write relation for power drawn in three phase star connected load
Answers
Power in three phase circuit is given by
√3×line voltage×line current
In star connected load, line voltage is sqrt(3) of phase voltage and line current is same as phase current.
Delta connected load, it is opposite to star connection that is, phase voltage = line voltage, line current = √3×phase current
So power in each phase = phase voltage × line current.
Power in each phase in star connected load
= (Line voltage/√3) × line current
In Delta connected load = line voltage × line current/√3
Power in each phase so in all three phases is equal in both cases of star and delta connection.
Let understand this problem from further prospective, if equal resistance R are connected in star and then delta from supply of line voltage V
Star connection: phase voltage = V/√3
Phase current = phase voltage/R = V/√3 R
Star connection line current = phase current = V/√3 R
So total power = √3 x V x V/√3 R = V^2/R
Delta connection: line voltage = phase voltage = V
Phase current = phase voltage/R = V/R
Power consumption = √3 line voltage x line current = √3 V x V/R = √3V^2/R
Ratio of power in delta to star connected resistance R = √3.
Contradictory results from both statement
Why?
Write your answer in comment box.
Answer
That's right when same impedance are connected in star and delta then equivalent impedance will change and due this line current in star and delta will be different so P=√3Vline×Iline will also be different.
This can be further understand by star delta transformation.
Power in delta connected circuit is 3 time power in star connected circuit.
What is the power ratio between power in Star and Delta circuit?
Power in three phase circuit is given by
√3×line voltage×line current
In star connected load, line voltage is sqrt(3) of phase voltage and line current is same as phase current.
Delta connected load, it is opposite to star connection that is, phase voltage = line voltage, line current = √3×phase current
So power in each phase = phase voltage × line current.
Power in each phase in star connected load
= (Line voltage/√3) × line current
In Delta connected load = line voltage × line current/√3
Power in each phase so in all three phases is equal in both cases of star and delta connection.
Let understand this problem from further prospective, if equal resistance R are connected in star and then delta from supply of line voltage V
Star connection: phase voltage = V/√3
Phase current = phase voltage/R = V/√3 R
Star connection line current = phase current = V/√3 R
So total power = √3 x V x V/√3 R = V^2/R
Delta connection: line voltage = phase voltage = V
Phase current = phase voltage/R = V/R
Power consumption = √3 line voltage x line current = √3 V x V/R = √3V^2/R
Ratio of power in delta to star connected resistance R = √3.
Contradictory results from both statement
Why?
Write your answer in comment box.
Answer
That's right when same impedance are connected in star and delta then equivalent impedance will change and due this line current in star and delta will be different so P=√3Vline×Iline will also be different.
This can be further understand by star delta transformation
For same power in both star delta connection phase impedance need to be converted as shown in above diagram.
For more about basic electrical engineering principle you can read
by Deltoro
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Power in delta connected circuit is 3 time power in star connected circuit
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An example of this is a 3 phase motor with 2400 volt windings. When connected in delta, the line voltage to operate it must be 2400 volts. The same motor when connected in Wye (Star) now must operate at 4160 volts. (delta volts * Square root of 3.) The motor will still produce the same brake horsepower at the shaft (or kW).
If you could vary the voltage, the power into the motor varies as the square of the voltage change. Like on a reduced voltage starter, the 65% tap drops the voltage OK to 0.65 x line voltage, but more importantly drops the power by the square of the voltage. Comes from combing Ohm’s law and the power formula.
