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 can handle a maximum of 200 V. The supply voltage is 1200 V. To ensure no capacitor exceeds its voltage rating, the 1200 V must be distributed across capacitors in series. Let \( n \) be the number of capacitors in series, then: \[ n \times 200\,\text{V} \geq 1200\,\text{V} \Rightarrow n \geq \frac{1200}{200} = 6 \] So, we must use at least 6 capacitors in series to tolerate the full 1200 V. Step 2: Capacitance of Series Combination
Capacitance of \( n = 6 \) capacitors in series (each of 1 \(\mu\text{F}\)) is: \[ C_{\text{series}} = \frac{1}{\frac{1}{1} + \frac{1}{1} + \cdots + \frac{1}{1}} = \frac{1}{6} \mu\text{F} \] Step 3: Achieving 2 \(\mu\text{F}\) Equivalent Capacitance
Let \( m \) be the number of such series combinations connected in parallel. Parallel combination adds capacitances: \[ C_{\text{eq}} = m \times \frac{1}{6} = 2 \Rightarrow m = 12 \] Step 4: Total Capacitors Required
Each row (series) has 6 capacitors and there are 12 such rows in parallel. Total capacitors = \( 12 \times 6 = 72 \) % Final Answer Statement Answer: Minimum 72 capacitors are required, arranged as 12 rows of 6 capacitors in series, connected in parallel.