Handbook of Electrical Engineering

216 HANDBOOK OF ELECTRICAL ENGINEERING

iii) Hence the voltage across the capacitance is,

Vc=Vs−Vsc

= 19456. 7 +j 0. 0 − 508. 35 −j 336. 28

= 18948. 35 −j336.28 volts/phase

The charging currentIcfor the capacitance is,

Ic=

Vc

Xc

=

18949. 35 −j 336. 28

−j 530. 52

= 0. 634 +j35.716 amps

DeductIcfromIsto findIr,

Ir=Is−Ic= 317. 92 −j 80. 691 − 0. 634 −j 35. 716

= 317. 286 −j116.41 amps

The volt-drop in the right-hand side of the cable isVcr,

Vcr=Ir

(

R

2

+j

X 1

2

)

=( 317. 286 −j 116. 41 )( 1. 25 +j 1. 375 )

= 556. 671 +j 290 .756 volts/phase

iv) Hence the voltage received at the load is,

Vr=Vc−Vcr

= 18948. 35 −j 336. 28 − 556. 671 −j 290. 756

= 18391. 68 −j 627 .04 volts/phase

|Vr|=18402.36 volts/phase

The total actual volt-drop

V=

|Vs|−|Vr|

|Vs|

× 100

=

( 19456. 7 − 18402. 36 ) 100

19456. 7

= 5 .419%

The receiving end volt-drop with respect to the nominal system voltage isVn,

Vn=

|Vn|−|Vr|

|Vn|

× 100

=

( 19052. 6 − 18402. 36 )

19052. 6

100 = 3 .413%