Let's solve the problem step by step. --- ## **Q4: Firing Angle(s) of the SCRs** **Given:** - Grid voltage: \( V_{rms} = 220 \) V - Frequency: \( f = 50 \) Hz - Load power \( P = 2000 \) W - Load: Series R-L - \( R = 19.12\ \Omega \) (since \( 1912/100 = 19.12 \)) - \( \frac{R}{L} = 637.333 \ \left(\frac{637333}{1000}\right) \) - Two firing angles required (for firing at different periods in the cycle) --- ### **Step 1: Find L** \[ \frac{R}{L} = 637.333 \Rightarrow L = \frac{R}{637.333} \] \[ L = \frac{19.12}{637.333} = 0.03\ \text{H} \] --- ### **Step 2: Load Impedance** \[ \omega = 2\pi f = 2\pi \times 50 = 314.16\ \text{rad/s} \] \[ Z = \sqrt{R^2 + (\omega L)^2} \] \[ \omega L = 314.16 \times 0.03 = 9.4248\ \Omega \] \[ Z = \sqrt{19.12^2 + 9.4248^2} = \sqrt{365.4144 + 88.832} = \sqrt{454.2464} = 21.32\ \Omega \] --- ### **Step 3: Find Load Current** \[ I_{rms} = \frac{P}{V_{rms} \cdot \text{pf}} \] But we don't know power factor yet. Alternatively, for full conduction (no phase control): \[ I_{rms} = \frac{V_{rms}}{Z} = \frac{220}{21.32} = 10.32\ \text{A} \] But since the **load absorbs only 2000 W**, let's verify: \[ P = V_{rms} \cdot I_{rms} \cdot \cos\phi \] \[ I_{rms} = \frac{P}{V_{rms} \cdot \cos\phi} \] \[ \cos\phi = \frac{R}{Z} = \frac{19.12}{21.32} = 0.897 \] \[ I_{rms} = \frac{2000}{220 \times 0.897} = \frac{2000}{197.34} = 10.14\ \text{A} \] This is consistent with our calculation above. --- ### **Step 4: Output Voltage for 2000 W** \[ P = \frac{V_{o,rms}^2}{Z} \] \[ V_{o,rms} = \sqrt{P \times Z} = \sqrt{2000 \times 21.32} = \sqrt{42640} = 206.5\ \text{V} \] --- ### **Step 5: Firing Angle (for AC/AC Controller with Phase Control)** For a resistive load: \[ V_{o,rms} = V_{m} \sqrt{ \frac{1}{2\pi} \int_{\alpha}^{\pi} \sin^2 \omega t \, d(\omega t) } \] But for an R-L load, the calculation is more complex due to the phase shift and current continuity. For simplicity, let's use the *approximate* formula (for an R load): \[ V_{o,rms} = V_{rms} \cdot \sqrt{1 - \frac{\alpha}{\pi} + \frac{\sin 2\alpha}{2\pi}} \] But our load is R-L, so this is an approximation. Let's proceed: \[ 206.5 = 220 \cdot \sqrt{1 - \frac{\alpha}{\pi} + \frac{\sin 2\alpha}{2\pi}} \] \[ \frac{206.5}{220} = \sqrt{1 - \frac{\alpha}{\pi} + \frac{\sin 2\alpha}{2\pi}} \] \[ 0.939 = \sqrt{1 - \frac{\alpha}{\pi} + \frac{\sin 2\alpha}{2\pi}} \] \[ 0.882 = 1 - \frac{\alpha}{\pi} + \frac{\sin 2\alpha}{2\pi} \] \[ -0.118 = - \frac{\alpha}{\pi} + \frac{\sin 2\alpha}{2\pi} \] \[ \frac{\alpha}{\pi} - \frac{\sin 2\alpha}{2\pi} = 0.118 \] \[ \alpha - \frac{\sin 2\alpha}{2} = 0.118 \pi \] \[ 0.118 \pi = 0.370 \text{ radians} \] So, \[ \alpha - \frac{\sin 2\alpha}{2} = 0.370 \] Let’s solve this numerically for \( \alpha \). --- ### **Step 6: Solve for \( \alpha \) (First Quadrant)** Let’s try values for \( \alpha \): #### Try \( \alpha = 0.4 \): \[ \sin(0.8) = 0.717 \] \[ 0.4 - 0.717/2 = 0.0415 \] #### Try \( \alpha = 0.5 \): \[ \sin(1.0) = 0.841 \] \[ 0.5 - 0.841/2 = 0.0795 \] #### Try \( \alpha = 0.6 \): \[ \sin(1.2) = 0.932 \] \[ 0.6 - 0.932/2 = 0.134 \] Now, 0.5 gives 0.08, 0.6 gives 0.134. Target is 0.37. #### Try \( \alpha = 1.0 \): \[ \sin(2.0) = 0.909 \] \[ 1.0 - 0.909/2 = 1.0 - 0.454 = 0.546 \] So, between 0.6 and 1.0. Now, interpolate: From 0.6 (0.134) to 1.0 (0.546): \[ \frac{0.37 - 0.134}{0.546 - 0.134} = \frac{x - 0.6}{1.0-0.6} \] \[ \frac{0.236}{0.412} = \frac{x-0.6}{0.4} \] \[ 0.573 = \frac{x-0.6}{0.4} \] \[ x-0.6 = 0.229 \] \[ x = 0.829 \] So, \( \alpha \approx 0.83 \) radians. Convert to degrees: \[ 0.83 \times \frac{180}{\pi} = 47.6^\circ \] --- ### **Step 7: Second Firing Angle (Symmetry)** For a full-wave controller, the other firing angle is: \[ \alpha_2 = \pi + \alpha = 180^\circ + 47.6^\circ = 227.6^\circ \] But usually, the firing angles are symmetric in the positive and negative half cycles, i.e., \[ \alpha_1 = \alpha = 47.6^\circ \] \[ \alpha_2 = 180^\circ - \alpha = 132.4^\circ \] **But** the question asks for two firing angles. In a full-wave controller, you fire at \( \alpha \) in the positive half and at \( \pi + \alpha \) in the negative half. So, \[ \boxed{47.6^\circ\ \text{and}\ 227.6^\circ} \] But if they want both firing points in the first half cycle: \[ \alpha_1 = 47.6^\circ \] \[ \alpha_2 = 132.4^\circ \] But in an AC/AC controller (bidirectional), the usual answer is: \[ \boxed{47.6^\circ \text{ and } 227.6^\circ} \] --- ## **Q5: Power Factor** Power factor is: \[ \text{Power factor} = \frac{P}{V_{rms} \cdot I_{rms}} \] Recall: \[ P = 2000\ \text{W} \] \[ V_{rms} = 220\ \text{V} \] \[ I_{rms} = \frac{P}{V_{rms} \cdot \cos\phi} \] But we've previously calculated: \[ I_{rms} = 10.14\ \text{A} \] \[ \text{Power factor} = \frac{2000}{220 \times 10.14} = \frac{2000}{2230.8} = 0.897 \] But this is the displacement power factor for the load; for a phase-controlled AC/AC controller, the power factor is further reduced due to the firing angle. The **input power factor** for a phase-controlled circuit is: \[ \text{pf} = \frac{\text{Real Power}}{\text{Apparent Power}} = \frac{P}{V_{rms} I_{rms}} \] First, calculate \( I_{rms} \) for the controlled voltage (\( V_{o,rms} \)): \[ I_{rms} = \frac{V_{o,rms}}{Z} = \frac{206.5}{21.32} = 9.69\ \text{A} \] \[ \text{pf} = \frac{2000}{220 \times 9.69} = \frac{2000}{2131.8} = 0.939 \] But this is an approximation. For the actual circuit, the power factor is: \[ \text{pf} = \text{Load Power Factor} \times \text{Control Factor} \] But since the controller reduces the conduction period, the actual power factor is slightly lower; but with the calculation above, the **input power factor** is: \[ \boxed{0.939} \] --- ## **Summary of Answers** ### **Q4:** **Firing angles:** \[ \boxed{47.6^\circ\ \text{and}\ 227.6^\circ} \] ### **Q5:** **Power factor:** \[ \boxed{0.939} \]