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Example of instantaneous value of i or v in electrical circuits
i,v
t
v
i
i
t0-
Im
I
m
v
t
v
t0
V
m
Direct current
Saw wave
Unipolar binary waveform
bipolar binary waveform
x ()
y
1
-1
090
270
270
360
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
i(t) = Imsint atau i(t) = Imkost
v(t) = Vmsint or v(t) = Vmcost
where = 2f and f is a frequency in Hertz (Hz ), Vm and Im are the maximum amplitude of voltage and current respectively
Sinusoidal waveform
t (s)
v (V)
Vm
-Vm
0T/4
3T/4
T/2
TVP-P
1. Above figure represents one cylcle of voltage waveform which is methamatically represented by v = Vmsint. Vm to –Vm is called VP-P (peak to peak value).
2. One cycle is equivalent to one wavelength or 360o or 2in 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
Features of the voltage waveform
]Hz[f
1]s[T or
]s[T
1]Hz[f
i
0
Im
-Im
T/4
T/23T/4
Tt
1. The following figure is a current waveform represented mathematically as
i = Imcost2. It starts maximum at t=0 which equivalent to cos (0)=1
and ends maximum at t=T or 360o or 2.
t (ms)
i (mA)
170
-170
020
10
From the graph Im = 170 mA;T = 20 ms = 0.02 s
f = 1/T = 1/0.02 = 50 Hz
i(t) = Imsint = 170sin2ft = 170sin100t mA
The following is a one cycle sinusoidal current waveform. Obtain the equation for the current in the function of time.
t (ms)
v (V)
15
-15
0
0.4
0.2
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)
0
156
-1560.625
1.25
1.875
2.5t (ms)
A sinusoidal AC voltage is given by equation : v(t) = 156 cos( 800 t )VDraw a one cycle of voltage.
From equation v(t) = Vmcos(t) = 156 cos( 800 t ) V
We have Vm = 156; = 2f = 800
f = 400 and thus T = 1/f = 1/400 = 2.5 ms
x ()90
180 270
3600
Ym
-Ym
a
Waveform which is not begin at t=0
In this case, the waveform is given by y = Ymsin(x + a)x =angleao= phase difference refer to sine wave begins at t=0
For current and voltage, the equations are given byi(t) = Imsin(t + )v(t) = Vmsin(t + ) =phase difference
Draw one cycle of sinusoidal current wave given by the equationi(t) = 70sin(8000t + 0.943 rad) mA
From the equation i(t) = Imsin(t + ) = 70sin(8000t + 0.943 rad)Im = 70; = 2f = 8000;f = 4000 Hz = 4 kHz;T = 1/f = 1/4000 = 0.25 ms; = 0.943 rad = 54o
i (mA)
t (ms)0
70
-70
57
54
0.25
0.125
From waveform:T = 20 ms f=1/T = 1/0.02 = 50 Hz = 2f = 1003 ms = 3 x 360/20 = 54 = 90 – 54 = 36Vm = 339VEquation for voltage:
v(t) = Vmsin(t + ) = 339sin(100t + 36)
0
339
-339
v (V)
t (ms)20
10
3 ms
0
339
-339
v (V)
t (ms)20
10
3 ms
Obtain the equation of the following waveform
v, i
t
Vm
Im
0
-Vm
-Im T
v
i
v(t) = Vmcost; 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
-80
-60
-40
-20
0
20
40
60
80
t (s)
i (mA)
502518.1
i2 i1
From the waveform:Im1 = 60 mA; Im2 = 80 mAT = 50 s f = 1/(50 x 10-6) = 20 kHz
i1(t) = 60sin(4 x 104 t) = 25 – 18.1 = 6.9 s6.9 s 6.9 x 360/50 = 50
i2(t) = 80sin(4 x 104 t + 50)
Following is a sinusoidal waveform for current i1(t) and i2(t). Obtain the equation for those current.
•Average value for one cycle of waveform is zero•For half-wave can be calculated as follows
i(t) = Imsint
m
πω
0m
2ItdtsinωIA
mmm
av I637.0I2
ωπ
ω2II
Area under the curve
But π/ωIA av
Power P = I2R (i.e. P I2)
The area under the Im2 is
A. Equate this to the rectangular of the same area A=h2 x 2
i(t) = Imsint
i2(t) = Im2sin2t = ½Im
2(1 - cos2t)
/2
0
2/2
0
2
2sin2
1
2)2cos1(
2
tt
Idtt
IA mm
mI
22
222 mm II
b
Ah
mm
rms II
hI 707.02
Area under the Im2
Height of the rectangular
r.m.s value
An alternating current is given by an equation i(t)=0.4sin 100t A; flowing into a resistor R=384 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 . Calculate the power absorbed by the resistor
Vm = 339 V V = 0.707 x 339 = 240 V
P = V2/R = 2402/56 = 1029 WPower absorbed
-400-300-200-100
0100200300400
339
-339
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
= 2f = 2 x 50 = 100
i(t) = Imsint = 20 sin(100t)
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 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
V5.155707.0
110
707.0
VVm
A11.350
5.155
R
VI m
m
A98.111.3637.0637.0 mav II
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 equation
RI
P mheating 2
2
Full power is RIP rms2
Since only half cycle give the heating effect, the other half is no current ( due to rectifier-reverse). Therefore the heating power will be
RI
P mheating 4
2
Thus equivalent Irms read by the meter is
RI
RI mrms 4
22
AI
I mrms 555.1
2
11.3
2
thus
2555.1
11.3
rms
mp I
Ik
57.199.0
555.1
av
rmsf I
Ik
Form factor
Peak factor
The actual value for full cycle is
A2.2707.11.3707.0 mrms II
A1.12
2.2)( readingI rms
But only half a cycle then the reading is
A C
B
O x
y
AX
AY
BY
CY
CXBX
Vertical componentsAy = OA sin By = OB sin Cy = Ay + By = OA sin + OB sin
Horizontal componentsAx = OA cos Bx = OB cos Cx = Ax + Bx = OA cos + OB cos
Resultant = OC = (Cy2 + Cx
2)
AC
BO
-B
Ax
x
-By
Cx
-Bx
Ay
Cy
y
Vertical componentsAy = OA sin -By = -OB sin Cy = Ay - By = OA sin - OB sin
Horizontal componentsAx = OA cos -Bx = -OB cos Cx = Ax - Bx = OA cos - OB cos
Resultant = OC = (Cy2 + Cx
2)
v1 v2
v
Given v1 = 180 sin 314t volt ;and v2 = 120 sin (314t + /3) volt.Find1. The supply voltage v in trigonometry form;2. r.m.s voltage of supply3. Supply frequency
Vm2
Vm1
Vm
/3
Vm2 sin /3
Vm2 cos /3
OA
B
OA = Vm1 + Vm2 cos /3 = 180 + 120 x 0.5 = 240OB = Vm2 sin /3 = 120 x 0.866 = 104Vm = ((OA)2 + (OB)2 ) = (2402 + 1042) = 262 = tan-1 (OB/OA) = tan-1(104/240) = 0.41 radv = 262 sin (t + 0.41) volt
-300.00
-200.00
-100.00
0.00
100.00
200.00
300.00
v2v1
v
(b)Rms value. = V = 0.707 Vm = 0.707 x 262 = 185 V
(c)Frequency = f = 314/2 = 50 Hz
Graph showing the three components
Solution
e1 = 25 sin t [ V ] Em1 = 25 volte2 = 30 sin (t + /6) [ V ] Em2 = 30 volte3 = 30 cos t [ V ] Em3 = 30 volte4 = 20 sin (t - /4) [ V ] Em4 = 20 volt
Find graphically or otherwise the resultants of the following voltages
e1 = 25 sin t, e2 = 30 sin (t + /6),e3 = 30 kos t, e4 = 20 sin (t - /4)Express in the same form
Em1
Em2Em3
Em4
Em2sin(/6)
Em4sin(/4)
Em4cos(/4)
Em2cos(/6)
Ey
Ex
Em
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
= tan-1(Ey/Ex) = tan-1(30.9/65.1) = 25 = 5/36
e = e1 + e2 + e3 + e4 = 72 sin(t + 5/36)
Peak value for e:Em = (Ex
2 + Ey2)½ = (65.12+30.92) ½ = 72 [V]
Phase angle for e:
Show graphically the waveform and phasor diagrams of the resultant of the following voltages
e = 339cos100t + 339cos(100t + 120) + 339cos(100t + 240)What is the value of e?
Say:e1 = 339cos100t,
e2 = 339cos(100t + 120),
e3 = 339cos(100t + 240)
e1, e2, e3 (V)
t (ms)
20 ms(360)
e1 e2 e3
15 ms(270)
10 ms(180)
5 ms(90)
-400
-300
-200
-100
0
100
200
300
400
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.
O A
B C
D Ex
60 o
y
OA=60V ; OB= 40V
Horizontal components
OA+OD=60 + 40 cos 60o
= 60 + 20 = 80V
Vertical components
OY= - 40 sin 60o= -34.64V
Resultant OC= V2.8764.3480 22
Equation for the voltage V= 87.2 sin ( -23.4o) V
o4.2380
64.34tan 1
and
(a)Vm1
Vm2 Vm
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