It isn't!
star delta motors start as a Y for 3 phase so the windings that normally see 480 volts see 277V a relay switches to delta then it runs as a normal delta motor the reduces the starting current by about 60% for 3 phase
Higher capacity (higher KW rating) motors take huge starting current to get over the starting torque. To avoid the high inrush of current, star delta starter is used. When started, the motor gets connected in star for few seconds and switches over to Delta and continues to run in delta. In this case motor is designed to run in Delta.
In 3phase induction motor, stator winding splited into 3windings (Each winding conductor size, no of turns etc are equal to each other). So 3 windings have 6terminal. Generally all 6 terminals are taken out for termination on 5HP and above rated 3phase motors. So, that motor can be started with star / delta starters to reduce starting current (DOL starter does not reduce starting current). Major advantage is It is easy to change the connection (STAR / DELTA) acoording to the purpose going to be used. Also starting and ending terminal to be marked properly.
150000va divided by (600volts X 1.732) = 144 amps
The start up current should be listed on the motor nameplate as FLA , full load amps.
when starting a 3 phase induction motor,first start with star connection,because reduced current is being applied and after start change it to delta connection then. full line current is applied to the motor.
Using star delta method of wiring you can run high rated motor with out any high starting current. Star connection can provide 3 phase power as well as single phase load
Ohm's Law: Current = Voltage divided by resistance 9 volts divided by 3 ohms = 3 amperes.
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
If a motor is connected in star, the current in the motor winding will be equivalent to the line current. If the motor is connected in delta, the current will be 1/sqrt(3) of the line current. If three phase CTs are connected in Delta, their secondary current will be sqrt(3) higher than the CT ratio implies by the line current.
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star delta motors start as a Y for 3 phase so the windings that normally see 480 volts see 277V a relay switches to delta then it runs as a normal delta motor the reduces the starting current by about 60% for 3 phase
1.current ratio:It is referred by current asset divided by the current liabilities. 2.quick ratio: It is referred bi the current assets minus inventory divided by the current liabilities. 3.cash ratio: It is referred by the cash in hand ,bank balance ,temporary investnebts divided by the current liabilities.
One need to know the voltage, the power factor and starting torque, to derive the starting current. The information provided in the question is not sufficient.
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.
Higher capacity (higher KW rating) motors take huge starting current to get over the starting torque. To avoid the high inrush of current, star delta starter is used. When started, the motor gets connected in star for few seconds and switches over to Delta and continues to run in delta. In this case motor is designed to run in Delta.
DC Current divided by 1.225