Locus Diagram

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LOCUS DIAGRAM

Sometimes we are interested in

knowing the current (or voltage) in a

circuit (or a part of it) when one of

the circuit elements (R, L, or C.) is

varied over a certain range.

Let us consider a simple R-L circuit

as shown in Fig 2.1(a)

Fig. 2.1a

If we vary, say, the resistance R from

O to w, keeping the voltage and

frequency constant, what would be thelocus of the current ? Let us first study

the variation of the impedance.

Imz. locus

Fig. 2.1b

V = Z I

or I = Y V

= Y ( Per unit ) taking the voltage

to be 1.0 p.u

So the current locus is the locus of

Y (with some scale). Since the

locus of Z is a straight line the locus

of Y will be the inverse of a straight

I. INVERSE Of A STRAIGHT LINE

Let AB be a straight line, OP1 is

the perpendicular drawn from theorigin. Q1 and R 1 are any two points on

Fig. 2.2a

OQ1. OQ2 = OP1. OP2

We want to invert OP1,OR 1 and OQ1.

Let P2, R 2, Q2 be the points of

inversion.

@ OP1. OP2 = OR 1. OR 2 = OQ1.OQ2=1

@P1 P2 Q1 Q2 are con cyclic.

Now OP1Q1 = 900

@ Q1Q2 P1 = 900

Similarly R 2 = 900

@ All the points corresponding to the

inverses of vectors whose locus is the

straight line A B will lie on the circle

with diameter OP2.

Thus the inversion of a straight line is a

circle passing through the origin(Fig. 2.2b)

Fig. 2.2b

Y ± locus - circle

P1Z locus

st.line

and conversely the inversion of a

circle passing through the origin is a

st.line.

II. INVERSION OF A CIRCLE

ABOUT THE ORIGIN IS ALSO

A CIRCLE.

Let us consider a circle P1QP2R with

centre at C (away from the origin ),

as shown in Fig. 2.3a and we want to

invert it.

Fig. 2.3a

is any chord which passes

through the origin.

Let P1/ , P2/ be the inversions of

P1 and P2 about the origin.

@ OP1. OP1/ = OP2 . OP2

For a given location of the circle,

OP1. OP2 = k 2 (= OQ2) = constant.

@ OP1/.OP2

/ = = constant2

@ P1/ . P2

/ and similar points of

inversion of the original circle lie on

a circle P1

as shown.When we consider vectors

[or phasors], i.e., OP1, OP2 etc. are

considered with their angles, the

inversion will result in a reversal of

the angle also and hence the final

result will be a circle in the fourth

quadrant as shown in Fig. 2.3(b).

-Z Locus

Y - Locus

Fig. 2.3 (b)

For simplicity of drawing and

transferring points, we shall.

(i) frequently change the scale.

(ii) bring all the diagrams in the same

first quadrant.

(iii) perform addition and

subtraction in the same way (scaled

values will not involve

mistakes), and

(iv) write scales, and image or true

ones against each value.

EXAMPLE 2.1

(i) Find the current locus for the case

shown in Fig. 2.4

5 ± 50;

Fig. 2.4

(ii) Also find the p. f and power for the

two cases ± R = 5 ; and R = 50 ;.

(iii)Determine the current in each case.

Let us first draw the locus (st. line) for

z which is R + j4 , and mark the two

points

P and Q corresponding to

R =5 and R =50 (Fig.2.5)

Fig. 2.5

R = 50 Z - Locus

OP and OQ denote the

corresponding impedances 5+j4 and

50+j4 respectively.

The minimum value of the

impedance can be O+j4

corresponding to OR, which is

perpendicular to the line.

Therefore the locus of

Y (= ) will be a circle with

as the diameter and it will

pass through the origin.

¸©ª

We take a scale factor of 40 and

show the locus of 40 Y by the dottedsemi circle .

Thus 40Y gives a diameter of 40 units = 10 units current for R =5,

40100)(voltageOPof lengthI

Similarly, for R = 50

Power transferred for (R = 5)

= I 1cos U1 V

voltageOQof length /v

COSvoltage.OP

Similarly, for R =

50, power transferred

Scale for mho and ohm can be same or

different.

CosvoltageOQ

EXAMPLE 2.2

Find the current loci for the circuits

shown in Fig. 2.6(a) and 2.6(b)

Fig. 2.6(a)

Fig.2.6(b)

find C. when the total current ishaving a p.f 0.4 (lagging) in 2.6 b

SOLUTION

We shall first consider the impedance

and hence admittance locus for the

circuit in Fig. 2.6a.

SinceI =

, for a given value of V

(=200V ), the same admittance locus

will give the current locus with a scalefactor of 200.

Fig. 2.6(c)

I 2= -j10

20 Real

Ref Voltage

Locus of I1 = 200 Y1

Locus for Z1 (Fig. 2a)

Locus of I = I1 + I2

Z1min = 10 ohm = OQ.

@ Y1max = mho

= 0.1 mho

@ I1m = 200 Y1m= 20A

I2 = 200 Y2 = -j 10 A

I = I1+ I2

If Cos U = 0.4 ( lagging)

then U = Cos-1 (0.4) OP1 is the

total current I . A vertical line drawn

from P1 determines the corresponding

current I 1 = OP2 at an angle F.

If now a line is drawn at - F, from o,we get the location of Z1

(= OR) on the Z1- locus and QR is the

corresponding reactance.

@The required value of QRw.

EXAMPLE 2.3Find C for maximum current in the

circuit ( Fig.2.6a)

~ Z123

0.5 + 0.5 j

3/ 2/ 1/

3 2 1z3

Fig. 2.7(a)

Fig. 2.7b

Im Y12 Locus (= Y1 + Y2)

2z123 = 2 [z12 +z3]

O/ P2P3

@ 2Z12 semi circle shown by

firm line.

min12max

5.00.2

The total admittance of Z1 and Z2, isY12 whose locus is shown by the

vertical line through Q.

The locus of 2Z12 is the semi circle

shown,which when added with

2Z3 [= 2 (0.5 + 0.5 j ) ] is 2Z123 andis shown by the shifted semi circle

with centre P2. Join OP2 to obtain the

minimum value Z123. Through P3 a

line P3 P4 is drawn parallel to 00/.

P5 is the image of P4 ( to obtain the

point P6); by extending the line

OP5, we obtain P6.Q

wc for minimum overall impedance

Exercise 2.1

Draw the locus of current I , for the

circuit shown below when XL is

varied. Determine the value of L

when the power factor is unity(Fig 2.8)

~ 200V

Fig 2.8

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