Transcript Cct component - Universiti Sains Malaysia

Single Phase System
I
R-L parallel circuit
IR
IL
VL = VR = V
V  p2
V
V
IL 


p
XL XL  2
XL
IR 
V
R
IL lags V by p/2
IR and V are in phase
IR
V

From the phasor diagram
p/2
I T  (I R  I L )
2
2
IT
IL
 (V / R)2  (V / XL )2
IT
1
2
2
 (1 / R)  (1 / X L )   Y
V
Z
IL
1 R
1 R
  tan
 tan
 tan
IR
XL
L
1
or
IR
1 Z
  cos
 cos
IT
R
1
Admittance triangle
For parallel circuit we look at admittance
G
1
 Y  admittance [ S ]
Z
V
1
 G  conductanc e[ S ]
R
1
 BL  susceptance[ S ]
XL
Y  G 2  BL  (1 / R) 2  (1 / L) 2
2
 = tan-1 (BL/G) = tan-1 (1/LG)
cos  = G/Y = Z/R

-p/2
Power factor
BL
Y
R-L parallel circuit
I
IR
VC = VR = V
V p2
V
V
IL 


p
XC
XC   2
XC
V
IR 
R
IC leads V by p/2
IR and V are in phase
IC
IC
IT
From the phasor diagram
p/2

IT  ( I R  I C )
2
2
 (V / R) 2  (V / X C ) 2
I
Y   (1 / R) 2  (1 / X C ) 2
V
 (1 / R )  (C )
2
IR
V
Admittance triangle
G
V
2

-p/2
 (G) 2  (BC ) 2
 = tan-1 (BC/G) = tan-1 (C/G)
cos  = G/Y = Z/R power factor
BL
G  1/ R
Y
BC  C
A circuit consists of a 115 resistor in parallel
with a 41.5 mF capacitor and is connected to a
230 V, 50Hz supply. Calculate:
(a) The branch currents and the supply current;
I
(b) The circuit phase angle;
(c) The circuit impedance
V
IR
230V
V 230
IR  
 2. 0 A
50Hz
R 115
115
XC 
1
1

 76.7
2p f C 2p  50 41.5 106
V
230
IC 

 3.0 A
X C 76.7
IC
41.5mF
continue
IT  ( I R  I C ) 
2
  cos
1
2
2.0
2

 3.02  3.6 A
IR
1 2.0
 cos
 56.3o
I
3.0
V 230
Z

 63.956.3o 
IT 3.6
Three branches, possessing a resistance of 50W, an inductance
of 0.15H and a capacitor of 100mF respectively, are connected
in parallel across a 100V, 50Hz supply. Calculate:
(a) The current in each branch;
(b) The supply current;
(c) The phase angle between the supply current and the supply
I
voltage
V
100V
50Hz
IR
50
IC
mF
IL
0.15H
IC
solution
IT
IC-IL
V 100
IR  
 2.0 A
R 50
V
100
IL 

 2.12A
X L 2p  50 0.15

IR
IL
V
IC 
 100 2p  50100106  3.14A
XC
I T  I R  I C  I L   2.02  3.14  2.12  2.24A
2
2
IR
2
cos  

 0.893
I
2.24
2
  26o 45'
V
Parallel impedance circuits
Impedance sometime has the magnitude and phase . For
example we combine the resistance and inductance such in
inductor. In practical inductor has resistance and inductance.
If we have two impedances in parallel the current passing
through impedance 1 will be I1 and in impedance 2 will be
I2. To solve this, these current components can be resolve
into two components, i.e active and reactive , thus
I  I1  I 2
I cos  I1 cos1  I 2 cos2
(active)
I sin   I1 sin 1  I 2 sin 2
(reactive)
I  ( I1 cos 1 ) 2  ( I 2 sin 2 ) 2
I1 sin 1  I 2 sin 2
  tan
I1 cos1  I 2 cos2
1
  cos
1
I1 sin 1  I 2 sin 2
I
A parallel network consists of branches A,B and C. If
IA=10/-60oA, IB=5/-30oA and IC=10/90oA, all phase angles,
being relative to the supply voltage, determine the total
supply current.
IC
I
V
IB
IC
IA
o
o
10/-60oA 5/-50 A 10/90 A
C
A
I  I A  IB  IC
I cos  I A cosA  I B cosB  I C cosC
I sin   10sin A  5 sin B  10sin C
B

IB
V
I
IA
IA +I
B
I cos  10cos(60)  5 cos(30)  10cos(90)  9.33A
I sin   10sin(60)  5 sin(30)  10sin(90)  1.16A
I  ( I cos 2   I sin 2  )

9.33   1.16 
2
2
 9.4 A
I sin 
1  1.16
  tan
 tan
 7.1o
I cos
9.33
1
I  9.4  7.1
o
The current is 7.1o lags the voltage
A coil of resistance 50W and inductance 0.318H is connected in parallel
with a circuit comprising a 75W resistor in series with a 159mF
capacitor. The resulting circuit is connected to a 230V, 50Hz a.c
supply. Calculate:
(a) the supply current;
(b)the circuit impedance; resistance and reactanceI.
X L  2p f L  2p  50 0.318  100
V
R1
I1
50
L
0.318H C
R2
I2
75
Z1  R1  X L  50 2  100 2  112 
2
2
V 230
I1 

 2.05A
Z1 112
1  cos1
R1
50
 cos1
 cos1 0.447  63.5o
Z1
112
I1  2.05  63.5o Lags the voltage
VL
V
1
IR1
1
VR1
159mF
VR2
1
1
XC 

 20
6
2p f C 2p  5015910
2
I2
Z 2  R2  X C  752  202  77.7
2
2
V
V
230
I2 

 2.96A
Z 2 77.7
2  tan1
XC
20
 tan1
 tan1 0.267  15o
R2
75
VC
Leads the voltage
I 2  2.9615o
I  I1  I 2
I cos  I1 cos1  I 2 cos2
I2
2
1
V

I
 2.05cos(63.5o )  2.96cos(15o )  3.77A
I1
I sin   I1 sin 1  I 2 sin 2
 2.05sin(63.5o )  2.96sin(15o )  1.07A
I
I cos  2  I sin  2

3.77 2   1.07 2
 3.9 A
(b)
Z
V 230

 58.7
I 3.9
I cos 
3.77
R  Z cos   Z .
 58.7.
 56
I
3.9
I sin 
 1.07
X  Z sin   Z .
 58.7.
 16
I
3.9
More capacitive
Resonance circuits
I
From phasor diagram, since the voltage
VL (BO) and VC (CO) are in line, thus the
resultants for these two component is DO
(BO-CO) which is not involved in phase.
AO is the voltage across R. Thus
EO2= AO2 + DO2

I 
2
2

V  RI    2p f LI 
2p f C 

I
V
R
V
L
C
B
2
V

2
Z


1

R 2   2p f L 
2p f C 

D
O
C
E
A
I
Therefore the impedance is

V
1 
2

Z   R   2p f L 
I
2p f C 

Resultant reactanceX  2p f L -
X L  XC
tan  
R
When resonance
1
 r L 
r C
and Z=R
2
1
 X L  XC
2p f C
X
sin  
Z
RI R
cos  

ZI Z
X L  XC
1
r 
LC
therefore
I=V/R ;
or
fr 
 =0
1
2p LC
Current (I) and impedance (Z) vs frequensi (f) in
R, L & C serial circuit
Z []
I [A]
Z
R
I
0
fr
f [Hz]
At f=fr
increasing of voltage occured where
VL > V
or
VL = QV
VC > V
or
VC = QV

Q = VL/V = VC/V
In serial circuit:
VL = IXL; VC = IXC and V = IR

Q = IXL/IR = IXC/IR
Q = XL/R
= XC/R
Substitute XL and XC
Q = rL/R
= 1/rCR
where r = 2pfr
1
Q
R
L
C
Q is a circuit tuning quality
A circuit having a resistance of 12 , an inductance of 0.15H and a
capacitance of 100 mF in series, is connected across a 100V,
50Hz supply. Calculate:
(a)The impedance;
(b)The current;
(c)The voltage across R, L and C;
(d)The phase difference between the current and the supply voltage
(e)Resonance frequency
(a)

1 
2

Z  R   2p f L 
2p f C 

2

1
 122   2p  500.15 
2p  50100106

 144  47.1  31.85   19.4
2



2
note
XL=47.1
XC=31.85
(b)
I
V 100

 5.15 A
Z 19.4
VL=242.5
V=100
(c) VR  IR  5.1512  61.8V
VL-VC
=78.5
VL  IX L  5.15 47.1  242.5

VC  IXC  5.15 31.85  164V
(d)
(e)
  cos 1
fr 
VR
61.8
 cos 1
 51o50'
V
100
I
VR=61.8
VC=164
1
1

 41.1Hz
6
2p LC 2p 0.1510010
IL
IC
Z L  R 2  X L2
  tan
1
XL
R
ZL
XL
sin  
ZL
ZL=ZL/
V
V
IL 

 
Z L  Z L
ZC  X C   p / 2
V
V
IC 

p / 2
X C  p / 2 X C
Red notation for impedance triangle
XL
ZL
V VC
VL
IC

R
VR
IL
Not at resonance
Y  YL  YC
YL 
IC
1
1

 
Z L  Z L
 I
1
1
YC 

p / 2
X C  p / 2 X C
V
1
1
1
G  Y cos 
cos   
cosp / 2 
cos  
ZL
XC
ZL
B  Y sin  
1
1
1
1
sin    
sin p / 2 
sin    
ZL
XC
ZL
XC
Y  (G  B )
2
2
B
  tan
G
1
VC
Y  Y
IL
At resonance
At resonance =0 , thus sin  =0
1
1
1  XL  1
 
 
B0
sin    

ZL
XC ZL  ZL  XC
2
ZL  X L X C  0
IC
I
V
R2  X L2  X L X C  0
IL
R  (ωr L)  L/C  0
2
2
L
r L   R 2
C
2 2
2
 1
R
 r  
 2

LC
L


1
fr 
2p
2
 1
R

 2
 LC L 


When R is so small we can put R=0 then
fr 
1
2p LC
Previously we have   tan
1
XL
R
X L r L
tan  

Q
R
R
Q factor of the circuit