What is the formula use for calculation voltage and current in star and delta?
Balanced Star (Wye) Connected Systems:
- Line Voltage = 1.732 x Phase Voltage
- Line Current = Phase Current
Balanced Delta Connected Systems:
- Line Voltage = Phase Voltage
- Line Current = 1.732 x Phase Current
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
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.
What is the relationship between phase and line values of current and voltage quantities for star and delta connected systems?
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.
The question is unclear. If the answer does not address the correct issue, please restate the question with more details. Three phase voltage can be measured in star, each leg to neutral, or in delta, each leg to a different leg. Assuming that neutral is balanced in the center of the phase triangle, the star voltage is the delta voltage divided by 1.732. 1.732 is the square root of 3, and comes from the phasor… Read More
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.
To answer this question a voltage must be stated.
Transistors, at least the typical bi-junction transistor, actually amplify current. We set them up in a voltage divider circuit that converts current gain into voltage gain. The simple explanation is that a small delta current on base-emitter causes a larger delta current on collector-emitter. The gain is either hFe or collector resistance divided by emitter resistance, whichever is less.
What is the current difference in a 3 phase motor running continuously in star connection and delta connections?
The voltage difference between star and delta is 1.732, the square root of 3, or, going from delta to star, 1 over 1.732, which is 0.5774. That is the same as the current difference, neglecting motor efficiency at the two different operating voltages. The power difference between star and delta is 1.7322, or 3, and between delta and star is 0.57742, or 0.3333. <<>> In a wye connection, the line current is equal to the… Read More
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.
Line current = 1.732 x Phase Current Comment Only for balanced loads.
To answer this question the voltage of the motor must be stated.
The voltage of a winding that is connected in star is 1 over the square root of three, 1 / 1.732, or 0.5774, of the voltage if that winding were connected in delta. This results in a star power of 1/3, or 0.3333, of the delta power. Similarly, if you know the winding current in delta, you can calculate the star current by multiplying by 1.732. This all presumes that the neutral point of the… Read More
starter is used to limit the starting current to save the motor. for star-delta first star mode is enabled so the applied voltage is reduced to safe value after the motor catches the rated speed the mode is changed to the delta mode. hence the full supply voltage is applied to the motor.
A delta starter supplies what ever the supplied voltage to the line side of the contactor is. The inrush current is usually 300% of the full load amperage of the motor. Edit to above answer: Actually the question is for a Wye-Delta Starter (Wye=Star). There are two contactors, one in Wye configuration and one with Delta. The Wye is used to start and the Delta for run. This is a reduced voltage start scheme. Though… Read More
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.
A disadvantage with Open Delta is there are no phase-ground voltage measurements, which is only a disadvantage if directional relaying or distance relaying is needed. Some directional and voltage based ground fault detection relays don't work without the phase to ground input. Since most industrial systems MV are high resistance grounded and ground fault protection is provided by overcurrent, there isn't a need for the phase-ground measurements. Why will the directional or distance relaying not… Read More
A motor draws less current in star than in delta because there is less voltage in star, less by a factor of the square root of three, or about 1.732.
Full load current ofthe motor x 0.58
No. Line to line voltage is 1.732 (square root of 3) times the line to ground voltage on a properly balanced system.
Dr. Emmanuel Uduaghan is the current governor of Delta state.
star connection like y whle delta connection like a triangle. in star connection there is adifference between line voltage and phase voltagewhi delta connection line voltage and phase voltage are same. in star connectin line voltage is always greater than phase voltage
Richard H. Anderson is the current CEO of Delta Air Lines.
Can a 480 to 240 3 phase delta to delta step down transformer be used to step up voltage in the reverse?
Yes, a 480 to 240 3 phase delta to delta step down transformer can be used to step up voltage in the reverse. <<>> However transformers should always be used with the correct voltage on each winding.
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.
delta connection current rating
In Delta connected tertiary winding reduces Zero sequence impedance and allow adequate current for operation of protective device in order to limit voltage imbalance which may be produced when in balance in load.
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 detail Whenever 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… Read More
The main difference between star and delta is that in star, the phase voltage is root 3 multiplied by the line voltage making the speed of the machine to run at low speed while in delta, the phase voltage is equal to the line voltage making the machine to run at high speed.
For the same amount of power, a star or delta connection will use the same line current. However a delta connection supplies sqrt(3) times less current to the device on each of its connections because the voltage is sqrt(3) times more. The apparent contradiction comes about because in a delta connection each connection to the line is shared between two connections to the device (Historikeren 20-07-2015).
The use of a star delta starter is for motor control only and not resistive heating loads. This type of starter is used to dampen the inrush current by using a lower voltage to start the motor.
What is the continuous current of 10HP star delta motor
In a delta connection, the line voltage is equal to the phase voltage. In a star connection, the line voltage is 1.732 larger than the phase voltage.
The voltage between each lead of a 480 volt three phase delta secondary is 480 volts.
Gently we are using star delta starter for hiegh rating motors,,in star connection the line current is equal to the phase current and incase delta connection the line and phase voltages are equal,,at starting the current will be maximum.
For the same load, delta provides higher current than wye, or star. Wye voltage is 1 / 1.732, or 0.577 that of delta. (1.732 is the square root of 3.) This gives a power of 0.333 for wye, as opposed to delta. (0.5772 = 0.333) This is why (no pun intended) we use wye-delta starters on large motors - it allows them to come up to speed at 1/3 power, before switching to full power… Read More
Yes, Richard H. Anderson is the current CEO of Delta Airlines as of April 2011.
Commissioner of education in delta state nigeria
Motors with star delta starter, starts with star mode, avoiding inrush of current for a relatively higher starting load, as the speed picks up, it goes to Delta mode with full phase voltage applied across the winding and continues to cater to the designed load.
in star-star neutral transformer at unbalanced load condition zerosequnce current will be flows in secondary side but this current can not be balanced by primary side because its not having neutral to circulate zerosequence current.due to this unbalance voltage will appeared in primary side but the tertiary delta winding allows circulate the zerosequence current so the primary voltage will get stabilized.
Probably the reason is in the different voltages between the star and delta connection voltages. An electromagnetic brake designed for the lower voltage of the star connection would burn out or open when introduced to the higher voltage of the delta connection.
because the phase voltage of the delta connection is more the the phase voltage is stare connection by root 3 so if we have 380 at primary u will get 220 at secondary ( i think )
Reverse any two leads of the supply voltage before the star delta contactor.
As usual you need to know the voltage and the kVA. For a balanced load, find the line-to-neutral voltage, which is the nominal voltage between lines divided by sqrt(3). Each line supplies one-third of the kVA, so the current in each line is equal to the kVA times 1000 divided by the line-to-neutral voltage. As an example, 50 kVA on a 208 v three-phase supply would have a line current of 50/3x1000/120 or 139 amps… Read More
When an induction motor starts up, it draws a large current, 6-8 times for direct on line starting and around 3 times for star delta or soft starting. This results in the voltage dipping. VSD controlled induction motors which start from zero speed do not result in voltage dips.
Heat flow per area per time = (delta T) / R delta T = difference in temperature between the two sides of the wall area = area of the wall or part of the wall under consideration 'R' = "R"-value of the construction materials and/or insulation that makes up the wall. Intreresting how it's the direct analog of Ohm's Law for electrical voltage (delta T), current (heat flow per time), and resistance ('R').
A 415 v three phase system has a line-to neutral voltage of 240 v on each of the three phase wires. Each wire supplies 250,000/3 or 83,333 VA so the current is 83,333 / 240 or 347 amps. The current in each live wire is 347 amps. If a balanced load was delta-connected to it, the load current would be 200 amps at 415 volts. Another Answer I suspect that you are really asking what… Read More
There is no 'return path' for current in a delta connection, because the phasor sum of the three line currents is always zero.