A 6\text{ }\mu\text{F} capacitor is connected in series with a parallel arrangement of a 2… — JAMB UTME

A $6\text{ }\mu\text{F}$ capacitor is connected in series with a parallel arrangement of a $2\text{ }\mu\text{F}$ capacitor and a $1\text{ }\mu\text{F}$ capacitor. If the entire circuit is connected across a $30\text{ V}$ d.c. power supply, what is the electric charge stored on the $2\text{ }\mu\text{F}$ capacitor?

A
$20\text{ }\mu\text{C}$
B
$30\text{ }\mu\text{C}$
$40\text{ }\mu\text{C}$ Cevap
D
$60\text{ }\mu\text{C}$

Cevap

The charge stored on the

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

capacitor is

40\text{ }\mu\text{C}

To find the charge on the

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

capacitor, we first combine the parallel capacitors (

2\text{ }\mu\text{F} + 1\text{ }\mu\text{F} = 3\text{ }\mu\text{F}

). Next, we combine this in series with the

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

capacitor to get an overall capacitance

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

. The total charge drawn from the

30\text{ V}

supply is

Q = 2\text{ }\mu\text{F} \times 30\text{ V} = 60\text{ }\mu\text{C}

. This total charge enters the parallel combination, giving a potential drop across the parallel branch of

V_p = 60\text{ }\mu\text{C} / 3\text{ }\mu\text{F} = 20\text{ V}

. Therefore, the charge on the

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

capacitor is

2\text{ }\mu\text{F} \times 20\text{ V} = 40\text{ }\mu\text{C}

Adım Adım Çözüm

Calculate the equivalent capacitance of the parallel section

C_{p} = 2\text{ }\mu\text{F} + 1\text{ }\mu\text{F} = 3\text{ }\mu\text{F}

Capacitors in parallel add directly.

Calculate the total equivalent capacitance of the network

C_{eq} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2\text{ }\mu\text{F}

The

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

capacitor and the

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

parallel equivalent are in series.

Find the total charge supplied by the

30\text{ V}

source

Q_{total} = C_{eq} \times V = 2\text{ }\mu\text{F} \times 30\text{ V} = 60\text{ }\mu\text{C}

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

Determine the potential difference across the parallel branch

V_{p} = \frac{Q_{total}}{C_{p}} = \frac{60\text{ }\mu\text{C}}{3\text{ }\mu\text{F}} = 20\text{ V}

The total charge flows through the series combination, creating a potential drop across the parallel combination equal to

Q_{total} / C_{p}

Calculate the charge on the

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

capacitor

Q_{2\mu\text{F}} = C_{2\mu\text{F}} \times V_{p} = 2\text{ }\mu\text{F} \times 20\text{ V} = 40\text{ }\mu\text{C}

The charge on a specific capacitor in parallel is the product of its capacitance and the voltage across the parallel branch.

Anahtar Kavram

Charge and voltage distribution in mixed capacitor networks

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