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Transcript of 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.
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Let us consider a simple R-L circuit
as shown in Fig 2.1(a)
~ A
R L
Fig. 2.1a
I
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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.
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Rl
zn
z3
z2
z1
Imz. locus
Fig. 2.1b
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V = Z I
or I = Y V
= Y ( Per unit ) taking the voltage
to be 1.0 p.u
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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.
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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
A B.
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Fig. 2.2a
B
Real
lmA
Q2 P2
R 2
R 1
P1
Q1
O
Q2
P2P1
Q1
0
OQ1. OQ2 = OP1. OP2
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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
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@ 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.
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Thus the inversion of a straight line is a
circle passing through the origin(Fig. 2.2b)
Fig. 2.2b
Y ± locus - circle
P2
P1Z locus
st.line
O
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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 ),
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as shown in Fig. 2.3a and we want to
invert it.
Fig. 2.3a
R /
Y
O X
C/
P1/
P2/
Q/
P2
P1
Q
R
C
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P1
P2
is any chord which passes
through the origin.
Let P1/ , P2/ be the inversions of
P1 and P2 about the origin.
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@ OP1. OP1/ = OP2 . OP2
/ = 1
For a given location of the circle,
OP1. OP2 = k 2 (= OQ2) = constant.
@ OP1/.OP2
/ = = constant2
22kOP
1
OP
1.
1!
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@ 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
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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).
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-Z Locus
Im
Rl
U2
U2
Y - Locus
U1
U1
Fig. 2.3 (b)
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For simplicity of drawing and
transferring points, we shall.
(i) frequently change the scale.
(ii) bring all the diagrams in the same
first quadrant.
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(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.
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EXAMPLE 2.1
(i) Find the current locus for the case
shown in Fig. 2.4
~
5 ± 50;
Fig. 2.4
I
j 4;
100V
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(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
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P and Q corresponding to
R =5 and R =50 (Fig.2.5)
Fig. 2.5
Q
R = 50 Z - Locus
R E
RealO
4
R
Im
1040Y
P/
P
R =5
Q/
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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.
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Therefore the locus of
Y (= ) will be a circle with
as the diameter and it will
pass through the origin.
¹ º
¸©ª
¨! 4
1
OR
1
Z
1
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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,
4
1v
40100)(voltageOPof lengthI
/
1
!v!
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Similarly, for R = 50
Power transferred for (R = 5)
= I 1cos U1 V
402
voltageOQof length /v
! I
1
2/
COSvoltage.OP
v!
40
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Similarly, for R =
50, power transferred
NOTES
Scale for mho and ohm can be same or
different.
2
2
/
CosvoltageOQ
vv!
40
.
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EXAMPLE 2.2
Find the current loci for the circuits
shown in Fig. 2.6(a) and 2.6(b)
200V
50 Hz
10 I
C
~
Fig. 2.6(a)
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50 Hz
200V
j 20
C
10
Fig.2.6(b)
~
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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.
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SinceI =
Y.V
, for a given value of V
(=200V ), the same admittance locus
will give the current locus with a scalefactor of 200.
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Fig. 2.6(c)
I 2= -j10
0 F
U
P1
-10 Q
20 Real
Ref Voltage
Locus of I1 = 200 Y1
P2
Im
R
Locus for Z1 (Fig. 2a)
F
F
Locus of I = I1 + I2
10 Q
P1
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Z1min = 10 ohm = OQ.
@ Y1max = mho
= 0.1 mho
@ I1m = 200 Y1m= 20A
10
1
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Y2 =
I2 = 200 Y2 = -j 10 A
I = I1+ I2
20
1
j
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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.
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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.
1!C
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EXAMPLE 2.3Find C for maximum current in the
circuit ( Fig.2.6a)
~ Z123
0.5 + 0.5 j
z20.5
3/ 2/ 1/
3 2 1z3
Fig. 2.7(a)
z
c
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Fig. 2.7b
P6
Im Y12 Locus (= Y1 + Y2)
2z123 = 2 [z12 +z3]
P2
P3
1.0
P5
0
P4
2 Z12
2.0
Y2
Y1
Q Rl
O/ P2P3
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@ 2Z12 semi circle shown by
firm line.
min12max
121
Y Z !
5.00.2
1!!
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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
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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/.
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P5 is the image of P4 ( to obtain the
point P6); by extending the line
OP5, we obtain P6.Q
P6=
wc for minimum overall impedance
@w
QP6!C
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Exercise 2.1
Draw the locus of current I , for the
circuit shown below when XL is
varied. Determine the value of L
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when the power factor is unity(Fig 2.8)
~ 200V
50 Hz
I
-20 j
10;
Fig 2.8
jx