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If I have a three-phase Y-connected voltage source, and the voltages are

\$V_{a} = V_{m}cos(\omega t)\$

\$V_{b} = V_{m}cos(\omega t-120^{\circ})\$

\$V_{c} = V_{m}cos(\omega t-240^{\circ})\$

At a certain time, \$V_{b} = V_{c} = 60V\$

What is the RMS of the line voltage?

I understand that the three voltages are separated by a phase angle of \$120^{\circ}\$, and that at a time \$t\$ in the cycle the waveforms of \$V_{b}\$ and \$V_{c}\$ will interesect at 60V.

I know that if I find out \$t\$ then I can plug in \$V_b\$ \$V_c\$ and \$t\$ to solve for \$\omega\$ and \$V_m\$, and from there work out the RMS line voltage. However, I'm at a loss how to to get \$t\$.

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2 Answers 2

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I know that if I find out t then I can plug in Vb Vc and t to solve for ω and Vm, and from there work out the RMS line voltage.

You have already done the 50% of the solution.

Anyway, I'm not going to give you the full solution, but I'll show you the way.

Visualisation helps a lot, as you know. So let's have a look at how the voltages look like in a 3P system:

enter image description here

Image source

Let black curve (Phase1) be a, red curve (Phase 2) be b, and blue curve (Phase 3) be c.

Vb = Vc equality happens at \$\pi/2+N\ \pi\$ where N is an integer. And, at these locations, Va hits its peak (positive or negative, depending on N).


Now let's forget about the phase shifts for a moment and modify the curve pack above as following:

enter image description here

The purple vertical axis is our new zero. And the green numbers are our new angles. So you know what \$\sin(150°)=\sin(5\pi/6)\$ equals to. From there you can obtain the peak of the sine.

Once you have the peak, you can work out the phase and line voltages.

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  • \$\begingroup\$ Thanks for this graphical method. In addition I have worked out an algebraic method \$\endgroup\$
    – Kdwk
    May 12, 2023 at 13:07
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In addition to the graphical method, I have also figured out an algebraic method.

I thought I needed to find \$t\$, but it turns out if I treat \$\omega t\$ as one variable I can solve for \$\omega t\$ and find \$V_m\$ that way.

  1. Since I know \$V_b = V_c\$ at some point, I can just equate the two functions:

\$ V_mcos(\omega t-120^{\circ}) = V_mcos(\omega t-240^{\circ})\$

  1. Doing simplification and using compound angle formula to expand the \$cos\$ functions I get:

\$sin(\omega t) = -sin(\omega t)\$, which means \$sin(\omega t) = 0\$, and \$\omega t\ = 0\ or\ \pi\$.

  1. Plugging \$\omega t = \pi\$ and \$V_b=60V\$, I have:

\$60=V_mcos(180^{\circ}-120^{\circ})\$, from which I get \$V_m = 120V\$.

  1. In a three phase system, \$V_L = \sqrt3\ V_P\$

So peak line voltage = \$207.8V\$

RMS of line voltage = \$207.8/\sqrt2 = 146.9V\$

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