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__Three-phase fault - ____phase loss – failure caused by a burned resistor__

In a three-phase system, the following question arises often: What happens in case of a blown fuse, broken wire, worn contact or a resistor burned due to thermal overload?

We consider these error possibilities for star and delta connection at a usual line voltage of 480 V. All string resistors have the same value - thus the connected load is symmetrical. To keep the calculation simple, all string resistors have a value of 10Ω.

__Take a closer look at the star connection__

__Take a closer look at the star connection__

In general, the following applies to the star connection: It does not matter whether a resistor of the load (string) or a fuse in the supply line blows - the consequence is the same, because in the star connection the current in the line is equal to the current through the resistor (Line current I_{Line} = resistor current I_{Load}).

#### 3 Phase fault - Star connection with one phase loss

If you redraw the star connection, you can see that it is basically a parallel connection of two resistors, each connected to 480 V/√3 = 277 V.

I case of a phase fault, the new power is reduced to:

**P _{new} = 2/3 P_{Original}**

Since the remaining 2 currents no longer cancel each other out in the star point, the load now becomes asymmetrical, i.e. a current flows through the neutral conductor.

#### 3 Phase fault - Star connection with loss of 2 out of 3 phases

Now that two of the three resistors are gone, only 1/3 of the original power remains. The neutral conductor now carries the same current as the remaining phase.

**P _{neu} = 1/3**

**P**

_{original}#### 3 Phase fault - Star connection with Phase loss and loss of the neutral line

From the equivalent circuit you can see that now the circuit has changed to a series connection of two resistors, which are connected to a common voltage of 480 V. The total power is now calculated to:

P_{orginal} = 3 × (480V)^{2} / 10Ω = 23.04 kW

P_{new} = V^{2} / R_{tot} = (480V)^{2} / 20Ω = 11.52 kW

Thus: **P _{new} = 0.5 P_{orginal}**

__Take a closer look at the Delta connection__

__Take a closer look at the Delta connection__

With the Delta connection, the voltage at the resistor is equal to the line voltage. A neutral conductor is omitted for the delta connection. We consider possible errors:

#### 3 Phase fault - Delta connection - failure of a resistor

The load resistors are independent of each other, i.e. if one resistor fails, the others are not affected.

Thus, the following applies: **P _{new} = 2/3 P_{original}**

#### Phase fault - Delta connection - phase failure with defect resistor connected to

The remaining resistor is still connected to 480 V line voltage. This means that the current through this resistor also remains the same.

Thus, the following applies: **P _{new} = 1/3 P_{original}**

#### 3 Phase fault - Delta connection - failure of one phase

The equivalent circuit shows that a resistor network of three resistors is connected to a phase voltage, here 480 V. We calculate:

R_{tot} = (R1 + R2) II R3 = 20Ω x 10Ω / (20Ω + 10Ω) = 6.67 Ω

Thus the total power: P = V^{2}/R_{tot} = (480V)^{2} / 6,67 Ω = 34.54 kW

Original power: Porginal = 3 x V^{2}/R = 3 x (480V)^{2} / 10Ω = 69.12 kW

Thus: **P _{new} = 1/2 P_{original}**

#### 3 Phase fault - Delta connection - phase failure with defect resistor connected to

As the equivalent circuit shows, there is only one current-carrying resistor, which is connected to the line voltage. This gives the new power:

**P _{neu} = 1/3 P_{original}**

#### 3 Phase fault - Delta connection - phase failure with defect resistor not connected to

Two resistors in series are connected to an external conductor voltage of 400 V. This results in a total power:

P_{new} = V^{2} / R_{new} = 480^{2}V^{2} / 20Ω = 11.52 kW

R_{new} = R_{1 }+ R_{2} = 10Ω + 10Ω = 20Ω

**=> P**_{neu }**= 1/6 P**_{original}

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