Let's solve the problem step by step as you requested: --- ### **STEP 1: Introduction to the Concept** This problem deals with the charging of a battery using a **3-phase semi-converter** from a 3-phase AC supply. The semi-converter is a type of controlled rectifier that allows us to control the average output voltage by adjusting the firing angle (α). The main concepts involved are: - 3-phase controlled rectifiers (semi-converter) - Firing angle and its relation to output voltage - Power factor in AC-DC conversion --- ### **STEP 2: Given Data** Let's summarize the given data: - 3-phase supply voltage, \( V_{ph} = 230\,V \) (assuming line-to-neutral) - Frequency, \( f = 50\,Hz \) - Battery emf, \( E = 190\,V \) - Battery internal resistance, \( r = 0.5\,\Omega \) - Charging current (ripple-free), \( I_{dc} = 20\,A \) --- ### **STEP 3: Output Voltage of 3-Phase Semi-Converter** The **average output voltage** of a 3-phase semi-converter is given by: \[ V_{O_{avg}} = \frac{3\sqrt{6}}{\pi} V_{rms} \cos \alpha \] where \( V_{rms} = 230\,V \) (since 230V is line-to-neutral voltage). \[ V_{O_{avg}} = \frac{3\sqrt{6}}{\pi} \times 230 \times \cos \alpha \] \[ V_{O_{avg}} = 1.17 \times V_{m_{ph}} \cos \alpha \] where \( V_{m_{ph}} = \sqrt{2} \times V_{ph} = \sqrt{2} \times 230 = 325.27\,V \) So, \[ V_{O_{avg}} = 1.17 \times 325.27 \times \cos \alpha = 380.56 \cos \alpha \] --- ### **STEP 4: Find Firing Angle α** The output voltage must overcome the battery emf and the voltage drop across the internal resistance: \[ V_{O_{avg}} = E + I_{dc} \times r \] \[ 380.56 \cos \alpha = 190 + 20 \times 0.5 \] \[ 380.56 \cos \alpha = 190 + 10 = 200 \] \[ \cos \alpha = \frac{200}{380.56} = 0.5256 \] \[ \alpha = \cos^{-1}(0.5256) \approx 58.3^\circ \] --- ### **STEP 5: Supply Power Factor Calculation** The **supply power factor** for a fully controlled converter is \( \cos \alpha \), but for a semi-converter, it is slightly higher due to the freewheeling action. For simplicity, we use the fundamental displacement power factor: \[ \text{Power Factor} = \cos \alpha = 0.5256 \] --- ### **STEP 6: Final Short Answer** - **Firing angle delay (α):** \( 58.3^\circ \) - **Supply power factor:** \( 0.53 \) (rounded to two decimal places) --- ### **STEP 7: What We Did in Each Step** 1. **Introduced the concept** of 3-phase semi-converter and the problem context. 2. **Listed all given data** from the question. 3. **Set up the equation** for average output voltage of the converter. 4. **Calculated the firing angle (α)** by equating required output voltage to battery emf plus voltage drop. 5. **Found the supply power factor** using the relationship with firing angle. 6. **Summarized the final answers** briefly. 7. **Explained the approach** taken in each step for clarity. --- **In summary:** We calculated the firing angle needed to charge the battery at 20A to be approximately \( 58.3^\circ \), and the supply power factor is about 0.53. Each step involved identifying and applying relevant circuit and converter equations to solve for the required values.