single phase system. v l = v r = v irir ilil i l lags v by /2 i r and v are in phase r-l parallel...
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Single Phase System
VL = VR = V
I
IR IL
IL lags V by/2LLL
L X
V
X
V
X
VI 2
2
R
VIR IR and V are in phase
R-L parallel circuit
)II(I 2L
2RT
L
R
X
R
I
I
LR
L
111 tantantan
R
Z
I
I
T
R 11 coscos
IR
V
ILIT
/2
or
From the phasor diagram
2L
2 )X/V()R/V(
YZ
XRV
IL
T 1
)/1()/1( 22
][admittance1
SYZ
][econductanc1
SGR
][esusceptanc1
SBX LL
2222 )/1()/1( LRBGY L
VG
BLY
-/2
Admittance triangle
For parallel circuit we look at admittance
= tan-1 (BL/G) = tan-1 (1/LG)
cos = G/Y = Z/R Power factor
VC = VR = V
IC leads V by/2CCC
L X
V
X
V
X
VI 2
2
R
VIR IR and V are in phase
ICIR
I
R-L parallel circuit
CBC RG /1
)( 22CRT III VIR
IC IT
/2
From the phasor diagram
2C
2 )X/V()R/V(
2C
2 )X/1()R/1(V
IY
22 )()/1( CR
2C
2 )B()G(
VG
BL Y
-/2
= tan-1 (BC/G) = tan-1 (C/G)cos = G/Y = Z/R power factor
Admittance triangle
7.76105.41502
1
2
16 Cf
XC
AR
VIR 0.2
115
230
A circuit consists of a 115 resistor in parallel with a 41.5 F capacitor and is connected to a 230 V, 50Hz supply. Calculate: (a) The branch currents and the supply current;(b) The circuit phase angle;(c) The circuit impedance
I
IRIC
115 41.5 F
230V50Hz
V
AX
VI
CC 0.3
7.76
230
continue
AIII CRT 6.30.30.2)( 2222
o
TI
VZ 3.569.63
6.3
230
oR
I
I3.56
0.3
0.2coscos 11
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 voltage I
IR IC
50 F
100V50Hz
V IL
0.15H
solution
AR
VIR 0.2
50
100
AX
VI
LL 12.2
15.0502
100
AX
VI
CC 14.310100502100 6
AIIII LCRT 24.212.214.30.2 2222
'4526o893.024.2
2cos
I
IR
IC
IR
IC -I L
IL
IT
V
Parallel impedance circuits
2211 coscoscos III 21 III
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
2211 sinsinsin III
(active)
(reactive)
I
II 22111 sinsincos
2211
22111
coscos
sinsintan
II
II
222
211 )sin()cos( III
CCBBAA IIII coscoscoscos CBA IIII
ICIAIB
V
I
10 /-60 oA 5/-50 oA 10 /9 0oA
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.
CBAI sin10sin5sin10sin
V
IC
IIB
IA
I A +I B
A
B
C
AI 33.9)90cos(10)30cos(5)60cos(10cos
o
I
I1.7
33.9
16.1tan
cos
sintan 11
)sincos( 22 III
A4.916.133.9 22
oI 1.74.9
AI 16.1)90sin(10)30sin(5)60sin(10sin
The current is 7.1o lags the voltage
100318.05022 LfX L
AZ
VI 05.2
112
230
11
11210050 222211 LXRZ
o
Z
R5.63447.0cos
112
50coscos 11
1
111
V
I
75
159 F0.318H
50
CL
R 2R 1
I1 I2
A coil of resistance 50W and inductance 0.318H is connected in parallel with a circuit comprising a 75W resistor in series with a 159F capacitor. The resulting circuit is connected to a 230V, 50Hz a.c supply. Calculate:
(a) the supply current;(b)the circuit impedance; resistance and reactance.
o5.6305.2 1I Lags the voltage
V L
VR1IR1
V
1
V L
VR1I 1
V
1
7.772075 222222 CXRZ
2010159502
1
2
16 Cf
XC
AZ
VI 96.2
7.77
230
22
oC
R
X15267.0tan
75
20tantan 11
2
12
o1596.2 2I
Leads the voltage
I
I2
V2
I1
1
VC
VR2
I2
V
2
21 III
A
IIIoo 77.3)15cos(96.2)5.63cos(05.2
coscoscos 2211
AIII 9.307.177.3sincos 2222
7.589.3
230
I
VZ
569.3
77.3.7.58
cos.cosI
IZZR
169.3
07.1.7.58
sin.sinI
IZZX
A
IIIoo 07.1)15sin(96.2)5.63sin(05.2
sinsinsin 2211
More capacitive
(b)
Resonance circuits
V
L
C
R
I
2
22
22
Cf
ILIfRIV
Z
V
CfLfR
VI
2
2
21
2
O A I
ED
B
C
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
Z
R
ZI
RIcos
Z
Xsin
CL XX Cf2
1-Lf2XreactanceResultant
R
XX CL tan
2
2
2
12
CfLfR
I
VZ
Therefore the impedance is
LCr
1
CL
rr
1
CL XX
LCfr 2
1
When resonance
or
Z=Rand therefore I=V/R ; =0
Current (I) and impedance (Z) vs frequensi (f) in R, L & C serial circuit
f [Hz]
Z []
Z
I
0
R
fr
I [A]
At f=fr increasing of voltage occured whereVL > V or VL = QVVC > 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 = 2fr
C
L
R
1Q Q is a circuit tuning quality
A circuit having a resistance of 12 , an inductance of 0.15H and a capacitance of 100 F 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
2
2
2
12
CfLfRZ
2
62
10100502
115.050212
(a)
4.1985.311.47144 2
note
XL=47.1XC=31.85
'5051100
8.61coscos 11 oR
V
V
VIXV CC 16485.3115.5
VIRVR 8.611215.5
AZ
VI 15.5
4.19
100
5.2421.4715.5 LL IXV
VR=61.8I
V=100
VC=164
V L-V C
V L=242.5
=78.5
(b)
(c)
(d)
HzLC
fr 1.411010015.02
1
2
16
(e)
IC
IL
ZL
2/ CC XZ
R
X L1tan
LL
L Z
V
Z
VI
22LL XRZ
2/2/
CC
C X
V
X
VI
Red notation for impedance triangle
ILVR
VLV
XLZL
R
VC
IC
ZL=ZL/
L
L
Z
Xsin
LL ZZ
11LY
CL YYY
2/1
2/
1
CC XXCY
cos1
2/cos1
cos1
cosLCL ZXZ
YG
)( 22 BGY
V
IC
IL
I VC
Not at resonance
CLCL XZXZ
YB1
sin1
2/sin1
sin1
sin
G
B1tan YY
022 CLL XXXR
02 CLL XXZV
IC
IL
I
At resonance
At resonance =0 , thus sin =0
CL
L
LCL XZ
X
ZXZB
111sin
10
022 L/CL)(ωR r
222r R
C
LL
2
2
r1
L
R
LC
2
2
r1
2
1f
L
R
LC
QR
L
R
X rL tan
LC2
1fr
When R is so small we can put R=0 then
R
X L1tan Previously we have
Q factor of the circuit