**Delta Connection (Δ)**

**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.

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

### 1. **Line Voltages and Phase Voltages in Delta Connection**

_{RY}

_{YB}

_{BR}

_{RY}leads V

_{YB}by 120° and V

_{YB}leads V

_{BR}by 120°.

_{RY}= V

_{YB}= V

_{BR}= V

_{L}…………… (Line Voltage)

_{L}= V

_{PH}

### 2. **Line Currents and Phase Currents in Delta Connection**

- 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}

*Click image to enlarge*

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,

_{R}= I

_{Y}= I

_{B}= I

_{PH}…. The phase currents

_{L}or I

_{1}= 2 x I

_{PH}x Cos (60°/2)

_{PH}x Cos 30°

_{PH}x (√3/2) …… Since Cos 30° = √3/2

_{PH}

_{2}= I

_{Y}– I

_{R}… Vector Difference = √3 I

_{PH}

_{3}= I

_{B}– I

_{Y}… Vector difference = √3 I

_{PH}

_{1}= I

_{2}= I

_{3}= I

_{L}

_{PH}

- 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**

_{PH x }I

_{PH}x CosФ

_{PH x }I

_{PH}x CosФ ….. (1)

_{PH}= I

_{L}/ /√3 ….. (From IL = √3 I

_{PH})

_{PH}= V

_{L}

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 …

_{PH x }I

_{PH}x CosФ …. or

_{L}x I

_{L}x CosФ

**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:**

_{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.

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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.

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