Why does a 480 volts 3 phase power supply read 277 across legs?
In: 2
Because 480/sqrt (3) =480/1.73=277
The reason it is sqrt (3)=1.73… is that the 3 phases are 120 deg and VAB is 480, VA and VB is 277. You can just measure the relative lengths and get that the magnitude of a single phase. When you go in reverse you need to divide with it.
You can look at the voltage in [this phasor diagram](https://i.stack.imgur.com/QAGTe.jpg) and VAB is 480, VA and VB is 277. You can just measure the relative lengths and get that relationship.
You can also do the trig in the image and get constant as 2 * cos (30) = sqrt (3)
Because 480/sqrt (3) =480/1.73=277
The reason it is sqrt (3)=1.73… is that the 3 phases are 120 deg and VAB is 480, VA and VB is 277. You can just measure the relative lengths and get that the magnitude of a single phase. When you go in reverse you need to divide with it.
You can look at the voltage in [this phasor diagram](https://i.stack.imgur.com/QAGTe.jpg) and VAB is 480, VA and VB is 277. You can just measure the relative lengths and get that relationship.
You can also do the trig in the image and get constant as 2 * cos (30) = sqrt (3)
In a 480-volt, 3-phase power supply, the voltage between any two phases is equal to the voltage of each phase multiplied by the square root of 3. This is because the voltage of each phase is in phase with a third of the cycle of the power supply, and the three phases are 120 degrees out of phase with each other.
The voltage between any one phase and the neutral in a delta configuration is the difference between the voltage of that phase and the voltage of the phase that is 240 degrees ahead in the cycle. The difference in voltage is due to the fact that the voltage of each phase is not in phase with the neutral, but rather has a phase shift of 30 degrees.
Using trigonometry and the principles of electromagnetism, we can calculate that the voltage between any one phase and the neutral in a delta configuration is approximately 0.577 times the voltage between any two phases. For a 480-volt, 3-phase power supply, this results in a voltage of approximately 277 volts between any one phase and the neutral.
This calculation is based on the assumption that the system is balanced, meaning that each phase is carrying the same amount of current. If the system is unbalanced, the voltage readings may be different.
In a 480-volt, 3-phase power supply, the voltage between any two phases is equal to the voltage of each phase multiplied by the square root of 3. This is because the voltage of each phase is in phase with a third of the cycle of the power supply, and the three phases are 120 degrees out of phase with each other.
The voltage between any one phase and the neutral in a delta configuration is the difference between the voltage of that phase and the voltage of the phase that is 240 degrees ahead in the cycle. The difference in voltage is due to the fact that the voltage of each phase is not in phase with the neutral, but rather has a phase shift of 30 degrees.
Using trigonometry and the principles of electromagnetism, we can calculate that the voltage between any one phase and the neutral in a delta configuration is approximately 0.577 times the voltage between any two phases. For a 480-volt, 3-phase power supply, this results in a voltage of approximately 277 volts between any one phase and the neutral.
This calculation is based on the assumption that the system is balanced, meaning that each phase is carrying the same amount of current. If the system is unbalanced, the voltage readings may be different.
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