Electrical Power Systems

128 Electrical Power Systems
rom KCL, we can write,
\
I(x + Dx) = I(x) + y. Dx.V(x + Dx)
Ix ( + Dx)
- Ix ( )
= y.V(x + Dx) ...(6.21)
Dx
As Dx ® 0
dI( x)
= y.V(x) ...(6.22)
dx
Differentiating eqn. (6.20) and substituting from eqn. (6.22), we get,
\
2
d V( x)
dI( x)
= z × = z. yV(x)
2
dx dx
2
d V( x)
– zy V(x) = 0 ...(6.23)
dx
2
Let g 2 = zy ...(6.24)
Therefore,
2
d V( x)
2
– g V(x) = 0 ...(6.25)
dx
The solution of the above equation is
2
V(x) = C1e g x –g x
+ C2 e
where, g, known as the propagation constant and is given by,
...(6.26)
g = a + jb = zy ...(6.27)
The real part a is known as the attenuation constant, and the imaginary part b is known as
the phase constant. b is measured in radian per unit length.
rom eqn. (6.20), the current is,
I(x) = 1
Z
×
dV( x)
dx
\ I(x) = g
z (C 1 e gx – C 2 e –gx )
\ I(x) =
y
z (C 1 e gx – C 2 e –gx )
\ I(x) = 1
(C1 e
ZC gx – C2 e –gx ) ...(6.28)
where, Z C is known as the characteristic impedance, given by
Z C =
z
y
...(6.29)