Question

Two capacitors of 4 $\mu$F and 6 $\mu$F are connected in series. Their equivalent capacitance is:

• A. 10 \(\mu\)F
• B. 5 \(\mu\)F
• C. 2.4 \(\mu\)F
• D. 1.6 \(\mu\)F

Hint:

For capacitors in series, always use the reciprocal rule:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \] The result is always less than the smallest individual capacitor.

Answer 1

The Correct Option is C

To determine the equivalent capacitance \( C_{\text{eq}} \) for capacitors connected in series, the following formula is applied:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \] The provided values are:
\( C_1 = 4 \, \mu\text{F} \) and \( C_2 = 6 \, \mu\text{F} \).
Procedure:
1. Value Substitution
Substitute the given capacitances into the formula:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{4} + \frac{1}{6} \] To solve for \( \frac{1}{C_{\text{eq}}} \), find a common denominator:
\[ \frac{1}{C_{\text{eq}}} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \] 2. Reciprocal Calculation
Invert the result to obtain the equivalent capacitance:
\[ C_{\text{eq}} = \frac{12}{5} = 2.4 \, \mu\text{F} \]