Let's break down the two questions step by step. --- ## **Question 4: Find the firing angle(s) of the SCRs** ### **Given:** - Grid voltage: \( V_{rms} = 220\,V \) at \( f = 50\,Hz \) - Load: \( P = 2000\,W \) - Load is series R-L: \( R = 19.12\,\Omega \) - \( \frac{R}{L} = 637.333\,s^{-1} \) - AC/AC controller (assume single-phase full-wave phase control) - Find **two firing angles** (\( \alpha_1 \) and \( \alpha_2 \)) that provide 2000 W to the load. --- ### **Step 1: Find \( L \) from \( R/L \) ratio** \[ \frac{R}{L} = 637.333 \implies L = \frac{R}{637.333} = \frac{19.12}{637.333} = 0.03\,H \] --- ### **Step 2: Find the Peak and Average Load Voltage** \[ V_{rms} = 220\,V \implies V_{m} = V_{peak} = \sqrt{2} \times 220 = 311.13\,V \] --- ### **Step 3: Find the Power for Given Load** \[ P = 2000\,W \] --- ### **Step 4: For AC phase controller with RL load, average power:** For a **resistive load** (R only): \[ P = V_{rms}^2 \frac{1}{R} \cdot \frac{1}{2\pi} \int_{\alpha}^{\pi} \sin^2(\omega t) d(\omega t) \] But for R-L load, the **expression is more complex** due to phase lag and continuous current. But **since only R and L are given, and the power is specified, let's proceed:** --- #### **Calculate Load Current for 2000 W** \[ I_{rms, load} = \frac{P}{V_{rms, load}} \] But \( V_{rms, load} \) is **not** 220 V because the output is phase-controlled. Let’s denote \( \theta = \arctan \left( \frac{\omega L}{R} \right) \). \[ \omega = 2\pi f = 314.16\, rad/s \] \[ \omega L = 314.16 \times 0.03 = 9.425\,\Omega \] \[ \theta = \arctan \left( \frac{9.425}{19.12} \right) = \arctan(0.493) = 26.2^\circ \] --- #### **Total impedance** \[ Z = \sqrt{R^2 + (\omega L)^2} = \sqrt{19.12^2 + 9.425^2} = \sqrt{366.58 + 88.84} = \sqrt{455.42} = 21.35\,\Omega \] **If there was no phase control:** \[ I_{rms, max} = \frac{220}{21.35} = 10.3\,A \] \[ P_{max} = I_{rms}^2 \cdot R = (10.3)^2 \times 19.12 = 106.09 \times 19.12 = 2028\,W \] So, **full conduction** gives about 2028 W (close to 2000 W). But with phase control, you get less power as \( \alpha \) increases. --- #### **General Output Power Formula for Phase Angle Control** For RL load (with continuous conduction, which is likely at this R/L): \[ P = \frac{V_{rms}^2}{Z^2} \cdot R \cdot \left[ 1 - \frac{\sin 2\alpha + m \sin 2(\alpha - \theta) }{2\pi} \right] \] But this is an approximation. For simplicity, let's use the **resistive case formula** to estimate \( \alpha \), since R is much larger than \(\omega L\): \[ P = \frac{V_{rms}^2}{R} \cdot \frac{1}{2\pi} (\pi - \alpha + 0.5 \sin 2\alpha) \] Given that \( P_{max} \approx 2028\,W \) at \( \alpha = 0^\circ \), let's use this to estimate the angles for 2000 W: \[ \frac{P}{P_{max}} = \frac{(\pi - \alpha + 0.5 \sin 2\alpha)}{\pi} \] \[ \frac{2000}{2028} = \frac{\pi - \alpha + 0.5 \sin 2\alpha}{\pi} \] \[ 0.9878 = 1 - \frac{\alpha}{\pi} + \frac{0.5 \sin 2\alpha}{\pi} \] \[ \frac{\alpha}{\pi} - \frac{0.5 \sin 2\alpha}{\pi} = 0.0122 \] \[ \alpha - 0.5 \sin 2\alpha = 0.0122 \pi \] \[ \alpha - 0.5 \sin 2\alpha = 0.03834 \, \text{(in radians)} \] --- ### **Step 5: Solve for \( \alpha \)** Let’s try values close to zero, and close to pi (as there are two solutions). #### **First Solution: \( \alpha \approx 0 \)** Try \( \alpha = 0.04 \) rad: \[ \sin(2 \times 0.04) = \sin(0.08) \approx 0.0799 \] \[ 0.04 - 0.5 \times 0.0799 = 0.04 - 0.03995 = 0.00005 \] That’s too small. Try \( \alpha = 0.05 \) rad: \[ \sin(0.1) = 0.0998 \] \[ 0.05 - 0.5 \times 0.0998 = 0.05 - 0.0499 = 0.0001 \] Try \( \alpha = 0.04 \) rad: \[ 0.04 - 0.03995 = 0.00005 \] Try \( \alpha = 0.04 \) rad, but the right side is 0.03834, so let's increase \( \alpha \). Try \( \alpha = 0.04 + 0.038 = 0.078 \) rad: \[ \sin(2 \times 0.078) = \sin(0.156) = 0.155 \] \[ 0.078 - 0.5 \times 0.155 = 0.078 - 0.0775 = 0.0005 \] Still too small. Try \( \alpha = 0.1 \) rad: \[ \sin(0.2) = 0.1987 \] \[ 0.1 - 0.09935 = 0.00065 \] Still too small. Try \( \alpha = 0.04 + 0.03834 = 0.078 \) rad. Let's try larger values. Try \( \alpha = 0.04 + 0.03834 \times 10 = 0.423 \) rad: \[ \sin(0.846) = 0.749 \] \[ 0.423 - 0.5 \times 0.749 = 0.423 - 0.3745 = 0.0485 \] Try \( \alpha = 0.1 \) rad: 0.1 - 0.0993 = 0.0007 Try \( \alpha = 0.2 \) rad: \[ \sin(0.4) = 0.3894 \] \[ 0.2 - 0.5 \times 0.3894 = 0.2 - 0.1947 = 0.0053 \] Try \( \alpha = 0.3 \) rad: \[ \sin(0.6) = 0.5646 \] \[ 0.3 - 0.2823 = 0.0177 \] Try \( \alpha = 0.4 \) rad: \[ \sin(0.8) = 0.7174 \] \[ 0.4 - 0.3587 = 0.0413 \] Our target is 0.03834. So at \( \alpha = 0.4 \) rad, result is 0.0413, at \( \alpha = 0.3 \) rad, result is 0.0177. Let's interpolate: \[ \alpha = 0.3 + (0.03834 - 0.0177) \times \frac{0.4 - 0.3}{0.0413 - 0.0177} \] \[ = 0.3 + (0.02064) \times \frac{0.1}{0.0236} = 0.3 + (0.02064 / 0.0236) \times 0.1 \] \[ = 0.3 + 0.875 \times 0.1 = 0.3 + 0.0875 = 0.3875\, \text{rad} \] So, **first solution**: \[ \alpha_1 \approx 0.39\,rad \] \[ \alpha_1 \approx 0.39 \times \frac{180}{\pi} = 22.4^\circ \] --- #### **Second Solution: \( \alpha \) close to \( \pi \) (\( 3.14 \) rad)** Try \( \alpha = 3.1 \) rad: \[ \sin(6.2) = \sin(6.2 - 2\pi) = \sin(-0.0832) = -0.0831 \] \[ 3.1 - 0.5 \times (-0.0831) = 3.1 + 0.0416 = 3.1416 \] Try \( \alpha = 3.14 \) rad: \[ \sin(2 \times 3.14) = \sin(6.28) = 0 \] \[ 3.14 - 0 = 3.14 \] But our target is \( 0.03834 \) less than \( \pi \), that is \( 3.14 - 0.03834 = 3.1017 \) rad. Set \( \alpha - 0.5 \sin 2\alpha = 0.03834 \). Wait, for second solution, the equation is: \[ \alpha - 0.5 \sin 2\alpha = 0.03834 \] But near \( \pi \), \( \alpha \) is big, so let’s try: Let’s try \( \alpha = 3 \) rad: \[ \sin(6) = -0.2794 \] \[ 3 - 0.5 \times (-0.2794) = 3 + 0.1397 = 3.1397 \] Try \( \alpha = 3.1 \) rad: \[ \sin(6.2) = -0.0831 \] \[ 3.1 + 0.0416 = 3.1416 \] But our equation wants \( \alpha - 0.5 \sin 2\alpha = 0.03834 \), but as \( \alpha \to \pi \), the values approach \( \pi \). Wait, let's look for the value where \( \alpha \) is just before \( \pi \): \[ \alpha = \pi - x \] Try \( x = 0.03834 \): \[ \alpha = 3.1416 - 0.03834 = 3.1033 \] \[ \sin(2 \times 3.1033) = \sin(6.2066) = \sin(6.2066 - 2\pi) = \sin(-0.0766) = -0.0765 \] \[ 3.1033 - 0.5 \times (-0.0765) = 3.1033 + 0.03825 = 3.1415 \] But we are using the form \( \alpha - 0.5 \sin 2\alpha = 0.03834 \). Wait, that's not matching. Actually, as per the original equation: \[ \alpha - 0.5 \sin 2\alpha = 0.03834 \implies \alpha \approx 0.39 \quad \text{(first solution)} \] But also, for \( \alpha \) near \( \pi \): At \( \alpha = \pi \): \[ \sin(2\pi) = 0 \] \[ \pi - 0 = 3.1416 \] At \( \alpha = \pi - 0.04 \): \[ \