Let's solve the problem **step by step**: ## **Step 1: Understanding the Problem** - We have a **single-phase AC voltage controller** with a resistive load (**R**) and two SCRs in inverse-parallel. - We are to find the **firing angle (α)** when the output power is **80%**, **50%**, and **30%** of the **maximum power**. ## **Step 2: Output Power Formula for a Phase-Controlled AC Circuit** For a resistive load, the **average power delivered** to the load is: \[ P = P_{max} \left[ 1 - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \right] \] Where: - \(P\) = Output power - \(P_{max}\) = Maximum power (at α = 0°) - \(\alpha\) = Firing angle (in radians) ## **Step 3: Express the Power Ratio** Let \( \frac{P}{P_{max}} = k \) So, \[ k = 1 - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] We are to find \(\alpha\) for \(k = 0.8, 0.5, 0.3\). --- ## **Step 4: Find \(\alpha\) for Each Power Level** ### **A. For \(k = 0.8\):** \[ 0.8 = 1 - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] Rearrange: \[ 0.8 - 1 = - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] \[ -0.2 = - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] Multiply both sides by -1: \[ 0.2 = \frac{\alpha}{\pi} - \frac{1}{2\pi} \sin(2\alpha) \] Multiply both sides by \(\pi\): \[ 0.2\pi = \alpha - \frac{1}{2} \sin(2\alpha) \] \[ \alpha - \frac{1}{2} \sin(2\alpha) = 0.2\pi \] Now, solve for \(\alpha\): \[ \alpha - 0.5 \sin(2\alpha) = 0.2\pi \] **From the question, \(\alpha = 60.5^\circ\) (convert to radians: \(60.5 \times \frac{\pi}{180}\))** --- ### **B. For \(k = 0.5\):** \[ 0.5 = 1 - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] \[ -0.5 = - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] \[ 0.5 = \frac{\alpha}{\pi} - \frac{1}{2\pi} \sin(2\alpha) \] \[ 0.5\pi = \alpha - 0.5\sin(2\alpha) \] \[ \alpha - 0.5 \sin(2\alpha) = 0.5\pi \] Given: \(\alpha = 90^\circ\) (which is \(\frac{\pi}{2}\) radians). --- ### **C. For \(k = 0.3\):** \[ 0.3 = 1 - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] \[ -0.7 = -\frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] \[ 0.7 = \frac{\alpha}{\pi} - \frac{1}{2\pi}\sin(2\alpha) \] \[ 0.7\pi = \alpha - 0.5 \sin(2\alpha) \] \[ \alpha - 0.5 \sin(2\alpha) = 0.7\pi \] Given: \(\alpha = 108.6^\circ\) (convert to radians: \(108.6 \times \frac{\pi}{180}\)) --- ## **Step 5: Summary Table** | Power Ratio \( \frac{P}{P_{max}} \) | Firing Angle (\(\alpha\)) | |:-----------------------------------:|:------------------------:| | 0.8 | \(60.5^\circ\) | | 0.5 | \(90^\circ\) | | 0.3 | \(108.6^\circ\) | --- ### **Conclusion:** - At **80%** power: **α = 60.5°** - At **50%** power: **α = 90°** - At **30%** power: **α = 108.6°** If you need the detailed calculation for the sine values or how to solve for α numerically, let me know!

Certainly! Let's delve into the detailed calculations to find the firing angles \(\alpha\) corresponding to the given power ratios. We'll use the main power control equation for a resistive load controlled by phase angle: \[ k = 1 - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] where: - \(k = \frac{P}{P_{max}}\), - \(\alpha\) is in radians. --- ### **Step-by-step calculations for each power level** --- ## **1. For \(k = 0.8\)** \[ 0.8 = 1 - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] Rearranged: \[ -0.2 = - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] Multiply through by \(\pi\): \[ -0.2 \pi = - \alpha + \frac{1}{2} \sin(2\alpha) \] Multiply both sides by \(-1\): \[ 0.2 \pi = \alpha - \frac{1}{2} \sin(2\alpha) \] --- ### **Numerical solution:** Since the equation involves both \(\alpha\) and \(\sin(2\alpha)\), we can attempt a numerical approach. **Given:** \[ \alpha - 0.5 \sin(2\alpha) = 0.2 \pi \approx 0.2 \times 3.1416 = 0.6283 \] **Initial estimate:** Let's check \(\alpha = 60^\circ = \pi/3 \approx 1.0472\) radians: \[ LHS = 1.0472 - 0.5 \times \sin(2 \times 1.0472) = 1.0472 - 0.5 \times \sin(2.0944) \] Calculate \(\sin(2.0944)\): \[ \sin(2.0944) \approx 0.8660 \] So, \[ LHS \approx 1.0472 - 0.5 \times 0.8660 = 1.0472 - 0.4330 = 0.6142 \] Compare with RHS = 0.6283. Since 0.6142 < 0.6283, increase \(\alpha\). Try \(\alpha = 65^\circ \approx 1.1345\) radians: \[ LHS = 1.1345 - 0.5 \times \sin(2 \times 1.1345) = 1.1345 - 0.5 \times \sin(2.269) \] \(\sin(2.269) \approx 0.769\) \[ LHS \approx 1.1345 - 0.5 \times 0.769 = 1.1345 - 0.3845 = 0.75 \] Now, 0.75 > 0.6283, so \(\alpha\) is between 60° and 65°. Try \(\alpha = 62.5^\circ \approx 1.0908\) radians: \[ LHS = 1.0908 - 0.5 \times \sin(2.1816) \] \(\sin(2.1816) \approx 0.823\) \[ LHS \approx 1.0908 - 0.5 \times 0.823 = 1.0908 - 0.4115 = 0.6793 \] Close to 0.6283 but slightly higher. Try \(\alpha = 60^\circ \approx 1.0472\) radians (already calculated): \[ LHS \approx 0.6142 \] Between 60° and 62.5°, the value crosses 0.6283. Interpolating: At 60°, LHS ≈ 0.6142; at 62.5°, LHS ≈ 0.6793. Desired LHS: 0.6283. Approximate \(\alpha\): \[ \alpha \approx 60^\circ + \left( \frac{0.6283 - 0.6142}{0.6793 - 0.6142} \right) \times (62.5^\circ - 60^\circ) \] \[ \approx 60^\circ + \left( \frac{0.0141}{0.0651} \right) \times 2.5^\circ \approx 60^\circ + 0.216 \times 2.5^\circ \approx 60^\circ + 0.54^\circ \approx 60.5^\circ \] **Result:** \[ \boxed{ \alpha \approx 60.5^\circ } \] which matches the given answer. --- ## **2. For \(k=0.5\)** \[ 0.5 = 1 - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] Rearranged: \[ -0.5 = - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] Multiply through by \(\pi\): \[ -0.5 \pi = - \alpha + \frac{1}{2} \sin(2\alpha) \] Multiply both sides by \(-1\): \[ 0.5 \pi = \alpha - \frac{1}{2} \sin(2\alpha) \] Calculate RHS for \(\alpha = 90^\circ = \pi/2 \approx 1.5708\): \[ LHS = 1.5708 - 0.5 \times \sin(2 \times 1.5708) = 1.5708 - 0.5 \times \sin(3.1416) \] \(\sin(3.1416) \approx 0\) \[ LHS \approx 1.5708 - 0 = 1.5708 \] which equals \(0.5 \pi \approx 1.5708\). Perfect match. **Result:** \[ \boxed{ \alpha = 90^\circ } \] which matches the given answer. --- ## **3. For \(k=0.3\)** \[ 0.3 = 1 - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] Rearranged: \[ -0.7 = - \frac{\alpha}{\pi} + \frac{1}{2\pi} \sin(2\alpha) \] Multiply through by \(\pi\): \[ -0.7 \pi = - \alpha + \frac{1}{2} \sin(2\alpha) \] Multiply both sides by \(-1\): \[ 0.7 \pi = \alpha - \frac{1}{2} \sin(2\alpha) \] Evaluate at \(\alpha \approx 108.6^\circ\): Convert \(108.6^\circ\) to radians: \[ \alpha = 108.6^\circ \times \frac{\pi}{180} \approx 1.896 \text{ radians} \] Calculate RHS: \[ \text{RHS} = 1.896 - 0.5 \times \sin(2 \times 1.896) = 1.896 - 0.5 \times \sin(3.792) \] \(\sin(3.792) \approx -0.588\) Thus: \[ \text{RHS} \approx 1.896 - 0.5 \times (-0.588) = 1.896 + 0.294 = 2.19 \] Calculate \(0.7 \pi \approx 0.7 \times 3.1416 \approx 2.199\). Close enough to 2.19, confirming the approximate \(\alpha\). --- ## **Final notes:** - The approximate angles are consistent with the provided values. - For precise control, numerical methods or graphing calculators are used to solve such transcendental equations accurately. --- **Summary:** | Power Ratio \( \frac{P}{P_{max}} \) | Firing Angle \(\alpha\) | Degrees | |:-----------------------------------:|:------------------------:|:--------:| | 0.8 | ~0.628 radians | ~60.5° | | 0.5 | \(\pi/2\) radians | 90° | | 0.3 | ~1.896 radians | ~108.6°| --- If you want, I can help you set up a calculator or software code to solve these equations numerically!