91Ó°ÊÓ

• .$V_1=A\text{sin}{\mathrm{Ó\neg }}_{d}t$
• .$V_2=A\text{sin}\left({\mathrm{Ó\neg }}_{d}t-{\text{120}}^{\text{0}}\right)$
• .$V_3=A\text{sin}\left({\mathrm{Ó\neg }}_{d}t-{\text{240}}^{\text{0}}\right)$

The three-phase generator has three ac voltages that are $120°$out of phase with each other. Calculate the potential difference between any two wires by using the trigonometric identity.

The formula is as follows:

$A-\sin B=2\sin \left[\frac{A-B}{2}\right]\cos \left[\frac{A+B}{2}\right]$

Consider different combinations for potential difference:

$$\begin{gathered} n{V}_{12}=V_1-V_2 \\ n{V}_{12}=A\text{sin}{\mathrm{Ó\neg }}_{d}t-A\text{sin}\left({\mathrm{Ó\neg }}_{d}t-{\text{120}}^{\text{0}}\right) \end{gathered}$$

By using the trigonometric identity,

$$\begin{gathered} n{V}_{12}=2A\sin \left(\frac{{\mathrm{Ó\neg }}_{d}t-{\mathrm{Ó\neg }}_{d}t+120°}{2}\right)\cos \left(\frac{{\mathrm{Ó\neg }}_{d}t+{\mathrm{Ó\neg }}_{d}t-120°}{2}\right) \\ n{V}_{12}=2A\sin \left(60°\right)\cos \left({\mathrm{Ó\neg }}_{d}t-60°\right) \\ n{V}_{12}=2A\left(\sqrt{\frac{3}{2}}\right){\mathrm{Ó\neg }}_{d}t-60° \\ n{V}_{12}=\sqrt{3}A\left({\mathrm{Ó\neg }}_{d}t-60°\right) \end{gathered}$$

Now,

$$\begin{gathered} n{V}_{13}=V_1-V_3 \\ n{V}_{13}=A\text{sin}{\mathrm{Ó\neg }}_{d}t-A\text{sin}\left({\mathrm{Ó\neg }}_{d}t-{\text{240}}^{\text{0}}\right) \end{gathered}$$.

By using the trigonometric identity,

role="math" localid="1663157004430" $$\begin{gathered} n{V}_{13}=2A\sin \left(\frac{{\mathrm{Ó\neg }}_{d}t-{\mathrm{Ó\neg }}_{d}t+240°}{2}\right)\cos \left(\frac{{\mathrm{Ó\neg }}_{d}t+{\mathrm{Ó\neg }}_{d}t-240°}{2}\right) \\ n{V}_{13}=2A\sin \left(120°\right)\cos \left({\mathrm{Ó\neg }}_{d}t-120°\right) \\ n{V}_{13}=2A\left(\sqrt{\frac{3}{2}}\right)\cos \left({\mathrm{Ó\neg }}_{d}t-120°\right) \\ n{V}_{13}=\sqrt{3}A\cos \left({\mathrm{Ó\neg }}_{d}t-120°\right) \end{gathered}$$

Similarly,

$$\begin{gathered} n{V}_{23}=V_2-V_3 \\ n{V}_{23}=A\text{sin}\left({\mathrm{Ó\neg }}_{d}t-{\text{120}}^{\text{0}}\right)-A\text{sin}\left({\mathrm{Ó\neg }}_{d}t-{\text{240}}^{\text{0}}\right) \end{gathered}$$

By using the trigonometric identity,

$$\begin{gathered} n{V}_{23}=2A\sin \left(\frac{{\mathrm{Ó\neg }}_{d}t-120°-{\mathrm{Ó\neg }}_{d}t+240°}{2}\right)\cos \left(\frac{{\mathrm{Ó\neg }}_{d}t-120°+{\mathrm{Ó\neg }}_{d}t-240°}{2}\right) \\ n{V}_{23}=2A\sin \left(60°\right)\cos \left({\mathrm{Ó\neg }}_{d}t-180°\right) \\ n{V}_{23}=2A\left(\sqrt{\frac{3}{2}}\right)\cos \left({\mathrm{Ó\neg }}_{d}t-180°\right) \\ n{V}_{23}=\sqrt{3}A\cos \left({\mathrm{Ó\neg }}_{d}t-180°\right) \end{gathered}$$

The expressions $n{V}_{12},n{V}_{13},n{V}_{23}$ are the sinusoidal function of $t$ angular frequency ${\mathrm{Ó\neg }}_d$.

Hence, the potential difference between any two of the wires oscillates sinusoidally with the angular velocity${\mathrm{Ó\neg }}_d$.

$$\begin{gathered} n{V}_{12}=\sqrt{\text{3}}A\left({\mathrm{Ó\neg }}_{d}t-{\text{60}}^{\text{0}}\right) \\ n{V}_{13}=\sqrt{\text{3}}A\text{cos}\left({\mathrm{Ó\neg }}_{d}t-{\text{120}}^{\text{0}}\right) \\ n{V}_{23}=\sqrt{\text{3}}A\text{cos}\left({\mathrm{Ó\neg }}_{d}t-{\text{180}}^{\text{0}}\right) \end{gathered}$$

Thus, all these expressions have an amplitude of $\sqrt{\text{3}}A.$

Hence, the potential difference between any two of the wires has an amplitude of $\sqrt{\text{3}}A.$

Showed that the potential difference between any two of the wires oscillates sinusoidally with the angular velocity ${\mathrm{Ó\neg }}_d,$and it has an amplitude of $\sqrt{\text{3}}A.$