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Single Phase System
Pure Resistive Circuit in Series i
v Rv = Vm sin t
Instantaneous voltage is given by
tv
i sinR
V
Rm
Instantaneous current is given by
Im = Vm/R
The maximum value for current Im and maximum value for voltage Vm can be related as
I = V/R
The rms value for current Irms (simply I) and rms value for voltage Vrms (simply V) can be related as
Vm
Im
voltage
current
+
-
time
In circuit contains resistor , the V and I are in phase as in diagram below
I
V
waveform
phasor
p = i2R = (Im2/R)sin2 t
p = v2/R = (Vm2/R) sin2 t
p = vi = VmIm sin2 t
Power dissipated in the resistor
/2
0
m/2
0
2m)cos1(
4
I)(sin
2
Idtt
Vdtt
VP
mmav
Average value for power
VIVV
ttV mmm
2
I.
22
I2sin
2
1
4
I mm/2
0
m
voltage v
current i
power p
Wave in pure resistance circuit
v
i
Pure Inductive Circuit in Seriesi = Im sin t;If the current is
dt
diLevL
)sin( tIdt
dL m
tLIm cos
)2/sin()2/sin( tVtLI mm
The voltage is leading the current by /2 rad (90o) OR current is lagging behind the voltage by /2
Thus Vm= L Im
t
v, i
t
Vm
Im
/2
vi
Voltage and current waveform in a purely inductive circuit
Maximum voltage: Vm = LIm
Voltage r.m.s value .: V = LI
V/I = Vm/Im = L = XL
XL is measured in ohm and called as inductive
reactance=L
where = 2f
V
E
I90
90
I, XL I X
L
f
Changes of I and XL with frequency
Phasor for 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 2t
Power dissipated in purely inductive circuit
0cos2ωo8π
IV)2(sin
4
I /20
mm/2
0
m
dtt
VP
mav
Average power
Half cycle has cancelled the other half cycle that is why the average power is zero.
t
v, i, p
vi
p
++
--
CURRENT, VOLTAGE AND AVERAGE POWER WAVEFORM IN A PURELY
INDUCTIVE CIRCUIT.
Pure capacitive Circuit in Series
v
iv = Vm sin t If the voltage is given as
)cos)sin( tCVtVdt
dC
dt
dvCi mm
Then
)2/sin()2/sin( tItCV mm
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
t
v iVm
Im
/2
AC VOLTAGE AND CURRENT WAVEFORM IN PURELY
CAPACITIVE CIRCUITMaximum current value Im = CVm
r.m.s value .: I = CVratio V/I = Vm/Im = 1/C = XC
XC = is measured in ohm and called as capacitive reactance =1/C
I
V
/2
Phasor diagram
f
I, Xc
Xc
I
Changes of I and XC with frequency
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 2t
Power dissipated by capacitor
0cos2ωo8π
IV)2(sin
4
I 2ππ/0
mm/2
0
m
dttV
Pm
av
Average power
Half cycle has cancelled the other half cycle that is why the average power is zero.
arus ivoltan v
Kuasa p
CURRENT, VOLTAGE AND POWER IN PURELY CAPACITIVE CIRCUIT
vR
vL
i
v
vR = iR = ImR sin t in phase with ivL = iXL = ImXLsin (t + /2) = LIm sin (t + /2) leading the i by /2v = vR + vL = ImR sin t + LIm sin (t + /2) = ImR sin t + LIm cos tv = Vm sin (t + ) (1)Where (2)and = tan-1(L/R) (3)
Ifi = Im sin t
Then
22mm )( RIV L
Current and Voltage waveform in L-R serial circuit
VR in phase with I VL is leading in phase with I by
t
v, i
v
ivR
vL
VL
VRI
V
/2
Phasor diagram for I, VR, VL & V in R & L serial circuit
•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
2L
2R VV(V RL
1 V/Vtan and
The representation of voltage can be written as V /
RI
XL Z
/2
Phasor diagram for I, R, XL & Z in R & L serial circuit
2222 )(( LRXRZ L
RLRX L /tan/tan 11
ZZImpedance is represented as
p = vi = Vm sin (t + ) Im sin t
= ½VmIm [cos - cos (2t - )]
= ½VmIm cos - ½VmIm cos (2t - )
component ½VmIm cos (2t - ) is zero
rmsrmsmm
mm IVIV
IVP 2
.2
cos2
1
Therefore the average value is only given by
CURRENT, VOLTAGE AND POWER WAVEFORMS FOR L-R SERIAL CIRCUIT
p
i
v
t
V,I
i
v
P
f (Hz)
R, XL, Z ()
R
Z
X
L
Phasor diagram
V
I
I cos
I sin
VR = V cos
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
Reactive power (VAR) = VLI = VI sin
cos = VR/V = R/Z.
L=31.8mHV=100V50Hz
R=7.0
V L
V R
I
AZ
VI 2.8
2.12
100
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
ooL orlagR
X5555
0.7
0.10tantan 11
2.120.100.7 21
21
2222LXRZ
0.10108.315022 3fLX L
L=318mHV (50Hz)
R=75
V L
V R=150V
I
12510075 21
21
2222LXRZ
VVR 150
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
VVVV LR 250200150 21
21
2222
VIXV LL 2001002
AR
VI 2
75
150
100103185022 3fLX L
check
VIZV 2501252
Z L=50 V
R=40
V L
V R
I=3A
VIRVR 120403
A coil, having both resistance and inductance, has a total effective impedance of 50 and the phase angle of the current 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
VIZV LrLr 150503
62500
lago25
707.01501202150120cos2 22222 LrLrRLrR VVVVV
904.0
250
707.0150120coscos
V
VV LrLrR
VV 250
V LrV
V R
VLr cosLr
I
R=12
L=0.1H
100V50Hz
IX LV
IRI
o69
A coil having a resistance of 12 W and a inductance of 0.1H is connected across a 100V, 50Hz supply. Calculate:
(a)The reactance and the impedance of the coil;(b)The current;(c)The phase difference between the current and the applied voltage:
617.212
4.31tan
R
X
AZ
VI 97.2
6.33
100
4.311.05022 LfX L
6.334.3112 2222LXRZ
A serial ac circuit ,ABCD, contains a resistor AB, an inductor BC with resistance R and inductance L and a resistor CD. When a current 6.5A flow in the circuit, voltage drops across those points are: VAB = 65 V; VBC = 124 V; VAC = 149 V. The supply voltage is 220 V at 50 Hz. Calculate
(a)Draw the circuit(b)Voltage drop VBC and its phase compared to the current(c)Resistance and inductance of the inductor
A B
CD
Current= I = 6.5 ASupply voltage = V = 220 V Z = V/I = 220/6.5 = 33.8 (1)VAB = 65 V; R1 = 65/6.5 = 10 (2)VBC = 124 V; ZL = 124/6.5 = 19.1 (3)VAC = 149 V; ZAC = 149/6.5 = 22.9 (4)
From Equation (3):-ZL
2 = RL2 + XL
2
19.12= 364.8 =RL2 + XL
2 (5)
From Equations (2) and (4):-ZAC
2 = (R1 + RL)2 + XL2
22.92=524.4= (10 + RL)2 + XL2
(6)From Equations (5) and (6):-
RL = 3.0 (7)
Substitute RL in (5)XL = 18.9
(8)Dari (1)
Z2 = (R1 + RL + R2)2 + XL2
33.82=(10+3.0 + R2)2 + 18.92
R2 = 15
VR1 = 6.5 x 10 = 65 VVR2 = 6.5 x 15 = 97.5 VVRL = 6.5 x 3.1 = 20.2 VVXL = 6.5 x 18.8 = 122.2 V
VVVVV XLRRLBD 7.1692.122)5.972.20()( 22222
(b)
o46)5.972.20(
2.122tan 1
(b) Resistance of inductor = RL = 3.0 XL = 18.9 = 2fLL = 18.9/2f = 18.9/2 x 50 = 60.1 mH
vR
vC
i
If i = Imsint
vR = iR = ImR sin t --- in phase with i
t)dt(C
Idt
Cm
c sini1
v
)2/sin()2/sin(cos
tXItC
It
C
ICm
mm
C
1XC
f2
v = vR + vC
= (Im/C) sin (t - /2) + ImR sin t
v = Vm sin (t - )
22m )C/1(RV )CR/1(tan 1
where
; and;
Vc is lagging by 90o refer to I
v
vC
vR
it
/2
/ 3/2 2/
CURRENT AND VOLTAGE WAVEFORMS IN R-C SERIAL CIRCUIT
VR
VC
V
I
-
/2
2C
2R VVV
= tan-1 (VC/VR)
VCan be written as
R
XC
Z
I
-
/2
2C
2 XRZ
22 )C/1(R
= 2f
Z-
Phasor diagram Impedance triangle
where
= tan-1 (XC/R)
p = vi = Vm sin (t - ) Im sin t
= ½VmIm [cos - cos (2t - )]
= ½VmIm cos - ½VmIm cos (2t - )]
POWER
The average of component [½VmIm cos (2t - )] is zero, therefore
coscos2
.2
cos21
rmsrmsmm
mm IVIV
IVP
p
i
v
t
CURRENT, VOLTAGE AND POWER WAVEFORMS FOR
R-L SERIAL CIRCUIT
Phasor diagram
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
Reactive power (VAR) = VCI = VI sin
Cos = VR/V = R/Z.Z
XC
R
f
I
I cos
I sin
V
230V
0.5A
C=8 F
R
A capacitor of 8.0F 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.
HzCX
fC
5.862301082
1
2
16
CfI
VX C 2
1230
0.1
230
(a)
21
224605.0
230CXR
I
VZ
398230460 2222CXZR
o
Z
R30
460
398coscos 11
(b)
(c)
Leading by 30o
A
V
WlampofcurrentRated 5.7
100
750
VVC 270
222 100230 CV
222CR VVV
230V60Hz
I
C
R
V C
100V
V RI
VV C
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
435.0230
100cos
V
VR
I
V
CfX C
C 2
1
'1264o
FVf
IC
C
7.73270602
5.7
2
(b)