**Three Phase Delta Connection: Line, Phase Current, Voltages & Power in Δ Configuration**

Table of Contents

**What is Delta Connection (Δ)?**

**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 while for distribution, Star connection is generally used.

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. 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** in Delta connection.

When a terminal is open in Δ, then there is no chance of flowing currents with basic frequency around the closed mesh.

Also Read:

- Star Connection (Y): Three Phase Power, Voltage & Current Values
- Difference between Star (Y) and Delta (Δ) Connections

**Good to Remember:** In Delta configuration, 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.

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

Now we will find the values of Line current, Line Voltage, Phase Current, Phase Voltages and Power in three phase Delta AC system.

**Line Voltages (V**_{L}) and Phase Voltages (V_{Ph}) in Delta Connection

_{L}) and Phase Voltages (V

_{Ph}) in Delta Connection

It is seen in 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 = V
_{RY} - Line 2 and Line 3 = V
_{YB} - Line 3 and Line 1 = V
_{BR}

Then, we see that V_{RY} leads V_{YB} by 120° and V_{YB} leads V_{BR} by **120°**.

Let’s suppose,

V_{RY} = V_{YB} = V_{BR} = V_{L} …………… (Line Voltage)

Then

**V _{L} = V_{PH}**

I.e. **in Delta connection, the Line Voltage is equal to the Phase Voltage**.

- Also read : Three Phase Current Values in a 3-Phase System

**Line Currents (I**_{L}) and Phase Currents (I_{Ph}) in Delta Connection

_{L}) and Phase Currents (I

_{Ph}) 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 in Delta connection **flowing through that line. i.e.;

- Current in Line 1= I
_{1}= I_{R}– I_{B} - Current in Line 2 =I
_{2}= I_{Y}– I_{R} - Current in Line 3 =I
_{3}= I_{B}– I_{Y}

{Vector Difference}

The current of Line 1 can be found by determining the vector difference between I_{R} and I_{B} and we can do that by increasing the I_{B} Vector in reverse, so that, I_{R} and I_{B} makes a parallelogram. The diagonal of that parallelogram shows the vector difference of I_{R} and I_{B} which is equal to current in Line 1= I_{1}. Moreover, by reversing the vector of I_{B}, it may indicate as (-I_{B}), therefore, the angle between I_{R} and -I_{B} (I_{B}, when reversed = -I_{B}) is 60°. If,

**I _{R} = I_{Y} = I_{B} = I_{PH}** …. The phase currents

Then;

The current flowing in Line 1 would be;

I_{L} or I_{1} = 2 x I_{PH} x Cos (60°/2)

= 2 x I_{PH} x Cos 30°

= 2 x I_{PH} x (√3/2) …… Since Cos 30° = √3/2

**I _{L}= √3 I_{PH}**

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.,

I_{2} = I_{Y} – I_{R} … Vector Difference = √3 I_{PH}

I_{3} = I_{B} – I_{Y} … Vector difference = √3 I_{PH}

As, all the Line current are equal in magnitude i.e.

**I _{1} = I_{2} = I_{3} = I_{L}**

Hence

**IL = √3 I _{PH}**

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.

Related Post: Star and Delta Connected Lighting Loads

**Power in Delta Connection**

We know that the power of each phase;

**Power / Phase = V _{PH x }I_{PH} x CosФ**

And the total power of three phases;

**Total Power = P = 3 x V _{PH x }I_{PH} x CosФ** ….. (1)

We know that the values of Phase Current and Phase Voltage in Delta Connection;

I_{PH} = I_{L} /√3 ….. (From I_{L} = √3 I_{PH})

**V _{PH} = V_{L} **

Putting these values in power eq……. (1)

P = 3 x V_{L} x ( I_{L}/√3) x CosФ …… (I_{PH} = I_{L} / /√3)

P = √3 x√3 x V_{L} x ( I_{L}/√3) x CosФ …{ 3 = √3x√3 }

**P = √3 x V _{L}x I_{L} x CosФ** …

Hence proved;

**Power in Delta Connection**,

**P = 3 x V _{PH x }I_{PH} x CosФ** …. or

**P = √3 x V _{L} x I_{L} x CosФ**

Where Cos Φ = Power factor = the phase angle between Phase Voltage and Phase Current (not between Line current and line voltage).

The same is explained in 3-Phase Circuit MCQs with explanatory Answer (MCQs No.1)

**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 V _{L} x I_{L} 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.

Also Read :

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.

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Whats your status now? Altaf Kah

need about high voltage

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

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

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

thank you for this great post

What happens when one of the phases of delta gets revese. what will be the resultant voltage?