# Y-connected 3-phase voltage source RMS voltage

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\$$.

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:

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:

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.

• Thanks for this graphical method. In addition I have worked out an algebraic method
– 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\$$