Electric Power Generation, Transmission, and Distribution

VAB–Ais calculated by integrating the electric field intensity, due to the charge on conductor A, on
conductor B fromrAtoD

VABA¼

ðD

rA

EAdx¼

q

2 p« 0

ln

D

rA



(13:65)

VAB–Bis calculated by integrating the electric field intensity due to the charge on conductor B fromDtorB

VABB¼

ðrB

D

EBdx¼

q

2 p« 0

ln

rB

D

hi

(13:66)

The total voltage is the sum of the generated voltagesVABAandVABB

VAB¼VABAþVABB¼

q

2 p« 0

ln

D

rA





q

2 p« 0

ln

rB

D

hi

¼

q

2 p« 0

ln

D^2

rArB



(13:67)

If the conductors have the same radius,rA¼rB¼r, then the voltage between conductorsVAB, and the
capacitance between conductorsCAB, for a 1-m line length are

VAB¼

q

p« 0

ln

D

r



ðÞV (13:68)

CAB¼

p« 0

ln

D

r

ðÞF=m (13:69)

The voltage between each conductor and ground (G) (Fig. 13.16) is one-half of the voltage between the two
conductors. Therefore, the capacitance from either line to ground is twice the capacitance between lines

VAG¼VBG¼

VAB

2

ðÞV (13:70)

CAG¼

q

VAG

¼

2 p« 0

ln

D

r

ðÞF=m (13:71)

CAG

A

CBG

VAG VBG

VAB

+

q−

q−

q+

q+

−

B

VBG

VAG

VAB

CAG

CBG

B

A

−

FIGURE 13.16 Capacitance between line to ground in a two-wire single-phase line.