A 3\text{ }\mu\text{F} capacitor and a 6\text{ }\mu\text{F} capacitor are connected in ser… — JAMB UTME

A $3\text{ }\mu\text{F}$ capacitor and a $6\text{ }\mu\text{F}$ capacitor are connected in series across a direct current voltage source. What is the total equivalent capacitance of the combination, in microfarads ( $\mu\text{F}$ )?

Answer: 2 µF

Answer

The total equivalent capacitance of the combination is

2\text{ }\mu\text{F}

For capacitors connected in series, the reciprocal of the total equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances. Substituting

3\text{ }\mu\text{F}

and

6\text{ }\mu\text{F}

gives

\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}\text{ }\mu\text{F}^{-1}

, which yields an equivalent capacitance of

2\text{ }\mu\text{F}

Step-by-Step Solution

State the formula for equivalent capacitance of two capacitors in series

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

Capacitors connected in series combine reciprocally, unlike resistors connected in series.

Substitute the values of

C_1

and

C_2

\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\text{ }\mu\text{F}^{-1}

Find a common denominator and add the fractions.

Calculate the reciprocal to determine

C_{eq}

C_{eq} = 2\text{ }\mu\text{F}

Inverting

\frac{1}{2}

yields the total equivalent capacitance.

Key Concept

Equivalent Capacitance in Series

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