Vector Analysis (Phasor-diagrams) of poly-phased circuits

Question

I have a poly-phased circuit of $q$ phase ($q$ input voltage in equilibrium) such that $$1\le i \le q, \quad V_i= V_{max}\sin\left(\omega t - (i-1)\cfrac {2\pi}{q}\right) $$

How can I use vector analysis to calculate $max(V_i-V_j)$ such that $i\ne j$ and $1\le i,j \le q$?

The method I used was calculus optimization which is, ironically non-optimal. My professor used phasor-diagram, vector analysis in a very fast and optimal way to calculate them but I couldn't get it. Any help will be appreciated.

Answer 1

Taking $i,j$ such that $$1\le i,j \le q, \quad V_i= V_{max}\sin\left(\omega t - (i-1)\cfrac {2\pi}{q}\right)\\ \quad\quad\quad\quad\quad\quad V_j= V_{max}\sin\left(\omega t - (j-1)\cfrac {2\pi}{q}\right)$$ We can write $V_i-V_j=V_{max}\left[\sin\left(\omega t - (i-1)\cfrac {2\pi}{q}\right)-\sin\left(\omega t - (j-1)\cfrac {2\pi}{q}\right)\right]$ in the form $2V_{max}\cos \theta\sin\varphi$ where $$\begin{align}&\theta+\varphi=\omega t - (i-1)\cfrac {2\pi}{q}\\&\theta-\varphi=\omega t - (j-1)\cfrac {2\pi}{q} \end{align}$$ Solving these equations yields $$\begin{align}&\theta=\omega t-(i+j-2)\cfrac{\pi}q\\&\varphi=(j-i)\cfrac{\pi}q\end{align} $$

Thus we have $$V_i-V_j=2V_{max}\cos \left(\omega t-(i+j-2)\cfrac{\pi}q\right)\sin\left((j-i)\cfrac{\pi}q\right)$$

It is clear to see that $$max(V_i-V_j)=2V_{max}\sin\left((j-i)\cfrac{\pi}q\right)$$

Answer 2

As the sine varies between $\pm 1$ the maximum will come when one is $-1$ and the other $+1$. This can only happen when the two waves are out of phase by $\pi$. So let $i=1, j=\frac q2 +1$ and the two waves will be opposite. Then choose $\omega t$ to be $-\frac \pi 2$ and you are there. If $i,j$ must be integers and $q$ isn't even, you can't quite have the two waves out of phase by $\pi$, but you can come close with $j=\frac {q\pm 1}2$