How many `1muF` capacitors must be connected in parallel to store a charge of 1 C with a potential of 110 V across the capacitors?

Options

A. 990
B. 900
C. 9090
D. 909

Answer

To solve the problem of how many `1 µF` capacitors must be connected in parallel to store a charge of `1 C` with a potential of `110 V` across the capacitors, we can follow these steps:

### Step 1: Understand the relationship between charge, capacitance, and voltage
The relationship between charge (Q), capacitance (C), and voltage (V) is given by the formula:
\[ Q = C \times V \]

### Step 2: Determine the equivalent capacitance for capacitors in parallel
When capacitors are connected in parallel, the equivalent capacitance (C_eq) is the sum of the individual capacitances. If we have `n` capacitors each of capacitance `C`, then:
\[ C_{eq} = n \times C \]

### Step 3: Substitute the values into the formula
We know:
- The charge \( Q = 1 \, C \)
- The voltage \( V = 110 \, V \)
- The capacitance of each capacitor \( C = 1 \, \mu F = 1 \times 10^{-6} \, F \)

Substituting these values into the charge formula:
\[ Q = C_{eq} \times V \]
\[ 1 = (n \times 1 \times 10^{-6}) \times 110 \]

### Step 4: Solve for n
Rearranging the equation to solve for \( n \):
\[ n = \frac{Q}{C_{eq} \times V} \]
\[ n = \frac{1}{(1 \times 10^{-6}) \times 110} \]

Calculating the denominator:
\[ n = \frac{1}{1.1 \times 10^{-4}} \]
\[ n = \frac{1}{1.1} \times 10^{4} \]
\[ n \approx 9.09 \times 10^{3} \]

Since \( n \) must be a whole number, we round it up to the nearest whole number:
\[ n \approx 9091 \]

### Step 5: Conclusion
Therefore, approximately **9091 capacitors** of `1 µF` must be connected in parallel to store a charge of `1 C` with a potential of `110 V` across the capacitors.

---