A-C Circu its

1 . 1 I NTRODUCTION

Network theory, in general, deals into three components, the network, usually consisting of a suitable combination of R,

L

and C, the source and the response. Mostly, it is desirable that whatever is the waveform of the source, same waveform should appear in the response (Excep­

tions to be ignored). We have various waveforms viz. square, triangular, trapezoidal, sinusoidal etc. Out of all these, if the circuit has resistance only, one can use any of the waveforms men­

tioned above and the response (v-i relation) will have the same waveform as the excitation (source). However, with

L

and C as the element, it is only sinusoidal waveform which satisfies the requirement of identical waveforms for source and response. If the source is voltage source the current through the inductor is proportional to the integral of the voltage wave and for the capacitor, it is the derivative of the voltage wave. It is a very useful and important property of sine waves that both it's derivative and integrals are also sinusoidal. The basis for standardi­

zation on the sine wave thus comes from both physics and calculus. Also, it is very easy to generate the sine waves using rotating machines and transmit and distribute electric energy using transformers. We next illustrate that a uniform circular motion or simple harmonic motion is a sinusoidal variation.

Sine wave is the basic waveform of a.c. circuit theory and it is necessary to learn how to handle such waves.

Let us relate a uniform circular motion or rotation to a sinusoidal wave form. We con­

sider a rotating voltage/current phasor that has a length

Vm!Im

and is coincident along x-axis Fig. 1.1. We start counting time and rotate the phasor anti-clockwise and at various instant of time we take the projection of the rotating phasor along y-axis.

At t ⁼ 0, the projection is zero and hence we plot a point on the right hand side along t(9) axis. After certain time say t the phasor rotates through an angle 8 and there is some finite projection along y-axis, mark this on the right hand side along t(9) axis as shown in Fig. 1 . 1 . As we rotate the phasor further when 8 = 90° the value of the projection i s

ViJim.

Mark it on the right hand side along t(8) axis as shown. Rotate phasor further and obtain the projection along y-axis, the projections decreases beyond 90° and continues to be along + y axis till 8 = 180° when projection is zero.

49

0 --- 0

/,, ... -

-P a

r' �

- - - -

;-.z�---

0

\ 0 /

\ /

\ /

... ... _ _ ,,,. ... ...

b

.. 0 / 0

\ /

\ ' ; /

.... ... _ _ _ ... b

_ _ _ _ _ _ _ _ 0

0

;---,

b

; '

- - - -;

,,,.---

^... b

; '

(

^{\ \ '}

�o

^{; ;}

^\:

^{/ oa e 0}

... _... -

-... p

� � - - - -^...

b

; '

/ '

(

^{g 0}

...... ^{_ _} ...

p

^e

Fig. 1 .1

Similarly if we rotate the phasor further through 27t radian or

360°

or one cycle, we obtain a complete sine wave as shown in Fig.

1.1.

Thus, if the speed of rotation is f revolutions/

sec the phasor has a frequency of f hertz (Hz). Hence the angular velocity is

co =

27t f rad/sec.

If the instantaneous value of time at any instant of rotation is t seconds the total angle in radians from time zero upto that point is

2nft

=

rot

The concept of rotating phasor is a function of time, we may substitute 2nft or wt for 8 in the equation.

V= Vm

sin 8

and now therefore, the instantaneous voltage in terms of time is

v = vm

sin 2nft

= vm

sin wt

when equation

1.1

is used, the axis of the waveform is not 8 but time t.

1 .2 PHAS E

... (1.1)

A sinusoidal waveform having a displacement of 8 to the left is shown in Fig. means that at t

= 0

the phasor has an initial displacement of 8 in counterclockwise direction.

1.2(a)

^which

Therefore, the initial instantaneous value of voltage/current at time t

= 0

^is

V = V,n

sin 8 putting wt

= 0.

This shift of the waveform is called a phase displacement and the angle 8 is called phase angle. Fig.

1.2

(b-d) show different phase displacements of the same sine wave. The sine wave at 'a' is said to be leading the wave at 'b' as

Va = Vm

sin (cot + 8) and at b

vb = vm

sin (wt

-

⁸⁾

Example 1.1.

The equation for a voltage wave is v

=

0.02 sin (4000t + 30°). Find the frequency, the instantaneous voltage when t

=

320 µ sec. What is the time represented by 30°

phase difference?

here

Solution: The general expression for voltage is given as

V = V m

sin (2rcft + 8) When

4000

2nf

= 4000

^or

^f=

^2;-

⁼ 637

^Hz.

t

= 320µ.

^sec.

.

(

^-6

180 )

v

= 0.02

^sm

4000

^x

320

^x

10

^x---;-+

30°

v

= 0.02

^sin

(73.3

⁺

30)

= 0.02

^sin

(103.3)

= 0.0195

^volt

f f

Fig. 1 .2 Displaced sinusoidal voltage waveforms:

(a) displacement of e to the left, (b) Displacement of e to the right,

(c) Displacement to the left of 90° (rr/2 rad ians), (d) Displacement of 1 80° (rr rad ians).

The time of 1 cycle is 1 ¹

f

637

which is equal to 360° electrical degrees. Hence for 30° the time taken is 30 1

360

x

₆₃₇ sec = 0. 131 m sec ^Ans.

1 .3 THE AVERAG E VALUE OF A WAVEFORM

In a.c. circuit applications we are interested in finding out the average value of a waveform, that wave could be sinusoidal, triangular trapezoidal or any other shape.

The average value of a cycle of a waveform is the area under the waveform divided by the length of one cycle.

M^at^he^mat^icall^y,

V ^au =

�

_T ^{U dt}

=

^_2_

_27t

^{U dt}

where T is the length of one cycle.

... (1.2)

We first consider average value of sinusoidal waves for which we consider the following different shapes.

These are all sine waves In (a) since it is a full sine wave, the total area under the curve

is zero as half of it is above ffit axis and equal half is below ffit-axis and hence the average area of the curve is zero and hence over the complete sine wave the average value is zero.

0

_{TC 2TC TC}

(a) (b) (C)

Fig. 1 .3 (a) Full sine wave (b) Half-rectified wave (c) Full rectified wave

However, in Fig. 1.3(b) the average value is

i .

In document Basic Electrical Engineering, 4th Edition - CL Wadhwa (Page 66-70)