Question

You are provided with a large number of 1 \(\mu\text{F}\) identical capacitors and a power supply of 1200 V. The dielectric medium used in each capacitor can withstand up to 200 V only. Find the minimum number of capacitors and their arrangement required to build a capacitor system of equivalent capacitance of 2 \(\mu\text{F}\) for use with this supply.

Hint:

When working with high voltage sources and limited voltage-rated capacitors, always split the voltage across series-connected capacitors. Then adjust the number of such strings in parallel to meet the desired total capacitance.

Answer 1

Step 1: Voltage Limit Per Capacitor
Each capacitor has a maximum voltage rating of 200 V. The supply voltage is 1200 V. To prevent any capacitor from exceeding its rating, the 1200 V must be divided among capacitors connected in series. Let \( n \) represent the number of capacitors in series. The following equation applies: \[n \times 200\,\text{V} \geq 1200\,\text{V} \Rightarrow n \geq \frac{1200}{200} = 6\] Therefore, a minimum of 6 capacitors in series is required to handle the full 1200 V. Step 2: Capacitance of Series Combination
The capacitance of \( n = 6 \) capacitors in series, each with a capacitance of 1 \(\mu\text{F}\), is calculated as follows: \[C_{\text{series}} = \frac{1}{\frac{1}{1} + \frac{1}{1} + \cdots + \frac{1}{1}} = \frac{1}{6} \mu\text{F}\] Step 3: Achieving 2 \(\mu\text{F}\) Equivalent Capacitance
Let \( m \) denote the number of these series combinations connected in parallel. Capacitances in parallel add directly: \[C_{\text{eq}} = m \times \frac{1}{6} = 2 \Rightarrow m = 12\] Step 4: Total Capacitors Required
Each series row contains 6 capacitors, and there are 12 such rows connected in parallel. The total number of capacitors is calculated as: \( 12 \times 6 = 72 \). % Final Answer Statement Answer: A minimum of 72 capacitors are required. These should be arranged as 12 parallel rows, with each row consisting of 6 capacitors connected in series.