LOCUS DIAGRAM

1

• 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.

2

• Let us consider a simple R-L circuit as shown in
  Fig 2.1(a)

R
L

A
I
Fig. 2.1a
3

• If we vary, say, the resistance R from O to ?,
  keeping the voltage and frequency constant, what
  would be the locus of the current ? Let us first
  study the variation of the impedance.

4
z. locus
Im
z1
z2
z3
zn
Rl
Fig. 2.1b
5

• V Z I
• or I Y V
• Y ( Per unit ) taking the voltage to be
  1.0 p.u

6

• 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
  line.

7

• I. INVERSE Of A STRAIGHT LINE
• Let AB be a straight line, OP1 is the
  perpendicular drawn from the origin. Q1 and R1
  are any two points on A B.

8
lm
Q2
A
Q1
Q1
P1
P2
0
P1
Q2
P2
R1
OQ1. OQ2 OP1. OP2
R2
O
Real
B
Fig. 2.2a
9

• We want to invert OP1,OR1 and OQ1. Let P2, R2,
  Q2 be the points of inversion.
• ? OP1. OP2 OR1. OR2 OQ1.OQ21
• P1 P2 Q1 Q2 are con cyclic.
• Now

10

• Q1Q2 P1 900
• Similarly R2 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.

11

• Thus the inversion of a straight line is a circle
  passing through the origin (Fig. 2.2b)

P1
Z locus st.line
P2
O
Y locus - circle
Fig. 2.2b
12

• 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 ),

13

• as shown in Fig. 2.3a and we want to invert it.

R
Y
C
P1
R/
P2
Q
C/
P2/
Q/
P1/
O
X
Fig. 2.3a
14

• P1 P2 is any chord which passes through the
  origin.
• Let P1/ , P2/ be the inversions of P1 and P2
  about the origin.

15

• ? OP1. OP1/ OP2 . OP2/ 1
• For a given location of the circle,
• OP1. OP2 k2 ( OQ2) constant.
• ? OP1/.OP2/ constant

16

• ? P1/ . P2/ and similar points of inversion of
  the original circle lie on a circle P1/. Q/ P2/
  R/ as shown.
• When we consider vectors or phasors, i.e.,
  OP1, OP2 etc. are

17

• 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).

18
Im
-Z Locus
?2
?1
?1
Rl
?2
Y - Locus
Fig. 2.3 (b)
19

• For simplicity of drawing and transferring
  points, we shall.
• frequently change the scale.
• (ii) bring all the diagrams in the same
  first quadrant.

20

• (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.

21

• EXAMPLE 2.1
• (i) Find the current locus for the case shown in
  Fig. 2.4

5 50 ?
j 4 ?
I

Fig. 2.4
100V
22

• (ii) Also find the p.f and power for the two
  cases R 5 ? and R 50 ?.
• 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

23

• P and Q corresponding to
• R5 and R50 (Fig.2.5)

10
40Y
Im
R5
R 50
Z - Locus
P/
P
R ?
R
Q
4
Q/
O
Real
Fig. 2.5
24

• OP and OQ denote the corresponding impedances
  5j4 and 50j4 respectively.
• The minimum value of the impedance can be Oj4
  corresponding to OR, which is perpendicular to
  the line.

25

• Therefore the locus of Y ( ) will be a circle
  with as the diameter and it will pass
  through the origin.

26

• We take a scale factor of 40 and show the locus
  of 40 Y by the dotted semi circle .
• Thus 40Y gives a diameter of 40 units
  10 units current for R5,

27

• Similarly, for R 50
• Power transferred for (R 5)
• I1cos ?1 V

28

• Similarly, for R 50, power transferred
• NOTES
• Scale for mho and ohm can be same or different.

29

• EXAMPLE 2.2
• Find the current loci for the circuits shown in
  Fig. 2.6(a) and 2.6(b)

10
I
C
200V

50 Hz
Fig. 2.6(a)
30

10
200V
C
j 20

50 Hz
Fig.2.6(b)
31

• find C. when the total current is having 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.

32

• Since I Y.V, for a given value of V (200V),
  the same admittance locus will give the current
  locus with a scale factor of 200.

33
Locus of I1 200 Y1
Im

P2
Ref Voltage
?
?
10 Q
10 Q
0
Real
-
20
?
?
P1
P1
Locus of I I1 I2
I2 -j10
R
Locus for Z1 (Fig. 2a)
Fig. 2.6(c)
34

• Z1min 10 ohm OQ.
• ?? Y1max mho
• 0.1 mho
• ? I1m 200 Y1m 20A

35

• Y2
• I2 200 Y2 -j 10 A
• I I1 I2

36

• If Cos ? 0.4 ( lagging)
• then ? Cos-1 (0.4) ? OP1 is the total current
  I. A vertical line drawn from P1 determines the
  corresponding current I1 OP2 at an angle ?.

37

• If now a line is drawn at -?, from o, we get
  the location of Z1( OR) on the Z1- locus and QR
  is the corresponding reactance.
• ?The required value of

38

• EXAMPLE 2.3
• Find C for maximum current in the circuit (
  Fig.2.6a)

3
2
1
z3
0.5 0.5 j
c

0.5
Z123
z2
z
3/
2/
1/
Fig. 2.7(a)
39
P6

Z123 / min oP3
Y12 Locus ( Y1 Y2)
Im
P2
P2
O/
P3
2z123 2 z12 z3
P3
P5
Y1
1.0
2.0
0
Rl
Y2
Q
P4
Fig. 2.7b
2 Z12
40

• ? 2Z12 ? semi circle shown by firm line.

41

• The total admittance of Z1 and Z2, is Y12 whose
  locus is shown by the vertical line through Q.
• The locus of 2Z12 is the semi circle shown,which
  when added with

42

• 2Z3 2 (0.5 0.5 j ) is 2Z123 and is 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/.

43

• P5 is the image of P4 ( to obtain the point P6)
  by extending the line OP5, we obtain P6. QP6
  wc for minimum overall impedance
• ?

44

• Exercise 2.1
• Draw the locus of current I, for the circuit
  shown below when XL is varied. Determine the
  value of L

45

• when the power factor is unity (Fig 2.8)

10 ?
I
200V
jx

50 Hz
-20 j
Fig 2.8