A 2.0\text{ }\mu\text{F} capacitor and a 3.0\text{ }\mu\text{F} capacitor are connected in… — JAMB UTME

A $2.0\text{ }\mu\text{F}$ capacitor and a $3.0\text{ }\mu\text{F}$ capacitor are connected in series across a $100\text{ V}$ d.c. power supply. What is the magnitude of the electric charge stored on the $2.0\text{ }\mu\text{F}$ capacitor?

A
$500\text{ }\mu\text{C}$
$120\text{ }\mu\text{C}$ Answer
C
$200\text{ }\mu\text{C}$
D
$80\text{ }\mu\text{C}$

Answer

The electric charge stored on the

2.0\text{ }\mu\text{F}

capacitor is

120\text{ }\mu\text{C}

The correct answer is

120\text{ }\mu\text{C}

. For capacitors connected in series, the equivalent capacitance

C_{eq}

is determined using

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

, giving

C_{eq} = \frac{2.0 \times 3.0}{2.0 + 3.0} = 1.2\text{ }\mu\text{F}

. Multiplying by the source voltage of

100\text{ V}

yields a total charge

Q = C_{eq} V = 120\text{ }\mu\text{C}

. Because capacitors in series store equal amounts of charge, the charge on the

2.0\text{ }\mu\text{F}

capacitor is

120\text{ }\mu\text{C}

Step-by-Step Solution

Calculate the equivalent capacitance

C_{eq}

of the series combination.

C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2.0 \times 3.0}{2.0 + 3.0} = 1.2\text{ }\mu\text{F}

Capacitors in series combine according to the reciprocal formula

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

Determine the total charge

Q

supplied by the

100\text{ V}

source.

Q = C_{eq} V = 1.2\text{ }\mu\text{F} \times 100\text{ V} = 120\text{ }\mu\text{C}

The total charge is the product of the equivalent capacitance and the total voltage.

Identify the charge on the individual

2.0\text{ }\mu\text{F}

capacitor.

Q_1 = Q = 120\text{ }\mu\text{C}

Components connected in series carry the exact same electric charge.

Key Concept

Equivalent Capacitance and Charge Distribution in Series Circuits

Estimated Time:1m 30s

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