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Following are the three-phase voltage signals, va= 100sin(100πt), vb= 100sin(100πt-2π/3), vc= 100sin(100πt+2π/3). These three-phase voltages are transformed into two-phase voltages in the αβ axes (vα,vβ), where the β-axis leads the α-axis by 90 degrees. va is aligned to the α-axis, and the positions of other phase voltages in the αβ plane are shown in the figure below. What is the value of vα at t=0.01 sec

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.

Following are the three-phase voltage signals, va= 100sin(100πt), vb= 100sin(100πt-2π/3), vc= 100sin(100πt+2π/3). These three-phase voltages are transformed into two-phase voltages in the αβ axes (vα,vβ), where the β-axis leads the α-axis by 90 degrees. va is aligned to the α-axis, and the positions of other phase voltages in the αβ plane are shown in the figure below. What is the value of vα at t=0.01 sec?

Explore the transformation of three-phase voltages into two-phase αβ voltages using trigonometric and electrical engineering concepts. Learn how to calculate vα at t=0.01 seconds for three voltage signals, va, vb, and vc, with detailed formulas and step-by-step explanation.

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