Let's break this problem down step by step:

Given:

• Line-to-line RMS voltage: $V_{LL} = 400 \, \text{V}$
• Frequency: $f = 50 \, \text{Hz}$
• Load power: $P_{\text{load}} = 1000 \, \text{W}$
• Ripple voltage requirement: $\Delta V_{\text{ripple}} = 15\% \times V_{\text{peak, LL}}$

Step 1: Calculate the peak line-to-line voltage

The line-to-line RMS voltage ($V_{LL}$) is related to the peak line-to-line voltage ($V_{\text{peak, LL}}$) by the following formula: $V_{\text{peak, LL}} = \sqrt{2} \times V_{LL}$ $V_{\text{peak, LL}} = \sqrt{2} \times 400 \, \text{V} \approx 565.7 \, \text{V}$

Step 2: Calculate the output DC voltage

For a three-phase full-bridge rectifier, the average DC output voltage $V_{\text{DC}}$ is approximately: $V_{\text{DC}} = \frac{3\sqrt{2}}{\pi} \times V_{\text{LL}}$ Substituting $V_{\text{LL}} = 400 \, \text{V}$: $V_{\text{DC}} \approx \frac{3\sqrt{2}}{\pi} \times 400 \, \text{V} \approx 548.7 \, \text{V}$

Step 3: Calculate the load current

From the power equation $P = V_{\text{DC}} \times I_{\text{DC}}$, we can solve for the load current: $I_{\text{DC}} = \frac{P_{\text{load}}}{V_{\text{DC}}}$ Substitute $P_{\text{load}} = 1000 \, \text{W}$ and $V_{\text{DC}} \approx 548.7 \, \text{V}$: $I_{\text{DC}} = \frac{1000}{548.7} \approx 1.82 \, \text{A}$

Step 4: Calculate the required capacitance

The ripple voltage $\Delta V_{\text{ripple}}$ is given by the formula: $\Delta V_{\text{ripple}} = \frac{I_{\text{DC}}}{f \cdot C}$ Where:

• $I_{\text{DC}}$ is the DC current,
• $f$ is the frequency of the ripple, which is $2f = 100 \, \text{Hz}$ (for a three-phase rectifier),
• $C$ is the capacitance.

Rearrange to solve for $C$: $C = \frac{I_{\text{DC}}}{\Delta V_{\text{ripple}} \cdot f}$

We are given that the ripple voltage is 15% of the peak line-to-line voltage: $\Delta V_{\text{ripple}} = 0.15 \times V_{\text{peak, LL}} = 0.15 \times 565.7 \, \text{V} \approx 84.86 \, \text{V}$

Substitute the values: $C = \frac{1.82}{84.86 \times 100} \approx \frac{1.82}{8486} \approx 2.14 \times 10^{-4} \, \text{F}$ $C \approx 214 \, \mu\text{F}$

Final Answer:

The minimum possible capacitance is approximately 214 μF.

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Would you like further details or explanations on any part of this solution?

Here are some related questions:

1. How does the frequency of the grid impact the output voltage ripple in a rectifier?
2. How do you calculate the peak voltage in a three-phase system?
3. What are the advantages of using a capacitor in a DC filter?
4. What is the impact of load resistance on the performance of a rectifier?
5. How would the output voltage change if the load power was increased to 1500 W?

Tip: When calculating the capacitance required for a specific ripple voltage, ensure that the frequency and the ripple requirement are carefully considered, as they directly affect the value.