Three phase power supply - what is line to line voltage

  • Well this seems like a basic principle, yet I can't seem to get it. (We're expect to "know" this already).

    In a three phase situation I'm given a source voltage of 230V. - So the waveform of each of the phases would be: \$ v_s = \sqrt2 \cdot 230 \cdot \sin(\omega t + \theta_i)\$

    Where \$\theta_i\$ is \$0, \tfrac{2}{3} \pi, \tfrac{4}{3} \pi\$ for each phase.

    So now I could calculate the line to line voltage by the formula: $$v_{ll} = 2 \cdot \left ( \sqrt2 \cdot 230 \cdot \sin(\tfrac{2}{3} \pi) \right)$$

    Is this correct?

  • No need for a complicated formula.

    If you have balanced three-phase power, where all three phase voltages are equal in magnitude and 120° apart in phase, then:

    $$ V_{L-L} = \sqrt{3} \times V_{L-N} $$

    To see why, consider the phasor diagram:

    enter image description here

    Applying some basic trig:

    enter image description here

  • Line to line voltage for a 3phase network (120deg separation) is sqrt(3)*phase voltage.

    So for a 230V 3ph network the line-line is 400V

  • Line-to-line voltage is the difference between line-to-neutral voltages on two phases: $$ v_{L-L} = v_{L-N} \cdot \left( sin(\omega t) - sin(\omega t - \frac{2 \pi}{3} )\right) $$

  • The key here is whether you have a delta or a Y configuration. For 230V line to line it is most likely a delta. Most (in the US at least), 3phase systems that are Y connected are 277/480 meaning 277VoltsRMS line to neutral,. and 480voltsRMS line to line. Since 230 is a delta, does it really make sense to define line to neutral?

    And, most residential is single phase with 230V line to line and 120V line to neutral to neutral of the center-tapped transformer secondary.

    230V L-N 400V L-L is *extremely* common in Europe, Australasia, and most everywhere that's not in north america.

  • Yes, it is correct for instantaneous difference for a 230V RMS system.

    230V RMS is $$ \sqrt2 \cdot 230V$$ line-to-line, and

    $$\sin(\tfrac{2}{3} \pi) = \tfrac{\sqrt3}{2}$$

    so $$ 2 \cdot \left ( \sqrt2 \cdot 230V \cdot \sin(\tfrac{2}{3} \pi) \right) = \sqrt3 \cdot \sqrt2 \cdot 230V$$

    So agrees with the simple phasor method of multiplying the voltage by \$\sqrt3\$ once you've converted RMS to line.

License under CC-BY-SA with attribution

Content dated before 6/26/2020 9:53 AM