Question

N capacitors,each with 1μF capacitance,are connected in parallel to store a charge of 1C. The potential across each capacitor is 100V. If these N capacitors are now connected in series,the equivalent capacitance in the circuit will be:

• A. 10^-4F
• B. 10^-6F
• C. 10^-10F
• D. 5x10^-8F
• E. 10^-2F

Answer 1

The Correct Option is C

The correct answer is (C): 10^-10F

Answer 2

Given,
- Capacitors are connected in parallel with capacitance \(C = 1 \mu F = 1 \times 10^{-6} F\).
- Total charge stored, Q = 1C.
- Potential across each capacitor, V = 100V.

1. Finding the number of capacitors N in parallel:
  We use the formula \(Q = C \cdot V\), where Q is the total charge, C is the capacitance, and V is the voltage.
 \(\begin{aligned}   Q & = C \cdot V \\   1 & = N \cdot (1 \times 10^{-6}) \cdot 100 \\   N & = \frac{1}{1 \times 10^{-6} \cdot 100} \\   & = \frac{1}{10^{-4}} \\   & = 10^4   \end{aligned}\)

So, there are \(N = 10^4\) capacitors connected in parallel.

2. Finding the equivalent capacitance when capacitors are connected in series:
When capacitors are connected in series, the equivalent capacitance \(C_{\text{eq}}\) is given by the reciprocal of the sum of the reciprocals of individual capacitances.
\(\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_N}\)

Substituting the values, since each capacitor has the same capacitance \(( C_1 = C_2 = \cdots = C_N = 1 \mu F )\):
\(\frac{1}{C_{\text{eq}}} = \frac{1}{1 \times 10^{-6}} + \frac{1}{1 \times 10^{-6}} + \cdots + \frac{1}{1 \times 10^{-6}} \quad (N \text{ times})\)

\(\frac{1}{C_{\text{eq}}} = N \times \frac{1}{1 \times 10^{-6}}\)

\(C_{\text{eq}} = \frac{1 \times 10^{-6}}{N}\)

\(C_{\text{eq}} = \frac{1 \times 10^{-6}}{10^4}\)
\(C_{\text{eq}} = 1 \times 10^{-10} F\)

So, the correct option is (C): \(10^{-10} F\)