Line and phase quantities in delta connection. Delta or Mesh Connection (Δ) System is also known as Three Phase Three Wire System (3-Phase 3 Wire) and it is the most preferred system for AC power transmission.

1. Presentation By :
Kartikeya(2K19/A7/35) &
Keshav(2K19/A7/36)
Submitted to:
Ankita Arora

2. The system which has three phases, i.e., the current will pass through the three wires,
and there will be one neutral wire for passing the fault current to the earth is known as
the three phase system. The sum of the line currents in the 3-phase system is equal to
zero, and their phases are differentiated at an angle of 120º
It has several advantages like it requires fewer conductors as compared to the single
phase system. It also gives the continuous supply to the load. The three-phase system
has higher efficiency ,minimum losses. & provides an uninterruptible power.

3. Delta or Mesh Connection (Δ) System is also known as Three Phase Three Wire
System (3-Phase 3 Wire) and it is the most preferred system for AC power
transmission.
In Delta (also denoted by Δ) system of interconnection, the starting ends of the three
phases or coils are connected to the finishing ends of the coil. and it looks like
a closed mesh or circuit as shown in fig.
In more clear words, all three coils are connected in series to form a close mesh
or circuit. Three wires are taken out from three junctions and the all outgoing currents
from junction assumed to be positive.

4. Relation between line current IL and
Iph
Here, IA ,IB ,IC are line current.
IAC,ICB,IBA are phase current.
vAB,vBC,vCA are phase voltages.
VAB,VBC,VCA are line voltages.
IA =IB =IC = IL
IAB=IBC=ICA=IPh
vAB = vBC = vCA = vPh
VAB=VBC=VCA=VL
We know,
IAB =IPh 0
IBC=IPh-120
ICA=IPh-240=IPh120

5. Applying KCL at Node A
IA = IAB - ICA
= IPh0 - IPh 120
=IPh( cos00 +j sin00) – IPh(cos1200 + j sin1200)
= IPh(1 +j0) - Iph(-0.5 +j 3/2)
=3 IPh(3/2 – j 1/2)
=3 IPh -30
Similarly, Applying KCL at node B
IB =IBC – IAB
=IPh-120 - IPh 0
= IPh(cos(-1200) + j sin(-1200)) - IPh(cos00 +j sin00)
= IPh(-0.5 - j3/2) - IPh(1)
= IPh(-3/2 - j3/2)
=3 IPh-1500 = IPh2100
Similarly, Applying KCL at node C
IC =ICA – IBC
=IPh120 - IPh -120
= IPh(cos1200 +j sin1200) - IPh(cos(-1200) + j sin(-1200))
= Iph(-0.5 +j 3/2) - IPh(-0.5 - j3/2)
= IPh(0 + j3)
=3 IPh900

6. PHASOR DIAGRAM
Leading
Lagging
IAB
IA
VAB
IAB
IA
VAB
300 -


300
300 +

7. RELATIONSHIP BETWEEN PHASE VOLTAGE
AND LINE VOLTAGE
It is clear from the figure that the
magnitude of
Line voltage is equal to phase voltage.
VAB =VPh 0
VBC=VPh-120
VCA=VPh-240 = VPh120
Therefore, VL =VPh

8. CONCLUSION
When a delta connected ,three phase system is connected to balanced
load then line voltage is equal to the phase voltage, VL =VPh
All line current lags behind their respective phase current by 300.
Line current is 3 times the phase current, IL = 3 Iph
The angle between line current and the corresponding line voltage is
300 - in case of leading phase current and 300 - in case of lagging
phase current.
In delta connection ,each winding receives 415V.

9. QUESTIONS
Q-1)A 415V, 3-phase a.c. motor has a power output of 12.75Kw and operates at a
power factor of 0.77 lagging and with an efficiency of 85 per cent. If the motor is
delta-connected, determine (a) the power input, (b) the line current and (c) the
phase current.
Answer)
Given: VL=415V, cos φ=0.77, =85%, Po =12.75Kw
(a) Efficiency=power output/power input.
Efficiency=85/100=0.85
Hence, Power input = 12.75×1000 × 0.85
= 15 000W or 15Kw
(b) Power, P=√3VLIL cos φ, hence
IL = P/ √3 (415) (0.77)
= 15000/ √3 (415) (0.77)
= 27.10A
(c) For a delta connection, IL =√3IP,
Hence,
Phase current, IP = IL/√3
= 27.10 /√3
= 15.65A

10. Question-2) A 400V, 3-phase star connected alternator supplies a delta- connected load, each
phase of which has a resistance of 30 and inductive reactance 40. Calculate (a) the current
supplied by the alternator and (b) the output power and the kVA of the alternator, neglecting
losses in the line between the alternator and load.
Answer-2)
(a) Considering the load
Phase current, IP =VP/ZP
VP =VL for a delta connection,
Hence, VP =400V.
Phase impedance,
ZP =√( X2
L +R2
L)
=√ (302 +402) =50
Hence IP =VP/ZP =400/50=8A.
For a delta-connection,
Line current, IL =√3 Ip =√3 (8) =13.86 A.
Hence, 13.86A is the current supplied by the alternator.
(b) Alternator output power is equal to the power Dissipated by the load
i.e. P =√3 VLIL cos φ, Where cos φ = RP/ZP = 30/50 = 0.6.
Hence P =√3 (400) (13.86) (0.6) = 5.76kW.
Alternator output kVA,
S =√3VL IL =√3 (400) (13.86)
= 9.60 kVA.