Understanding Capacitors in Series

This question involves calculating the capacitance of individual capacitors when they are connected in a series arrangement. We are given that there are ten capacitors, all having the same capacitance value, and their combined equivalent capacitance is 10 µF.

Formula for Capacitors in Series

When capacitors are connected in series, the reciprocal of the equivalent capacitance ($C_{eq}$) is equal to the sum of the reciprocals of the individual capacitances ($C_1, C_2, \dots, C_n$). The formula is:

$$ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n} $$

In this specific problem, all ten capacitors are identical, meaning $C_1 = C_2 = \dots = C_{10} = C$. So, the formula simplifies to:

$$ \frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \dots + \frac{1}{C} \quad (\text{10 times}) $$

This can be written as:

$$ \frac{1}{C_{eq}} = \frac{10}{C} $$

Rearranging this formula to find the capacitance of a single capacitor (C) gives:

$$ C = 10 \times C_{eq} $$

Calculating Individual Capacitance

We are given:

• Number of capacitors, $n = 10$.
• Equivalent capacitance, $C_{eq} = 10 \, \mu\text{F}$.

Now, we substitute the value of $C_{eq}$ into the rearranged formula:

$$ C = 10 \times (10 \, \mu\text{F}) $$

$$ C = 100 \, \mu\text{F} $$

Conclusion

Therefore, the capacitance of each individual capacitor is 100 µF.