Balanced Star (Wye) Connected Systems:
Balanced Delta Connected Systems:
normally delta connection wired in 3 phase induction motor. during starting wiring is in Star and after running normal speed changeover to delta .beacause starting time its phase voltage equals less root3 times of line voltage ,line current and phase current equals. in Delta phase voltage and line voltage equals, and phase current equals root3 times line current
Star voltage is 1 / 1.732, or 0.577 delta voltage. (1.732 is the square root of 3.) It follows, then, that star current is 0.577 delta current.Not asked, but answered for completeness, star power is 0.5772, or 0.3333 delta power. This is why large motors use a star-delta start sequence - to minimize startup thermal stress.
A star-delta motor should be supplied star-delta.Wye is simply another name for star. Star-delta motors use a star or wye configuration to start, and a delta configuration to run. This reduces the voltage on the windings during the high current starting time.
This configuration is used to reduce the starting current. Utility companies do not like large motor loads starting across the line. It dips the voltage level of the line. By reducing the starting current to a lower level also reduces the voltage dip in the supply lines.
A "star-delta" starter- also known as a "Y-delta" or "wye-delta" starter - is used to reduce the voltage on the motor windings to only 1/.'/'3 [one divided by root 3] times the incoming line voltage, to help reduce mechanical stress and in-rush current at start-up.More detailWhenever you start a big heavy electric motor, you need to start it slowly to prevent the rotor overheating and drawing an enormous current. When the windings of a 3-phase motor are connected in STAR the voltage is reduced to a fraction equal to one over root three (1 / 1.732) times the normal running voltage that is used when it is connected in DELTA, resulting in a reduction of nearly one half the normal current.Once the motor picks up speed, the connection is changed to DELTA so that the motor runs at full speed and torque from then on. It's a bit like using the gears of an automobile.For more information please see the answers to the Related Questions shown below.
The calculation will change daily. You will have to check with the website to find out the current calculations for your needs.
Formula for what?!
normally delta connection wired in 3 phase induction motor. during starting wiring is in Star and after running normal speed changeover to delta .beacause starting time its phase voltage equals less root3 times of line voltage ,line current and phase current equals. in Delta phase voltage and line voltage equals, and phase current equals root3 times line current
A: A DELTA transformer is a 1:1 voltage transfer delta to Y IS 1:2 voltage transfer. That is for 3 phase system, If the phases are not exactly matched or the voltage is not exactly right then on a Y setup there will be circulating current at the common node.
What is the continuous current of 10HP star delta motor
In star the voltage from line to neutral is 1/sqrt(3) times the nominal voltage, while the load current equals the line current. In delta the voltage between lines is the nominal voltage, while the load current is 1/sqrt(3) times the line current (for a balanced load). So a delta load needs 3 times the resistance compared to a star load of the same power.
Star voltage is 1 / 1.732, or 0.577 delta voltage. (1.732 is the square root of 3.) It follows, then, that star current is 0.577 delta current.Not asked, but answered for completeness, star power is 0.5772, or 0.3333 delta power. This is why large motors use a star-delta start sequence - to minimize startup thermal stress.
To answer this question a voltage must be stated.
Line current = 1.732 x Phase CurrentCommentOnly for balanced loads.
1.73 (the square-root of 3) is the ratio of line voltage to phase voltage in a star (wye) three-phase connection, and is the ratio of line to phase current ratio in a delta three-phase connection supplying a balanced load. In each case, it is derived from the phasor addition of two values displaced from each other by 120 electrical degrees.
in a delta configuration the current is split between the phases, as opposed to a wye configuration where the current will be equal on all phases. and vice versa for voltage.
V3= Voltage across the supply V2= voltage across inductive load V1= Voltage across known resistance R= Known resistance Cos (Phi)=(V32-V12-V22)/2v1v2 power = (V32-V12-V22)/2R