Question

The equivalent capacitance of n capacitors of equal capacitance when connected in series and parallel are respectively $0.4\ \mu F$ and $10\ \mu F$. The capacitance of each capacitor is:

• A. $2\ \mu F$
• B. $4\ \mu F$
• C. $5\ \mu F$
• D. $6\ \mu F$
• E. $1\ \mu F$

Hint:

The ratio $C_p/C_s = n^2$ is a useful identity for identical components.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves the formulas for calculating the equivalent capacitance for capacitors connected in series and in parallel. We have a system of two equations with two unknowns (the number of capacitors, n, and the individual capacitance, C).
Step 2: Key Formula or Approach:
Let C be the capacitance of each of the n identical capacitors. 1. Parallel Combination: The equivalent capacitance (\( C_p \)) is the sum of the individual capacitances. \[ C_p = C + C + \dots + C \text{ (n times)} = nC \] 2. Series Combination: The reciprocal of the equivalent capacitance (\( C_s \)) is the sum of the reciprocals of the individual capacitances. \[ \frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} + \dots + \frac{1}{C} \text{ (n times)} = \frac{n}{C} \implies C_s = \frac{C}{n} \] Step 3: Detailed Explanation:
We are given: - Equivalent capacitance in series, \( C_s = 0.4 \, \mu\text{F} \) - Equivalent capacitance in parallel, \( C_p = 10 \, \mu\text{F} \) Using the formulas from Step 2, we have two equations: (1) \( nC = 10 \) (2) \( \frac{C}{n} = 0.4 \) We need to find the value of C. We can solve this system of equations. A simple way is to multiply the two equations together: \[ (nC) \times \left(\frac{C}{n}\right) = 10 \times 0.4 \] The 'n' terms cancel out: \[ C^2 = 4 \] Take the square root to find C (capacitance must be positive): \[ C = \sqrt{4} = 2 \, \mu\text{F} \] We can also find the value of n for completeness, using equation (1): \[ nC = 10 \implies n(2) = 10 \implies n = 5 \] Let's check with equation (2): \( C/n = 2/5 = 0.4 \). This is consistent. Step 4: Final Answer:
The capacitance of each capacitor is 2 \( \mu \)F.