Let's break down the problem step by step: --- ### **Given** - Three-phase motor with **balanced phase currents** A, B, and C. - **Magnitude of each phase current:** \( I_A = I_B = I_C = 10\,A \) - **Parks (dq0) Transformation** to find **q-axis** quantity. - **d-axis leads the q-axis by 90° electrical**. - **Angle between q-axis and a-axis = 240°**. - **Scaling factor for q-axis = \( \frac{2}{3} \)**. --- ### **Step 1: Understand the dq0 (Parks) Transformation** The dq0 transformation (in the stationary reference frame) is usually: \[ \begin{bmatrix} I_d \\ I_q \\ I_0 \end{bmatrix} = \frac{2}{3} \begin{bmatrix} \cos\theta & \cos(\theta - 120^\circ) & \cos(\theta + 120^\circ) \\ -\sin\theta & -\sin(\theta - 120^\circ) & -\sin(\theta + 120^\circ) \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix} \begin{bmatrix} I_A \\ I_B \\ I_C \end{bmatrix} \] - \( \theta \) is **the angle between the a-axis and the d-axis**. But, **angle between q and a-axis is given as 240°**, which means: - The d-axis leads the q-axis by 90°, so if q-axis is at 240°, d-axis is at \( 240^\circ + 90^\circ = 330^\circ \) relative to the a-axis. But the standard dq0 transformation is referenced from the d-axis. So, for the **q-axis**, the angle from the a-axis is 240°. --- ### **Step 2: Substitute the values into the q-axis transformation** The **q-axis current** is: \[ I_q = \text{scaling} \times \left[ -\sin(\theta)I_A - \sin(\theta - 120^\circ)I_B - \sin(\theta + 120^\circ)I_C \right] \] - Here, \( \theta = 240^\circ \) (angle between q-axis and a-axis) - Scaling factor is \( \frac{2}{3} \) - \( I_A = I_B = I_C = 10\,A \) --- ### **Step 3: Calculate the individual sine terms** \[ \sin(240^\circ) = -\frac{\sqrt{3}}{2} \] \[ \sin(240^\circ - 120^\circ) = \sin(120^\circ) = \frac{\sqrt{3}}{2} \] \[ \sin(240^\circ + 120^\circ) = \sin(360^\circ) = 0 \] --- ### **Step 4: Plug into the formula** \[ I_q = \frac{2}{3} \left[ -\sin(240^\circ)I_A - \sin(120^\circ)I_B - \sin(360^\circ)I_C \right] \] \[ = \frac{2}{3} \left[ -(-\frac{\sqrt{3}}{2})\cdot 10 - (\frac{\sqrt{3}}{2})\cdot 10 - 0 \cdot 10 \right] \] \[ = \frac{2}{3} \left[ (\frac{\sqrt{3}}{2}) \cdot 10 - (\frac{\sqrt{3}}{2}) \cdot 10 \right] \] \[ = \frac{2}{3} \left[ 0 \right] = 0 \] --- ### **Step 5: Final Answer** \[ \boxed{I_q = 0} \] --- ## **Summary Table** | Step | Calculation | Value | |------|-------------|-------| | \( \sin(240^\circ) \) | | \( -\frac{\sqrt{3}}{2} \) | | \( \sin(120^\circ) \) | | \( \frac{\sqrt{3}}{2} \) | | \( \sin(360^\circ) \) | | \( 0 \) | | Scaling factor | | \( \frac{2}{3} \) | | \( I_A, I_B, I_C \) | | 10 A | --- ## **Conclusion** **The q-axis current \( I_q \) using Parks transformation for the given balanced 10A currents and the specified angle is:** \[ \boxed{0\ \text{A}} \] --- If you need the d-axis value or want the derivation for arbitrary angles, let me know!

Let's carefully analyze and solve this problem step by step. --- ### **Given Data:** - Balanced phase currents: \(I_A = I_B = I_C = 10\,A\) - Scaling factor for q-axis: \(\frac{2}{3}\) - The angle between **q-axis** and **a-axis**: \(240^\circ\) - The **d-axis leads the q-axis by 90°**: meaning the d-axis is at \(240^\circ + 90^\circ = 330^\circ\) relative to the a-axis. - The **angle between the q-axis and the a-axis** is \(240^\circ\). - The **angle between d-axis and q-axis**: \(90^\circ\). --- ### **Step 1: Understanding the coordinate system** - The **a-axis** is our reference (0°). - The **q-axis** is at \(240^\circ\) from the a-axis. - The **d-axis** is at \(330^\circ\) from the a-axis (since it leads q by 90°). --- ### **Step 2: Recall the q-axis current formula from Parks transformation** The **q-axis component** of the current is given by: \[ I_q = \frac{2}{3} \left[ -\sin \theta \cdot I_A - \sin(\theta - 120^\circ) \cdot I_B - \sin(\theta + 120^\circ) \cdot I_C \right] \] where: - \(\theta\) is the **angle of the q-axis** relative to the a-axis. In this problem: \[ \boxed{\theta = 240^\circ} \] --- ### **Step 3: Calculate the sine terms** \[ \sin(240^\circ) = -\frac{\sqrt{3}}{2} \] \[ \sin(240^\circ - 120^\circ) = \sin(120^\circ) = \frac{\sqrt{3}}{2} \] \[ \sin(240^\circ + 120^\circ) = \sin(360^\circ) = 0 \] --- ### **Step 4: Substitute into the formula** \[ I_q = \frac{2}{3} \left[ -\left(-\frac{\sqrt{3}}{2}\right) \times 10 - \frac{\sqrt{3}}{2} \times 10 - 0 \times 10 \right] \] Simplify each term: \[ I_q = \frac{2}{3} \left[ \frac{\sqrt{3}}{2} \times 10 - \frac{\sqrt{3}}{2} \times 10 \right] \] \[ I_q = \frac{2}{3} \times \left( 5 \sqrt{3} - 5 \sqrt{3} \right) = \frac{2}{3} \times 0 = 0 \] --- ### **Final answer:** \[ \boxed{ \boxed{I_q = 0\,A} } \] --- ### **Summary:** - The q-axis component of the current is **zero** for the given balanced currents and the specified angle. - The symmetry of the sinusoidal terms caused the positive and negative contributions to cancel out. --- ### **Note:** - The **d-axis current** can be computed similarly, but it's not asked here. - The key insight is that at \(\theta = 240^\circ\), due to symmetry, the q-axis component becomes zero in this specific case. --- **If you'd like, I can also help compute the d-axis component or analyze other configurations!**