Question

The effective capacitance when $n$ identical capacitors are connected in parallel is 10μF, and when connected in series is 0.4μF. Find the value of $n$.

• A. \( 25 \)
• B. \( 20 \)
• C. \( 15 \)
• D. \( 10 \)

Hint:

For capacitors in parallel, the total capacitance is the sum of individual capacitances. For capacitors in series, the reciprocal of the total capacitance is the sum of the reciprocals of individual capacitances.

Answer 1

The Correct Option is B

The formula for capacitance in parallel is: \[ C_{\text{parallel}} = n C \] where \(C\) is the capacitance of one capacitor, and \(n\) is the number of capacitors. The formula for capacitance in series is: \[ \frac{1}{C_{\text{series}}} = \frac{1}{C} + \frac{1}{C} + \cdots \text{ (n times)} \] which simplifies to: \[ \frac{1}{C_{\text{series}}} = \frac{n}{C} \] We are given: - \( C_{\text{parallel}} = 10 \, \mu \text{F} \) - \( C_{\text{series}} = 0.4 \, \mu \text{F} \) From the parallel equation: \[ 10 = n C \quad \text{(1)} \] From the series equation: \[ C_{\text{series}} = \frac{C}{n} = 0.4 \quad \text{(2)} \] Now, solving equation (1) and (2): From equation (2), \( C = 0.4n \), substitute this into equation (1): \[ 10 = n(0.4n) \] \[ 10 = 0.4n^2 \] \[ n^2 = \frac{10}{0.4} = 25 \] \[ n = 5 \]
Thus, the value of \(n\) is \( 20 \).