Handbook of Electrical Engineering

AUTOMATIC VOLTAGE REGULATION 95

And,

S 3 =

(

Va 31 −Va 32

Va 12 Vfd3pu

)

u=AeBVf d^3 (4.21)

Divide (4.21) by (4.20)

S 3

S 2

=eB(Vf d^3 −Vf d^2 )

From which,

B=

loge

(

S 3

S 2

)

Vfd 3 −Vfd 2

(4.22)

Also,

S 3

S 2

=

(Va 31 −Va 32 )

(Va 21 −Va 22 )u

Since

u=

Vfd 3

Vfd 2

by proportion from (4.18)

S 3 =AeBu Vf d^2

S 2 =AeBVfd^2

Hence,

logeS 3 =logeA+uBVfd 2 (4.23)

And,

logeS 2 =logeA+BVfd 2 (4.24)

Multiply (4.24) byu, and subtract from (4.23),

logeS 3 −ulogeS 2 =logeA−ulogeA

=( 1 −u)logeA

Therefore,

A^1 −u=

S 3

S 2 u

Hence,

A=

S 31 /(^1 −u)

S 2 u/(^1 −u)