ALTERNATING VOLTAGE AND CURRENT

Example of instantaneous value of i or v in electrical circuits

i,v

t

v

i

i

t0-

Im

Im

v

t

v

t0

Vm

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) = Imkostv(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[f1]s[T or

]s[T1]Hz[f

i

0

Im

-Im

T/4

T/2 3T/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

00.4

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

vi

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

0m2ItdtsinωIA

mmm

av I637.0I2ωπω2II

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

2sin21

2)2cos1(

2

ttIdttIA mm

mI

22

222 mm II

bAh

mm

rms IIhI 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

110707.0

VVm

A11.350

5.155

RVI m

m

A98.111.3637.0637.0 mav IIInitially 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

RIP 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

RIP mheating 4

2

Thus equivalent Irms read by the meter is

RIRI mrms 4

22

AII mrms 555.1

211.3

2

thus

2555.111.3

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

A1.122.2)( readingI rms

But only half a cycle then the reading is

e = Em sin t = Em sin

Em

e1 = Em sin te2 = Em sin (t + )

Here the magnitudes are same but the phases are different

A

B

e = Em sin ti = Im sin t

Here the phases are same but the magnitudes are different

i1 = Im1 sin ti2 = Im2 sin (t + )

Im1

Im2

i2 is leading the i1 by ori1 is lagging the i2 by

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

O A

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

E1 = 240 V

E2 = 240 V

E3 = 240 V

120

120

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.

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

OA=60V ; OB= 40V

Horizontal components

OA-OE=60 - 40 cos 60o=OD = 60 - 20 = 40V

Vertical components

OY= 40 sin 60o= 34.64V

Resultant OC= V9.5264.3440 22

Equation for the voltage V= 52.9 sin ( +40.9o) V

o9.4040

64.34tan 1

and

(b)

A

B

-B C

o DE 60 o

Y

Vm1

Vm2

-Vm2

Vm