380V ÷ √3 = 219.4
The term, 'unbalanced system' refers to an unbalanced load. Under normal circumstances, an unbalanced load leads to unbalanced line currents. The line voltages are determined by the supply and remain symmetrical, even when the load is unbalanced. As your question refers to a 'line to neutral' voltage (i.e. a phase voltage), you must be referring to a star (wye) connected load, in which case the phase voltage (line to neutral voltage) is 0.577 (the reciprocal of the square-root of 3) times the line voltage (line to line voltage).
In a 3 phase system, the voltage measured between any two phase is called line to line voltage.And the voltage measured between line to neutral is called phase to neutral (line to neutral) voltage.AnswerThere is no such thing as a 'phase-to-phase' or a 'phase-to-neutral' voltage. The correct terms are 'line-to-line' and 'line-to-neutral'.The voltage between any two line conductors is called a line voltage.In a three-phase, three-wire, system, the line voltage is numerically equal to the phase voltage.In a three-phase, four-wire, system, the voltage between any line conductor and the neutral conductor is called a phase voltage. The line voltage is 1.732 times larger than the phase voltage.
To match 2 phase line voltage it has to be the same voltage.
How do you calculate voltage drop for starting motor current
no load voltage - full load voltage by full load voltage
380V ÷ √3 = 219.4
Devide the wattage by the voltage
You can't "calculate" both. If you know the wattage & the line voltage then I = P / E and vice-versa.
First of all, there is no such thing as a 'phase-to-phase' voltage. The correct term is 'line-to-line' voltage. Secondly, without knowing what you mean by 'overall voltage', there is no way of answering your question.
The term, 'unbalanced system' refers to an unbalanced load. Under normal circumstances, an unbalanced load leads to unbalanced line currents. The line voltages are determined by the supply and remain symmetrical, even when the load is unbalanced. As your question refers to a 'line to neutral' voltage (i.e. a phase voltage), you must be referring to a star (wye) connected load, in which case the phase voltage (line to neutral voltage) is 0.577 (the reciprocal of the square-root of 3) times the line voltage (line to line voltage).
you calculate a voltage circuit by taking it apart and findng the circuit and calculate the voltage and then resible it.
formals to calculate exciation voltage of alternator
Phase to Phase voltageCorrection to the above answer:There is no such thing as a 'phase-to-phase' or 'phase-to-ground' voltage. The correct terms are 'line-to-line' (or 'line voltage') and 'line-to-ground' (or 'phase voltage'). Transmission-line voltages are line-to-line (or 'line') voltages.
It depends on the total load spread along the line and the voltage drop at full load at the end of the line. Generally up to 3 % voltage drop is considered ok. Anything beyond you need another transformer center.
A voltage is applied to a signal line. The voltage of the line changes gradually from 0 to +V. The "edge speed" is the rate of change of voltage of the line. A voltage is applied to a signal line. The voltage of the line changes gradually from 0 to +V. The "edge speed" is the rate of change of voltage of the line.
Ohm's Law - V = IR.