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

Given:

• N capacitors, each with capacitance \( C = 1 \, \mu\text{F} = 10^{-6} \, \text{F} \)
• Connected in parallel to store total charge \( Q_{\text{total}} = 1 \, \text{C} \)
• Potential across each capacitor \( V = 100 \, \text{V} \)

Step 1: Determine Number of Capacitors (N)

In parallel connection, total capacitance \( C_{\text{parallel}} = N \times C \).

Total charge stored: \( Q_{\text{total}} = C_{\text{parallel}} \times V \)

Substitute values:

\[ 1 = N \times 10^{-6} \times 100 \]

\[ N = \frac{1}{10^{-4}} = 10,000 \]

Step 2: Calculate Equivalent Capacitance in Series

When these N capacitors are connected in series:

\[ \frac{1}{C_{\text{series}}} = \sum_{i=1}^N \frac{1}{C_i} = N \times \frac{1}{C} \]

\[ C_{\text{series}} = \frac{C}{N} = \frac{10^{-6}}{10,000} = 10^{-10} \, \text{F} \]

Conclusion:

The equivalent capacitance in series is \( 10^{-10} \, \text{F} \).

Answer: \(\boxed{C}\)

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): \(10^{-10} F\)

Answer 3

1. Calculate the total capacitance when connected in parallel:

When capacitors are connected in parallel, the equivalent capacitance (C_p) is the sum of the individual capacitances. Since each capacitor has a capacitance of 1 μF:

\[C_p = N \times 1 \, \mu F = N \, \mu F\]

2. Calculate the total charge stored:

The potential across each capacitor in parallel is 100 V. The total charge (Q) stored is given by:

\[Q = C_p \times V = N \, \mu F \times 100 \, V = 1 \, C\]

Given that Q = 1 C, we can solve for N:

\[N \times 10^{-6} \, F \times 100 \, V = 1 \, C\]

\[N = \frac{1}{100 \times 10^{-6}} = 10^4\]

3. Calculate the equivalent capacitance when connected in series:

When capacitors are connected in series, the reciprocal of the equivalent capacitance (C_s) is the sum of the reciprocals of the individual capacitances:

\[\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_N}\]

Since all capacitors have the same capacitance (1 μF):

\[\frac{1}{C_s} = N \times \frac{1}{1 \, \mu F} \]

\[C_s = \frac{1 \, \mu F}{N} = \frac{1 \times 10^{-6} \, F}{10^4} = 10^{-10} \, F\]

Final Answer: The final answer is \(\boxed{C}\)