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Delta Connection (Δ): 3 Phase Power, Voltage & Current Values

Delta Connection  (Δ)

In this system of interconnection, the starting ends of the three phases or coils are connected to the finishing ends of the coil. Or the starting end of the first coil is connected to the finishing end of the second coil and so on (for all three coils) and it looks like a closed mesh or circuit as shown in fig (1).
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.
In Delta connection, the three windings interconnection looks like a short circuit, but this is not true, if the system is balanced, then the value of the algebraic sum of all voltages around the mesh is zero.
When a terminal is open, then there is no chance of flowing currents with basic frequency around the closed mesh.
Good to Remember: at any instant, the EMF value of one phase is equal to the resultant of the other two phases EMF values but in the opposite direction.
Delta or Mesh Connection System is also called Three Phase Three Wire System (3-Phase 3 Wire) and it is the best and suitable system for AC Power Transmission.
Click image to enlarge

Delta Connection (Δ) Three Phase Power, Voltage & Current Values

Voltage, Current and Power Values in Delta Connection (Δ)

1.   Line Voltages and Phase Voltages in Delta Connection

It is seen from fig 2 that there is only one phase winding between two terminals (i.e. there is one phase winding between two wires). Therefore, in Delta Connection, the voltage between (any pair of) two lines is equal to the phase voltage of the phase winding which is connected between two lines. Since the phase sequence is R → Y → B, therefore, the direction of voltage from R phase towards Y phase is positive (+), and the voltage of R phase is leading by 120°from Y phase voltage. Likewise, the voltage of Y phase is leading by 120° from the phase voltage of B and its direction is positive from Y towards B.
If the line voltage between;
Line 1 and Line 2 = VRY
Line 2 and Line 3 = VYB
Line 3 and Line 1 = VBR
Then, we see that VRY leads VYB by 120° and VYB leads VBR by 120°.
Let’s suppose,
VRY = VYB = VBR = VL   …………… (Line Voltage)
I.e. in Delta connection, the Line Voltage is equal to the Phase Voltage.

2.   Line Currents and Phase Currents in Delta Connection

It will be noted from the below (fig-2) that the total current of each Line is equal to the vector difference between two phase currents flowing through that line. i.e.;
  • Current in Line 1= I1 = IR – IB
  • Current in Line 2 =I2 = IY – IR
  • Current in Line 3 =I3 = IB – IY
{Vector Difference}
Click image to enlarge

Delta Connection (Δ): 3 Phase Power, Voltage & Current ValuesThe current of Line 1 can be found by determining the vector difference between IR and IB and we can do that by increasing the IB Vector in reverse, so that, IR and IB makes a parallelogram. The diagonal of that parallelogram shows the vector difference of IR and IB which is equal to Current in Line 1= I1. Moreover, by reversing the vector of IB, it may indicate as (-IB), therefore, the angle between IR and -IB (IB, when reversed = -IB) is 60°. If,

IR = IY = IB = IPH …. The phase currents
The current flowing in Line 1 would be;
IL or I1 = 2 x IPH x Cos (60°/2)
= 2 x IPH x Cos 30°
= 2 x IPH x (√3/2) …… Since Cos 30° = √3/2
= √3 IPH
i.e. In Delta Connection, The Line current is √3 times of Phase Current
Similarly, we can find the reaming two Line currents as same as above. i.e.,
I2 = IY – IR … Vector Difference = √3 IPH
I3 = IB – IY … Vector difference = √3 IPH
As, all the Line current are equal in magnitude i.e.
I1 = I2 = I3 = IL
IL = √3 IPH
It is seen from the fig above that;
  • The Line Currents are 120° apart from each other
  • Line currents are lagging by 30° from their corresponding Phase Currents
  • The angle Ф between line currents and respective line voltages is (30°+Ф), i.e. each line current is lagging by (30°+Ф) from the corresponding line voltage.

3.   Power in Delta Connection

We know that the power of each phase
Power / Phase = VPH x IPH x CosФ
And the total power of three phases;
Total Power = P = 3 x VPH x IPH x CosФ ….. (1)
We know that the values of Phase Current and Phase Voltage in Delta Connection;
IPH = IL / /√3   ….. (From IL = √3 IPH)
VPH = VL    
Putting these values in power eq……. (1)
P = 3 x VL x ( IL/√3) x CosФ …… (IPH = IL / /√3)
P = √3 x√3 x VL x ( IL/√3) x CosФ …{ 3 = √3x√3 }
P = √3 x VLx IL x CosФ   …
 Hence proved;
Power in Delta Connection,
P = 3 x VPH x IPH x CosФ …. or
P = √3 x VL x IL x CosФ
 As explained this in 3-Phase Circuit MCQs with explanatory Answer (MCQs No.1)
Good to Know: Where Cos Φ = Power factor = the phase angle between Phase Voltage and Phase Current and not between Line current and line voltage.
Good to Remember:
In both Star and Delta Connections, The total power on balanced load is same.
I.e. total power in a Three Phase System = P = √3 x VL x IL x CosФ
Good to know:

Balanced System is a system where:

  • All three phase voltages are equal in magnitude
  • All phase voltages are in phase by each other i.e. 360°/3 = 120°
  • All three phase Currents are equal in magnitude
  • All phase Currents are in phase by each other i.e. 360°/3 = 120°
  • A three phase balanced load is a system in which the load connected across three phases are identical.
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  1. Dear Admin,
    I need help from you & I know you will able to help me.

    I have applied for electrical supervisor license exam can I get the question papers related to this exam so that I can prepare accordingly.
    Earlier revert will be great for me.

  2. need about high voltage

  3. In transformer primary is delta connection and secondary is star connection , the primary v1=110kv ,v2=22kv and power is 16 mva how to calculate current i1 and i2,
    If im using the formual p= 3*vph*iph*cos fi what is the value of cos fi

  4. Please help calculate the pahse voltage across three delta connected resistors when the line voltage is 250 v – please show me the working out?

  5. 16mva is not ‘P’ that is ‘S’ so you can use the formula s=root3*v*i to find the currents

  6. thank you for this great post

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