We are tasked with finding the value of $v_\alpha$ at $t = 0.01$ seconds, given the three-phase voltage signals:

$v_a = 100 \sin(100 \pi t), \quad v_b = 100 \sin(100 \pi t - \frac{2\pi}{3}), \quad v_c = 100 \sin(100 \pi t + \frac{2\pi}{3})$

Step 1: Calculate $v_a$, $v_b$, and $v_c$ at $t = 0.01$

First, substitute $t = 0.01$ seconds into each voltage expression.

• $v_a = 100 \sin(100 \pi (0.01)) = 100 \sin(\pi) = 100 \times 0 = 0$
• $v_b = 100 \sin(100 \pi (0.01) - \frac{2\pi}{3}) = 100 \sin(\pi - \frac{2\pi}{3}) = 100 \sin\left(\frac{\pi}{3}\right) = 100 \times \frac{\sqrt{3}}{2} = 86.6$
• $v_c = 100 \sin(100 \pi (0.01) + \frac{2\pi}{3}) = 100 \sin(\pi + \frac{2\pi}{3}) = 100 \sin\left(\frac{4\pi}{3}\right) = 100 \times \left(-\frac{\sqrt{3}}{2}\right) = -86.6$

So:

• $v_a = 0$
• $v_b = 86.6$
• $v_c = -86.6$

Step 2: Transformation from Three-Phase to Two-Phase (αβ) System

To transform the three-phase voltages $v_a$, $v_b$, and $v_c$ into the two-phase $\alpha \beta$ components $v_\alpha$ and $v_\beta$, we use the following equations:

$v_\alpha = \frac{2}{3} \left( v_a - \frac{1}{2} v_b - \frac{1}{2} v_c \right)$ $v_\beta = \frac{2}{3} \left( \frac{\sqrt{3}}{2} (v_b - v_c) \right)$

Step 3: Calculate $v_\alpha$

Substitute the values of $v_a = 0$, $v_b = 86.6$, and $v_c = -86.6$ into the $v_\alpha$ formula:

$v_\alpha = \frac{2}{3} \left( 0 - \frac{1}{2} \times 86.6 - \frac{1}{2} \times (-86.6) \right)$ $v_\alpha = \frac{2}{3} \left( -43.3 + 43.3 \right) = \frac{2}{3} \times 0 = 0$

Thus, the value of $v_\alpha$ at $t = 0.01$ seconds is 0.

Would you like more details on any step or the derivation of the transformation? Here are some additional related questions:

1. How do we calculate $v_\beta$ at $t = 0.01$ seconds?
2. Can you explain the significance of the $\alpha \beta$ transformation in more detail?
3. How do phase shifts affect the $\alpha \beta$ components?
4. What is the physical interpretation of the $v_\alpha = 0$ result?
5. How are the axes aligned in the $\alpha \beta$ plane based on the diagram?

Tip: Always double-check angle conversions and trigonometric values when working with phase-shifted signals, especially in transformations.