Two 12\text{ }\mu\text{F} capacitors are connected in series, and this combination is plac… — JAMB UTME

Two $12\text{ }\mu\text{F}$ capacitors are connected in series, and this combination is placed in parallel with a third capacitor of capacitance $6\text{ }\mu\text{F}$ . What is the total equivalent capacitance of the arrangement?

A
$3.0\text{ }\mu\text{F}$
B
$4.8\text{ }\mu\text{F}$
$12.0\text{ }\mu\text{F}$ Answer
D
$30.0\text{ }\mu\text{F}$

Answer

The total equivalent capacitance of the network is

12.0\text{ }\mu\text{F}

Combining the two identical

12\text{ }\mu\text{F}

capacitors in series gives an equivalent capacitance of

6.0\text{ }\mu\text{F}

. Adding this in parallel with the

6.0\text{ }\mu\text{F}

capacitor yields

6.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}

Step-by-Step Solution

Calculate the equivalent capacitance (

C_s

) of the two

12\text{ }\mu\text{F}

capacitors connected in series.

C_s = \frac{12 \times 12}{12 + 12} = 6.0\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}

Calculate the total equivalent capacitance (

C_{eq}

) by adding

C_s

to the parallel

6.0\text{ }\mu\text{F}

capacitor.

C_{eq} = C_s + C_3 = 6.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}

Capacitors in parallel add directly:

C_{eq} = C_{series} + C_{parallel}

Key Concept

Equivalent capacitance of series-parallel capacitor networks

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