Download 8. Single Phase cct

Single Phase System
Pure Resistive Circuit in Series
i
Instantaneous voltage is given by
v = Vm sin t
v
Instantaneous current is given by
v Vm
i 
sin t
R
R
The maximum value for current Im and maximum value for
voltage Vm can be related as
Im = Vm/R
The rms value for current Irms (simply I) and rms value for
voltage Vrms (simply V) can be related as
I = V/R
R
In circuit contains resistor , the V and I are in phase as
in diagram below
waveform
+
Vm
Im
current
time
voltage
-
V
phasor
I
Power dissipated in the resistor
p = i2R = (Im2/R)sin2 t
p = v2/R = (Vm2/R) sin2 t
p = vi = VmIm sin2 t
Average value for power
Vm Im 2 / 
Vm Im 2 / 
2
Pav 
(sin t )dt 
(1  cos t )dt


0
0
2
4
2 / 
 Im Vm  1


t

sin
2

t

4  2
0

Vm Im Vm Im

.
 VI
2
2 2
power p
current i
voltage v
Wave in pure resistance circuit
Pure Inductive Circuit in Series
If the current isi = Im sin t;
di
vL  e  L
dt
d
 L ( I m sin t )
dt
i
v
 LI m cos t
 LI m sin( t   / 2)  Vm sin( t   / 2)
Thus Vm=  L Im
The voltage is leading the current by /2 rad (90o) OR
current is lagging behind the voltage by /2
v, i
Vm
v
i
Im
tt
/2
Voltage and current waveform in a
purely inductive circuit
Maximum voltage:
Vm = LIm
Voltage r.m.s value .:
V = LI
where  = 2f
V/I = Vm/Im = L = XL
XL is measured in ohm and called as inductive
reactance=L
Changes of I and XL with frequency
I,
XL
X
I
L
f
Phasor for purely inductive circuit
V
90
90
E
I
Power dissipated in purely inductive circuit
P = vi
= (Im sin t)(Vm sin (t + /2)
= VmIm sin t sin (t + /2)
= VmIm sin t [sin t cos /2 + cos t sin /2 ]
= VmIm sin t cos t
= ½VmIm sin 2t
Average power
Vm Im 2 / 
Vm I m
2 / 


Pav 
(sin
2

t
)
dt


cos2ωo
0
0

4 0
8π
Half cycle has cancelled the other half cycle that is why the
average power is zero.
v, i, p
p
v
i
+
+
t
-
CURRENT, VOLTAGE AND AVERAGE
POWER WAVEFORM IN A PURELY
INDUCTIVE CIRCUIT.
Pure capacitive Circuit in Series
If the voltage is given as
v = Vm sin t
i
Then
dv
d
iC
 C (Vm sin t )  CVm cos t )
dt
dt
v
 CVm sin( t   / 2)  I m sin( t   / 2)
Therefore Im=  C Vm
In this case the current is leading a voltage by /2 ( 90o) OR
voltage is lagging behind the current by /2.
v, i
Vm
Im
v
i
t
/2

AC VOLTAGE AND CURRENT
WAVEFORM IN PURELY
CAPACITIVE CIRCUIT
Maximum current value
Im = CVm
r.m.s value
.:
I = CV
ratio
V/I = Vm/Im = 1/C = XC
XC = is measured in ohm and called as capacitive reactance =1/C
Changes of I and XC with frequency
I, Xc
Xc
I
f
Phasor diagram
I
/2
V
Power dissipated by capacitor
p = vi
= (Vm sin t)(Im sin (t + /2))
= VmIm sin t sin (t + /2)
= VmIm sin t [sin t cos /2 + cos t sin /2 ]
= VmIm sin t cos t
= ½VmIm sin 2t
Average power
Vm Im 2 / 
Vm I m
2ππ/


Pav 
(sin
2

t
)
dt


cos2ωo
0
0

0
4
8π
Half cycle has cancelled the other half cycle that is why the
average power is zero.
voltan
v
arus i
Kuasa p
CURRENT, VOLTAGE AND POWER IN
PURELY CAPACITIVE CIRCUIT
i
If
i = Im sin t
Then
vR
v
vL
vR = iR = ImR sin t
in phase with i
vL = iXL = ImXLsin (t + /2)
= LIm sin (t + /2)
leading the i by /2
v = vR + vL = ImR sin t + LIm sin (t + /2)
= ImR sin t + LIm cos t
v = Vm sin (t + )
(1)
Vm  I m R 2  (L) 2
Where
(2)
and = tan-1(L/R)
(3)
Current and Voltage waveform in L-R serial circuit
v, i
v
i
vR
t
vL
VR in phase with I
VL is leading in phase with I by /2
Phasor diagram for I, VR, VL & V in R & L serial circuit
VL
V
/2

I
VR
•I and VR overlap to represent the in phase
•VL is vertical to represent the 90o leading out of phase
•Resultant between VL and VR give the value of V
V  (VR  VL
2
2
and
  tan1 VL / VR 
The representation of voltage can be written as V / 
XL
Phasor diagram for I, R,
XL & Z in R & L serial
circuit
Z
/2

I
Z  ( R 2  X L  R 2  (L) 2
2
  tan
1
 X L / R  tan L / R
Impedance is represented as
1
Z  Z 
R
p = vi = Vm sin (t + ) Im sin t
= ½VmIm [cos  - cos (2t - )]
= ½VmIm cos  - ½VmIm cos (2t - )
component ½VmIm cos (2t - ) is zero
Therefore the average value is only given by
1
Vm I m
P  VmI m cos  
.
 Vrms I rms
2
2 2
CURRENT, VOLTAGE AND POWER
WAVEFORMS FOR L-R SERIAL CIRCUIT
V,I
P
p
i
t
v
i
v
R, XL,
Z ()
Z
X
L
R
f
(Hz)
Phasor diagram
V
I cos 

I
VR = V cos 
I sin 
We can also calculate the power from
P = I2R or P = VR2/R or VRI
All in r.m.s values I= Irms , VR=VRrms
From phasor diagram P=VRI=VIcos (active power)
Cos  is a power factor
cos  = VR/V = R/Z.
Reactive power (VAR) = VLI = VI sin 
I
A resistance of 7.0 is connected in
series with a pure inductance of 31.8mH
and the circuit is connected to a 100V,
50Hz, sinusoidal supply. Calculate (a) the
circuit current. (b) the phase angle
V=100V
50Hz
X L  2fL  2  50  31.8 103  10.0
Z  R  X
2
2
L
  7.0
1
2
2
 10.0
2

1
2
 12.2
V 100
I 
 8.2 A
Z 12.2
X
10.0
  tan 1 L  tan 1
 55o lag or  55o
R
7.0
VL
L=31.8mH
VR
R=7.0
A pure inductance of 318mH is connected
in series with a pure resistance of 75.
The circuit is supplied from a 50Hz
sinusoidal source and the voltage across
the 75 resistor is found to be 150V.
Calculate the supply voltage
V 150
VR  150V
I 
 2A
R 75
I
V
(50Hz)
VL
VR
=150V
L=318mH
R=75
X L  2fL  2  50  318 103  100
VL  IX L  2 100  200V

Z  R
  150
 X   75
V  VR2  VL2
2
2
L
1
2
2
1
2
2

 100 
 2002
2
1
2
1
2
 250V
 125
check
V  IZ  2 125  250V
A coil, having both resistance and
inductance, has a total effective impedance
of 50 and the phase angle of the current V
through it with respect to the voltage across
it is 45o lag. The coil is connected in series
with a 40 resistor across a sinusoidal
supply. The circuit current is 3A; by
constructing a phasor diagram, estimate the
supply voltage and the circuit phase angle
VR  IR  3 40  120V
VLr  IZ Lr  3 50  150V
I=3A
VL
VR
ZL=50
R=40
V
VLr
VLr cosLr

VR
V 2  VR2  VLr2  2VRVLr cos Lr  1202  1502  2 120 150  0.707
 62500
V  250V
cos  
VR  VLr cos Lr 120  150  0.707 

 0.904
V
250
  25o lag
A coil having a resistance of 12 W and a inductance of 0.1H is
connected across a 100V, 50Hz supply. Calculate:
I
(a)The reactance and the impedance of the coil;
(b)The current;
(c)The phase difference between the current
100V
50Hz
and the applied voltage:
R=12
L=0.1H
X L  2 f L  2  50  0.1  31.4
Z  R 2  X L2  12 2  31.4 2  33.6
IXL
V
V
100
I 

 2.97 A
Z
33.6
X
31.4
tan  

 2.617
R
12
  69o

IR
I
i
If i = Imsint
vR = iR = ImR sin t --- in phase with i
vR
Im
1
v c   idt 
( sin t)dt

C
C
vC
Im
Im

cos t 
sin( t   / 2)  I m X C sin( t   / 2)
C
C
v = vR + vC
= (Im/C) sin (t - /2) + ImR sin t
Vc is lagging by
90o refer to I
v = Vm sin (t - )
where
1
XC 
C
;
  2f
2
2
; Vm  R  (1 / C)
and
  tan1 (1 / CR)
vR
/2

/
3/2
2/
v
vC
t
i
CURRENT AND VOLTAGE WAVEFORMS
IN R-C SERIAL CIRCUIT
Phasor diagram
Impedance triangle
I
I
VR
-
R
-
/2
/2
V
VC
V  VR  VC
2
Z
XC
Z  R 2  XC
2
2
 R 2  (1 / C) 2
 = tan-1 (VC/VR)
Can be written as
 = tan-1 (XC/R)
V  
where
 = 2f
Z-
POWER
p = vi = Vm sin (t - ) Im sin t
= ½VmIm [cos  - cos (2t - )]
= ½VmIm cos  - ½VmIm cos (2t - )]
The average of component [½VmIm cos (2t - )] is zero, therefore
Vm I m
P  Vm I m cos  
.
cos   Vrms I rms cos 
2 2
1
2
p
CURRENT, VOLTAGE AND
POWER WAVEFORMS FOR
t
R-L SERIAL CIRCUIT
i
v
Phasor diagram
I sin 
We can also calculate the power from
P = I2R or P = VR2/R or VRI
All in r.m.s values I= Irms , VR=VRrms
I

I cos 
From phasor diagram P=VRI=VIcos (active power)
V
Cos  is a power factor
Cos  = VR/V = R/Z.
Z
Reactive power (VAR) = VCI = VI sin 
R
XC
f
A capacitor of 8.0F takes a current of 1.0 a
when the alternating voltage applied
across it is 230 V. Calculate:
(a)The frequency of the applied voltage;
(b)The resistance to be connected in series
with the capacitor to reduce the current in
the circuit to 0.5A at the same frequency;
(c)The phase angle of the resultants circuit.
(a)

XC 
0.5A
230V
V 230
1

 230 
I 1.0
2 f C
1
1
f 

 86.5Hz
6
2  CX C 2   8 10  230
C=8F
R
(b)

V 230
Z 
 460  R 2  X C2
I
0.5

(c)

1
2
R  Z 2  X C2  4602  2302  398
R
1 398
  cos
 cos
 30o
Z
460
1
Leading by 30o
I
A metal-filament lamp, rated at 750W, 100V,
is to be connected in series with a
capacitor across a 230V, 60Hz supply.
Calculate:
(a)The capacitance required
(b)The phase angle between the current and
the supply voltage
230V
60Hz
C
VC
100V
R
VR

V 2  VR2  VC2
2302
 100  VC2
VC  270V
2
VC
V
750W
Rated current of lamp 
 7.5 A
100V
I
XC
VC
1


2 f C
I
I
7.5
C

 73.7 F
2 f VC 2  60  270
(b)
VR
100
cos  

 0.435
V
230
  64 12'
o
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
L
C
B
2

I 

V  RI    2 f LI 
2 f C 

V
V
I

2
Z


1

R 2   2 f L 
2 f C 

2
2
R
V
D
O
C
E
A
I
Therefore the impedance is

V
1 
2

Z   R   2 f L 
I
2 f C 

2
1
Resultant reactance X  2 f L  X L  XC
2 f C
X L  XC
tan  
R
RI R
cos  

ZI Z
X
sin  
Z
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
(a)

1 

Z  R 2   2 f L 
2 f C 

2

1
 122   2  50 0.15 
2  50 100 106




2
 144  47.1  31.85  19.4
2
note
XL=47.1
XC=31.85
(b)
V 100
I 
 5.15 A
Z 19.4
VL=242.5
(c) V  IR  5.15 12  61.8V
R
VL  IX L  5.15  47.1  242.5
V=100
VL-VC
=78.5

VC  IX C  5.15  31.85  164V
VR=61.8
(d)
VR
1 61.8
  cos
 cos
 51o50'
V
100
1
I
VC=164