Question

You have \(N\) identical capacitors, each with capacitance \(C\), connected in series. The equivalent capacitance of this system of capacitors is a) \(N C\). b) \(C / N\). c) \(N^{2} C\). e) \(C\). d) \(C / N^{2}\).

Short answer

Answer: b) C / N

Step-by-step solution

Understand the concept of capacitors in series

When capacitors are connected in series, the total or equivalent capacitance of the system is calculated using the following formula: \[\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_N}\] where \(C_{eq}\) is the equivalent capacitance and \(C_1 , C_2 , ... , C_N\) are the capacitances of the individual capacitors in the series.

Apply the concept to the given problem

In this problem, we have \(N\) identical capacitors with capacitance \(C\). We can use the formula for capacitors in series to find the equivalent capacitance. Since all capacitors are identical, the formula becomes: \[\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \cdots + \frac{1}{C}\] We have \(N\) identical capacitors, so there are \(N\) terms in the equation: \[\frac{1}{C_{eq}} = N \times \frac{1}{C}\]

Solve for \(C_{eq}\)

To find the equivalent capacitance \(C_{eq}\), we can rearrange the equation: \[C_{eq} = \frac{1}{N \times \frac{1}{C}}\] \[C_{eq} = \frac{C}{N}\]

Match the solution to the given options

The equivalent capacitance of the system of \(N\) identical capacitors with capacitance \(C\) connected in series is \(\frac{C}{N}\). This matches option b) \(C/N\). Therefore, the correct answer is b) \(C / N\).