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\$.


2 Answers 2


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.

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

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\$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.