Transcript 7. AC Current & Voltage

ALTERNATING
VOLTAGE AND
CURRENT
Example of instantaneous value of i or v in electrical
circuits
Direct current
Unipolar binary waveform
i,v
v
v
i
t
t
bipolar binary waveform
Saw wave
i
v
I
V
m0
-
m
0
t
Im
t
y
Sinusoidal waveform
1
0
270
360
x ()
90
-1
270
The instantaneous value varies with time following the sine or cosine
waveform. This is the common waveform for alternating current (AC).
Graphically can be represented by the following equations
v(t) = Vmsint or
v(t) = Vmcost
i(t) = Imsint atau
i(t) = Imkost
where  = 2f and f is a frequency in Hertz (Hz ), Vm and Im are the
maximum amplitude of voltage and current respectively
Features of the voltage waveform
v (V)
Vm
VP-P
3T/4
0
T/4
T
t (s)
T/2
-Vrepresents
m
1. Above figure
one cylcle of voltage waveform which is
methamatically represented by v = Vmsint. Vm to –Vm is called
VP-P (peak to peak value).
2. One cycle is equivalent to one wavelength or 360o or 2 in degree.
3. One cycle also is said to have a periodic time T (sec).
4. A number of cycle per sec is said to have a frequency f (Hz)
5. The relationship between T and f is
T[s] 
1
f [Hz ]
or
1
f [Hz ] 
T[s]
1. The following figure is a current waveform represented
mathematically as
i = Imcost
2. It starts maximum at t=0 which equivalent to cos (0)=1
and ends maximum at t=T or 360o or 2.
i
Im
T/2
3T/4
0
t
T/4
-Im
T
The following is a one cycle sinusoidal current waveform.
Obtain the equation for the current in the function of time.
i (mA)
170
20
0
t (ms)
10
-170
From the graph Im = 170 mA;T = 20 ms = 0.02 s
f = 1/T = 1/0.02 = 50 Hz
i(t) = Imsint = 170sin2ft = 170sin100t mA
A sinusoidal AC voltage has a frequency of 2500 Hz and a peak
voltage value is 15 V. Draw a one cycle of the voltage.
Vm = 15 V
T =1/f= 1/2500= 0.4 ms
Thus the diagram as follows
v
(V)
15
0.4
0
t (ms)
0.2
-15
A sinusoidal AC voltage is given by equation :
v(t) = 156 cos( 800  t )V
Draw a one cycle of voltage.
From equation v(t) = Vmcos(t) = 156 cos( 800  t ) V
Vm = 156;  = 2f = 800
We have
f = 400 and thus T = 1/f = 1/400 = 2.5 ms
v (V)
156
1.25
0
t (ms)
2.5
0.625
-156
1.875
Waveform which is not begin at t=0
a
Ym
180
0
90
270
360
x ()
-Ym
In this case, the waveform is given by y = Ymsin(x + a)
x =angle
ao= phase difference refer to sine wave begins at t=0
For current and voltage, the equations are given by
i(t) = Imsin(t + )
v(t) = Vmsin(t + )
=phase difference
Draw one cycle of sinusoidal current wave given by the equation
i(t) = 70sin(8000t + 0.943 rad) mA
From the equation i(t) = Imsin(t + ) = 70sin(8000t + 0.943 rad)
Im = 70;  = 2f = 8000;f = 4000 Hz = 4 kHz;
T = 1/f = 1/4000 = 0.25 ms;  = 0.943 rad = 54o
i (mA)
54
70
57
0.125
0
t (ms)
0.25
-70
Obtain the equation of the following waveform
v
(V)
339
10
0
-339 
3 ms
From waveform:
T = 20 ms f=1/T = 1/0.02 = 50 Hz
 = 2f = 100
3 ms = 3 x 360/20 = 54   = 90 – 54 = 36
Vm = 339V
Equation for voltage:
v(t) = Vmsin(t + ) = 339sin(100t + 36)
t
20 (ms)
v, i
v
Vm

i
Im
t
0
-Im
T
-Vm
v(t) = Vmcost;
i(t) = Imcos(t + )
The current i(t) is leading the voltage by  (the minimum
or maximum comes first.
The voltage v(t) is lagging the current by 
Following is a sinusoidal waveform for current i1(t) and i2(t).
Obtain the equation
for those current.
i
(mA)
80
i2
i1
60
40
20
0
-20
t (s)
-40
-60
-80
From the waveform: 18.1 25
50
Im1 = 60 mA;
Im2 = 80 mA
T = 50 s
f = 1/(50 x 10-6) = 20 kHz
i1(t) = 60sin(4 x 104 t)
 = 25 – 18.1 = 6.9 s
6.9 s  6.9 x 360/50 = 50
i2(t) = 80sin(4 x 104 t + 50)
•Average value for one cycle of waveform is zero
•For half-wave can be calculated as follows
i(t) = Imsint
Area under the curve
But
A  Iav  π/ω
πω
A  I m  sin ωtdt 
0

2I m

2I m
2I m ω
I av 

 0.637I m
πω

Power P = I2R (i.e. P  I2)
i(t) = Imsint

i2(t) = Im2sin2t
= ½Im2(1 - cos2t)
The area under the Im2 is
A. Equate this to the
rectangular of the same
area A=h2 x 2/
Area under the Im2
2
2 / 
Im
A
2 0
I m


Height of the rectangular
A I m  I m
h  

b
2 
2
2
2
2
r.m.s value
2 / 
Im 
1

(1  cos 2t )dt 
t
sin 2t 

2  2
0
2
I rms
Im
h
 0.707 I m
2
An alternating current is given by an equation i(t)=0.4sin 100t A;
flowing into a resistor R=384 W for 48 hours. Calculate the energy
in kWh consumed by the resistor.
Im = 0.4

I = 0.707 x 0.4 = 0.283 A
P = I2R = 0.2832 x 384 = 30.7 W
W = Pt = 30.7 x 48 = 1.474 kWh
A sinusoidal voltage as in figure is applied to a resistor 56
W. Calculate the power absorbed by the resistor
339
-339
400
300
200
100
0
-100
-200
-300
-400
Power absorbed
Vm = 339 V
V = 0.707 x 339 = 240 V
P = V2/R
= 2402/56
= 1029 W
A purely resistive resistor of 17 W dissipates 3.4kW when a
sinusoidal voltage of frequency 50Hz apply across it.Give an
equation for the current passing through the resistor in a function
of time.
P = I2R
or
I = (P/R)
= (3400/17) = 14.14 A
Im = I/0.707
= 14.14/0.707 = 20 A
 = 2f
= 2 x 50
= 100
i(t) = Imsint = 20 sin(100t)
A moving –coil ammeter, a thermal ammeter and a rectifier are
connected in series with a resistor across a 110V sinusoidal a.c.
supply. The circuit has a resistance of 50 W to current in one
direction and , due to the rectifier, an infinity resistance to
current in the reverse direction . Calculate :
(1)The readings on the ammeters;
(2)The form and peak factors of the current wave
Vm 155.5
V
110
 Im 

 3.11A
Vm 

 155.5V
R
50
0.707 0.707
I av  0.637 I m  0.637  3.11  1.98A
Initially the moving coil-ammeter will read the Iav for the first half
of a cycle. The second half , the value of current will be zero (due to
rectifier- reverse ) . For the whole cycle, it will read 1.98/2=0.99A
Thermal ammeter only response to the heat. This heat effect is
corresponding to power dissipated in the resistor and given by the
2
equation
I
Pheating 
Full power is
m
2
2
P  I rms R
R
Since only half cycle give the heating effect, the other half is
no current ( due to rectifier-reverse). Therefore the heating
power will be
2
Im
Pheating 
R
4 2
Im
2
thus
I rms R 
R
4
Thus equivalent Irms read by the meter is
I rms
I m 3.11


 1.555 A
2
2
The actual value for full cycle is
I rms  I m  0.707  3.11 .707  2.2A
But only half a cycle then the reading is
2.2
I rms (reading ) 
 1.1A
2
Form factor
I rms 1.555
kf 

 1.57
I av
0.99
Peak factor
Im
3.11
kp 

2
I rms 1.555
Em

e = Em sin t = Em sin
B
A
e1 = Em sin t
e2 = Em sin (t + )
Here the magnitudes are same but the phases are different
e = Em sin t
i = Im sin t
Here the phases are same but the magnitudes are different

Im1
Im2

i1 = Im1 sin t
i2 = Im2 sin (t + )
i2 is leading the i1 by  or i1 is lagging the i2 by 
y
CY
A
C
AY
Vertical components
Ay = OA sin 
BY
By = OB sin 
O
Cy = Ay + By = OA sin  + OB sin 
Horizontal components
Ax = OA cos 
Bx = OB cos 
Cx = Ax + Bx = OA cos  + OB cos 
Resultant = OC = (Cy2 + Cx2)

B

BX AX
CX
x
y
Ay
Cy
A
C

-Bx
Vertical components
Ay = OA sin 
-B
-By = -OB sin 
Cy = Ay - By = OA sin  - OB sin 
Horizontal components
Ax = OA cos 
-Bx = -OB cos 
Cx = Ax - Bx = OA cos  - OB cos 
Resultant = OC = (Cy2 + Cx2)
O
B

-By
Cx
Ax
x
v1
v2
v
Given v1 = 180 sin 314t volt ;and v2 = 120 sin (314t + /3) volt.
Find
1. The supply voltage v in trigonometry form;
2. r.m.s voltage of supply
3. Supply frequency
B
Vm2
Vm
/3
Vm2 sin /3

O
Vm1
A
Vm2 cos /3
OA = Vm1 + Vm2 cos /3 = 180 + 120 x 0.5 = 240
OB = Vm2 sin /3 = 120 x 0.866 = 104
Vm = ((OA)2 + (OB)2 ) = (2402 + 1042) = 262
 = tan-1 (OB/OA) = tan-1(104/240) = 0.41 rad
v = 262 sin (t + 0.41)
volt
Graph showing the three components
300.00
200.00
100.00
v
v1
v2
0.00
-100.00
-200.00
-300.00
(b)Rms value. = V = 0.707 Vm = 0.707 x 262 = 185 V
(c)Frequency = f = 314/2 = 50 Hz
Find graphically or otherwise the resultants of the following voltages
e1 = 25 sin t,
e3 = 30 kos t,
e2 = 30 sin (t + /6),
e4 = 20 sin (t - /4)
Express in the same form
Solution
e1
e2
e3
e4
=
=
=
=
25 sin t [ V ]
30 sin (t + /6) [ V ]
30 cos t [ V ]
20 sin (t - /4) [ V ]
Em1
Em2
Em3
Em4
=
=
=
=
25 volt
30 volt
30 volt
20 volt
Em
Ey
Em3
Em2
Em2sin(/6)
Em4sin(/4)
Em1
Ex
Em2cos(/6)
Em4
Em4cos(/4)
Horizontal components:
Ex = Em1 + Em2cos(/6) + Em4cos(-/4)
= 25 + (30 x 0.866) + (20 x 0.707) = 65.1
Vertical components:
Ey = Em3 + Em2sin(/6) + Em4sin(-/4)
= 30 + (30 x 0.5) + (20 x -0.707) = 30.9
Peak value for e:
Em = (Ex2 + Ey2)½ = (65.12+30.92) ½ = 72 [V]
Phase angle for e:
= tan-1(Ey/Ex) = tan-1(30.9/65.1) = 25 = 5/36
 e = e1 + e2 + e3 + e4 = 72 sin(t + 5/36)
Show graphically the waveform and phasor diagrams of
the resultant of the following voltages
e = 339cos100t + 339cos(100t + 120) + 339cos(100t + 240)
What is the value of e?
Say:
e1 = 339cos100t,
e2 = 339cos(100t + 120),
e3 = 339cos(100t + 240)
e1, e2, e3 (V)
400
e1
e2
e3
300
200
100
t (ms)
0
-100
-200
-300
-400
5 ms
(90)
10 ms
(180)
15 ms
(270)
20 ms
(360)
E3 = 240 V
120
E1 = 240 V
120
E2 = 240 V
e = 0 for all instants
The instantaneous values of two alternating voltages are represented
respectively by v1=60 sin volts and v2=40 sin(/3) volts.
Derive an expression for the instantaneous values of
(a)The sum
(b)The difference of these voltages.
First we consider  =0 or t=0 as reference in order to simplified
the phasor diagram. Thus v1 will be in the x-axis and v2 will be –
/3 or -60o behind (lagging) v1. Magnitude for v1 is 60V and v2 is
40V.
(a)
OA=60V ; OB= 40V
O
60o
Horizontal components
OA+OD=60 + 40 cos 60o
= 60 + 20 = 80V
A
D

Vm2
y
B
OY= - 40 sin 60o= -34.64V
802   34.642  87.2V
OC=
 34.64 
o
  tan  
and
  23.4
80 

Equation for the voltage V= 87.2 sin ( -23.4o) V
1
x
Vm1
Vertical components
Resultant
E
C
Vm
(b)
OA=60V ; OB= 40V
-Vm2
Y
-B
Horizontal components
OA-OE=60 - 40 cos 60o=OD
= 60 - 20 = 40V
E
o
C Vm

60o D
Vertical components
B
OY= 40 sin 60o= 34.64V
Resultant
and
402  34.642
OC=
 52.9V
 34.64 
o
  tan 
  40.9
 40 
1
Equation for the voltage V= 52.9 sin ( +40.9o) V
Vm2
Vm1
A