit, IT is the phase
o
o
o
V

R

I

470


0

9
.
09
mA

0

4
.
273
V

0
reference
VS and ZT always have the same
R1
1 and
T
phase!!!
30
Example Cont’d
VP  Z P  I T  1.28k64.8o  9.09mA0o  11.6464.8o
VP 11.6464.8o
o
IL 


8
.
23
mA


25
.
2
XL
141090o
I R2
VP 11.6464.8o
o



3
.
88
mA

64
.
8
R2
30000o
• Alternatively using a current divider equation
o
ZP
o 1.28k64.8
o
I L  IT
 9.09mA0

8
.
23
mA


25
.
2
XL
141090o
I R2
o
ZP
o 1.28k64.8
o
 IT
 9.09mA0

3
.
88
mA

64
.
8
R2
30000o
• Does IT = IL + IR2?
31
32
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
33