Introduction to Electric Circuits

114 Three-phase a.c. circuits

Delta system

EpH = :

c

Figure 5.12

A IA = IL X

> - Ic

:' la~

ZA

ZB

If, instead of connecting together the corresponding ends of the three coils (say
all the 'start' ends a, b and c, or all the 'finish' ends a', b' and c') the finish end of
one coil were connected to the start end of the next in order (a' to b; b' to c; c' to
a) we obtain the so-called delta connection shown in Fig. 5.12. In this case it is
clear that the line voltages are the same as the phase voltages because each
phase is connected directly between two lines:

= v, (5.3)

Although the phase voltages act around the delta they sum to zero at any

instant since

eAB 4- eBC + eCA -- E sin tot + E sin (tot- 2rr/3) + E sin (tot- 4~r/3) = 0

There is therefore no circulating current around the delta.

The line currents are different from the phase currents and applying KCL to

node X, for example, we see that the line current IA is the difference of the two

phase currents lc and Ia. This is shown in the phasor diagram of Fig. 5.13. In this

diagram we have added -I~ to Ic to get IA. NOW

IA = 2IAB COS 0 = 21AB COS 30 ~

(the geometry being the same as that in Fig. 5.9), so

IA : 2IABX/3/2- ~/3IAB

But IA = IL and Ic = IPH SO that

IL- X/31pH (5.4)

Figure 5.13

[A = IL

Ii \ ~>tIc=IpH

~I a ',^1 o

Ib