Download chapter 6-4 - UniMAP Portal

Lecture 04:
AC SERIES/
PARALLEL
CIRCUITS,
VOLTAGE/
CURRENT DIVIDER
OBJECTIVES
1. Compute the total circuit impedance for
series RLC circuits.
2. Explain the effects of frequency changes on
RLC circuit response.
3. Determine circuit voltages and currents in
series RLC circuits.
4. Compute voltage drops across components
in an RLC circuit using the voltage divider
formula.
Example #1:
Parallel Circuits
iS
iRL
iC
R=100
VS
XL=100
50V<0o
XC=150
(a) Find total impedance
• The circuit given is
already in freq
domain:
iS
iRL
iC
R=100
VS
XL=100
50V<0o
XC=150
 100 
o
Z C  150  90   j150 
o
 j100 
Z L  10090
Z R  1000
o
Circuit simplification
Z total  Z C  Z R  Z L   Z C  Z RL 
100  10090
 150  90 100  j100
 150  90 141.4245

150  90141.4245

 j150  100  j100
 150  90
21213  45
21213  45


100  j 50
111.8  26.57
 190  18
 180  j 60 
(b) Draw Impedance
Triangle
Z  190  18  (180  j 60)
o
R
j
R
q
ZT
-j
ZT
q
XC
(c) Find is, iC, iRL
vs
500
is 

 26318 mA
Z T 190  18
vS  vC  vR  vL   500 V
vS
500
 iC 

 33390 mA
Z C 150  90
 iRL
vS
500


 353  45 mA
Z RL 141.4245
(d) Using Current Divider

ZT
190  18
ic 
iS 
 26318 mA
150  90
ZC
 33390 mA
iRL

ZT
190  18

iS 
 26318 mA
10010090
Z RL
 353  45 mA
Example #2: Find i and vc
Elements in Frequency
domain
• =4 rad/s
Time domain
vs  10 cos 4t
R 5
C  0 .1 F
Freq domain
Vs  100
R 5
1
XC   j(
)   j 2.5
C
Circuit in Freq Domain
(a) The current, i
• The current in series circuit,
Vs
I
ZT
• The total impedance is;
ZT  5   j 2.5
 5  j 2.5  5.59  26.56 
Vs
100
I

ZT 5.59  26.56
 1.7926.56
 1.6  j 0.8 A
(b) The capacitor voltage, VC
VC  IZC
 1.7926.56   j 2.5
 1.7926.56  2.5  90
 4.48  63.44 V
Example #3: Find iX
(time domain)
Solution:
iX (t )  0.4607 cos2000t  52.63
Example #4: Find Z1 and Z2
Solution:
Z1  3  j 21
Z 2  1.535  j8.896
Example #5: Find i(t)
Different waveform of VS
 VS1 and VS2 are different:
vs1  20 cos4t  90
vs 2  10 sin 4t  15
 When analyzing ac circuit, it is expedient to
express both signals as either Sine or Cosine.
Change the waveform of
VS1
• Using identity:
 cost   sin t  90
 sin t   cost  90
 cost   cost  180
 sin t   sin t  180
Changing Cosine to Sine:
vs1  20 cos4t  90
 vs1  20 sin 4t  90  90
 vs1  20 sin 4t 
Example #6: (a) Find i(t)
Different waveform of VS
 VS1 and VS2 are different:
vs1  20 cos4t  90
vs 2  10 sin 4t  15
 When analyzing ac circuit, it is expedient to
express both signals as either Sine or Cosine.
Change the waveform of VS1
• Using identity:
 cost   sin t  90
 sin t   cost  90
 cost   cost  180
 sin t   sin t  180
Redraw the circuit
Elements in Frequency
domain
• =4 rad/s
Time domain
vs1  20 sin 4t
vs 2  10 sin 4t  15
Freq domain
Vs1  200
Vs 2  10  15
R  60 
R  60 
C  10 mF
1
XC   j(
)   j 25
C
L5H
X L  jL  j 20
Find The Current
VT
I
ZT
VT  vs1  vs 2 ZT  Z1  Z 2 Z3 
Solution:
VT  29.66  j 2.59  29.77  4.99 V
ZT  60  j90  108.1756.31 
I  0.28  61.3 A
 i (t )  0.28 sin 4t  61.3A
(b) Find the vc(t), vL(t)
Answer :
VC  VL  2828.7 V
 vC (t )  vL (t )  28 sin 4t  28.7V
(c) Find the ic(t), iL(t)
Answer :
I C  1.12118.7 A
I L  1.4  61.3 A
 iC (t )  1.12 sin 4t  118.7A
 iL (t )  1.4 sin 4t  61.3A
(d) Sketch the waveform for vc(t),
vL(t), ic(t), iL(t)
(e) Explain the phase relationship
between V and I for capacitor and
inductor.
Example #7: Find ZT, IT,
VC2, P
R2
R1
Xc2
Xc1
R1  10k
X C1  12k
R3
Vs
VS  10V0
Z2
R2  8k
X C 2  10k
R3  11k
Solution
• Find the total impedance:
Z 2  R3 R2  X C 2   6.56k  23.6
 ZT  R1  X C1  Z 2  21.7k  42.4
• Then, the total current is:
VS
10V0
IT 

 461A42.4
Z T 21.7 k  42.4
Solution
VR2C2 R3
VC2 
Z2
6.56k  23.6
 VS 
10V0  3.03V18.8
ZT
21.7k  42.4
Z C2
Z R2C2
VR2C2 R3
10k  90

3.03V18.8  2.36V  19.8
8k  j10k
P = VI cos (qT) = (10V)(461 A) cos(-42.4o)
= 3.40 mW
Ex: Find ZT, IL3, P
VS  10V0
R2
R1
XL1
R3
E
XL2
Z2
Z3
R4
XL3
R1  5k
X L1  4k
R2  6k
X L 2  2k
R3  7k
X L 3  3k
R4  8k
Z2, Z3, ZT, and IT
Z 2  3.61k18.38
Z 3  8.41k18.2
Z T  9.87 k31.4
VS
IT 
 1.01mA  31.4
ZT
IR3, IL3, P
Z2
3.61k18.4
I R3  IT 
1.01mA  31.4  435A  31.2
Z3
8.41k18.2
I L3 
Z L3R4
Z L3
2.81k69.4
I R3 
435A  31.2  407.3A  51.8
3k90
P = EI cos (qT) = (10V)(1.01 mA) cos(31.4o)
= 8.62 mW
TOOLS
• Note from the previous examples, all of
our tools from DC work:
–
–
–
–
–
Ohm’s Law
Voltage Divider
Current Divider
Kirchoff’s Current Law
Kirchoff’s Voltage Law
• They are the same – but we use
COMPLEX NUMBERS.