Page 182 - Electrician - TT (Volume 1)

ELECTRICIAN - CITS

The voltage and current relationships of the delta connection can be explained with the aid of an illustration. The
line voltages V , V and V are directly across the load resistors, and in this case, the phase voltage is the
VW
UV
WU
same as the line voltage. The phasors V , V and V are the line voltages. This arrangement has already been
UV
VW
WU
seen in relation to the delta connection.
Because of the purely resistive load, the corresponding phase currents are in phase with the line voltages. (Fig 6)
Their magnitudes are determined by the ratio of the line voltage to the resistance R.
On the other hand, the line currents I , I and I are now compounded from the phase currents. A line current is
V
U
W
always given by the phasor sum of the appropriate phase currents. This is shown in Fig 7. The line current I is
U
the phasor sum of the phase currents I and I . (See also Fig 7)
UV UW
Hence, I = I Cos 30° + I Cos 30°
UW
U
UV
3
But Cos 30° = .
2
3
Thus I = Iph
L
Thus, for a balanced delta connection, the ratio of the line current to the phase current is .
3
3
Thus, line current = x phase current.
Application of star and delta connection with balanced loads
An important application is the `star-delta change over switch’ or star-delta starter.
Application of star connection: Alternators and secondoary of distribution transformers, have their three, single-
phase coils interconnected in star.
Assignment: Three identical coils, each of resistance 10 ohms and inductance 20mH is delta connected across
a 400-V, 50Hz, three-phase supply. Calculate the line current.

Fig 6 Fig 6

Neutral in 3-phase system

Objectives: At the end of this lesson you shall be able to
• explain the current in neutral of a 3-phase star connection
• state the earthing the neutral.

Neutral: In a three-phase star connection, the star point is known as neutral point, and the conductor connected
to the neutral point is referred as neutral conductor (Fig 1).

Current in the neutral conductor: In a star-connected, four-wire system, the neutral conductor N must carry the
sum of the currents I , I and I . One may, therefore, get the impression that the conductor must have sufficient
U
W
V
area to carry a particularly high current. However, this is not the case, because this conductor is required to carry
only the phasor sum of the three currents.
I = phasor sum of I , I and I W
V
U
N
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CITS : Power - Electrician & Wireman - Lesson 26-29