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Step-by-Step Solution
Step 1: Identify the Requirements
We need a net capacitance of $2\ \mu\text{F}$ across a potential difference of $1000\ \text{V}$ (1.0 kV). We only have $1\ \mu\text{F}$ capacitors, each capable of withstanding up to $300\ \text{V}$.
Step 2: Determine the Voltage Constraint
Since each capacitor cannot withstand more than $300\ \text{V}$, we need to arrange them so that each branch in the circuit does not exceed this voltage limit. To handle $1000\ \text{V}$, we require multiple groups in series so that each group sees a fraction of the total voltage.
Step 3: Devise the Arrangement
• First, create parallel combinations of $1\ \mu\text{F}$ capacitors to increase their capacitance.
• Then, connect several of these parallel combinations in series to safely share the total voltage.
Step 4: Capacitance Computation
Suppose each group has 8 capacitors in parallel. The net capacitance of one group ($C_{\text{group}}$) is:
$$C_{\text{group}} = 8 \times 1\ \mu\text{F} = 8\ \mu\text{F}.$$
If we place 4 such groups in series, the equivalent capacitance $C_{\text{eq}}$ of the series combination is given by:
$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_{\text{group}}} + \frac{1}{C_{\text{group}}} + \frac{1}{C_{\text{group}}} + \frac{1}{C_{\text{group}}}
= 4 \times \frac{1}{8\ \mu\text{F}}
= \frac{4}{8\ \mu\text{F}}
= \frac{1}{2\ \mu\text{F}}.$$
Hence,
$$C_{\text{eq}} = 2\ \mu\text{F}.$$
Step 5: Voltage Across Each Series Group
With 4 groups in series, the total $1000\ \text{V}$ splits into 4 equal parts of $250\ \text{V}$ each, which is within the safe limit of $300\ \text{V}$ for every parallel group.
Step 6: Total Number of Capacitors
Each group has 8 capacitors, and there are 4 identical groups in series. Therefore, the total number of capacitors required is:
$$8 \times 4 = 32.$$
Step 7: Final Arrangement
The arrangement is 8 capacitors in parallel forming one branch, and 4 such branches connected in series. This ensures both the capacitance and voltage requirements are met.
