
325Electrical Incidents andLightning
en
dY
dy
aY
2
2
2
0
(10.25)
So
Yy Ae Ae
ay ay
()=+
−
(10.26)
Now solving for X
1
2
2
2
dX
+=σ
(10.27)
dX
Xa j
2
2
2
0+−=
σ
(10.28)
Let
baj
22
σ
(10.29)
en
dX
bX
2
2
2
0
(10.30)
So
jb
bx
−
(10.31)
By applying boundary conditions, we may solve for the current distribution.
At the origin, the current is 0 and at the X and Y boundaries, the current is maximum,
which for our purposes, shall be dened as 1.
Applying the boundary conditions and solving
h
Ae Ae
ah ah
1
1
2
2
2
=+ =
−