Two capacitors of 0.0003μF and 0.0006μF are connected in series, find their combined capacitance.

Analysis

To find the combined capacitance of capacitors connected in series, we use the reciprocal formula. This is similar to how resistors are calculated when connected in parallel.

Concept

When capacitors are connected in series, the total capacitance ($C_T$) is given by the formula:

Alternatively, for two capacitors, you can use the product-over-sum rule:

Solution

Given:

• $C_1 = 0.0003 \, \mu\text{F}$
• $C_2 = 0.0006 \, \mu\text{F}$

Using the product-over-sum formula:

1. Calculate the numerator: $0.0003 \times 0.0006 = 0.00000018$

2. Calculate the denominator: $0.0003 + 0.0006 = 0.0009$

3. Divide the results: $$C_T = \frac{0.00000018}{0.0009}$$To make this easier, move the decimal points:$$C_T = \frac{1.8 \times 10^{-7}}{9 \times 10^{-4}}$$ $$C_T = 0.2 \times 10^{-3} = 0.0002 \, \mu\text{F}$$

Conclusion

The combined capacitance is $0.0002 \, \mu\text{F}$. This matches option B.

Exam Tip

Always remember: When capacitors are in series, the total capacitance is always smaller than the smallest individual capacitor in the circuit. Since $0.0002 < 0.0003$, your answer is logically consistent!