Question

In Fig. 31-38, a three-phase generator G produces electrical power that is transmitted by means of three wires. The electric potentials (each relative to a common reference level) are ${\mathit{V}}_{\mathbf{1}}\mathbf{=}\mathit{A}\mathbf{}\mathbf{\text{sin}}\mathbf{}{\mathit{\omega }}_{\mathbf{d}}\mathit{t}$for wire 1, ${\mathit{V}}_{\mathbf{2}}\mathbf{=}\mathit{A}\mathbf{}\mathbf{\text{sin}}\mathbf{}\left({\omega }_{d}t-{\text{120}}^{\text{0}}\right)\mathbf{}$ for wire 2, and ${\mathit{V}}_{\mathbf{3}}\mathbf{=}\mathit{A}\mathbf{}\mathbf{\text{sin}}\mathbf{}\left({\omega }_{d}t-{\text{240}}^{\text{0}}\right)\mathbf{}\mathbf{}$for wire 3. Some types of industrial equipment (for example, motors) have three terminals and are designed to be connected directly to these three wires. To use a more conventional two-terminal device (for example, a lightbulb), one connects it to any two of the three wires. Show that the potential difference between any two of the wires (a) oscillates sinusoidally with angular frequency ${\mathit{\omega }}_{\mathbf{d}}$and (b) has an amplitude of$\mathit{A}\sqrt{\mathbf{\text{3}}}$.

Short answer

a. The potential difference between any two of the wires oscillates sinusoidally with the angular velocity ${\omega }_d$ .

b. The potential difference between any two of the wires has an amplitude of $A\sqrt{3}$.

Step-by-step solution

Step 1: Given Information

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

Determining the concept

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:

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

(a) Determining the potential difference between any two of the wires oscillating sinusoidally with the angular frequency ωd  

Consider different combinations for potential difference:

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

By using the trigonometric identity,

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

Now,

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

By using the trigonometric identity,

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

Similarly,

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

By using the trigonometric identity,

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

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

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

(b) Determining the potential difference between any two of the wires has amplitude of  A3

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

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 ${\omega }_d,$and it has an amplitude of $\sqrt{\text{3}}A.$