Two capacitors with capacitances of 3.0\text{ }\mu\text{F} and 6.0\text{ }\mu\text{F} are … — JAMB UTME

Two capacitors with capacitances of $3.0\text{ }\mu\text{F}$ and $6.0\text{ }\mu\text{F}$ are connected in parallel across a direct-current source. What is the equivalent capacitance of this combination?

$9.0\text{ }\mu\text{F}$ Answer
B
$2.0\text{ }\mu\text{F}$
C
$4.5\text{ }\mu\text{F}$
D
$18.0\text{ }\mu\text{F}$

Answer

The equivalent capacitance of the parallel combination is

9.0\text{ }\mu\text{F}

When capacitors are connected in parallel, each capacitor experiences the full potential difference of the voltage source, and the total charge stored is the sum of individual charges (

Q_{total} = Q_1 + Q_2

). Thus, the equivalent capacitance is the direct sum of the individual capacitances:

C_{eq} = 3.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 9.0\text{ }\mu\text{F}

Step-by-Step Solution

Identify the combination rule for parallel capacitors.

The total capacitance is given by

C_{eq} = C_1 + C_2

Capacitors in parallel share the same potential difference, so total charge stored is the sum of individual charges.

Substitute the given values into the parallel capacitance formula.

C_{eq} = 3.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 9.0\text{ }\mu\text{F}

Direct addition yields the total equivalent capacitance.

Key Concept

Parallel Combination of Capacitors

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