Two capacitors of capacitances 12.0\text{ }\mu\text{F} and 6.0\text{ }\mu\text{F} are conn… — JAMB UTME

Two capacitors of capacitances $12.0\text{ }\mu\text{F}$ and $6.0\text{ }\mu\text{F}$ are connected in parallel. This combination is then connected in series with a $9.0\text{ }\mu\text{F}$ capacitor across a $45.0\text{ V}$ d.c. power supply. What is the total charge supplied by the power source?

$270\text{ }\mu\text{C}$ Answer
B
$585\text{ }\mu\text{C}$
C
$1215\text{ }\mu\text{C}$
D
$125\text{ }\mu\text{C}$

Answer

The total charge supplied by the power source is

270\text{ }\mu\text{C}

Combining the parallel capacitors yields

18.0\text{ }\mu\text{F}

. Connecting this combination in series with the

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

capacitor gives an equivalent total capacitance of

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

. Multiplying total capacitance by the supply voltage of

45.0\text{ V}

gives

270\text{ }\mu\text{C}

Step-by-Step Solution

Calculate equivalent capacitance of the parallel branch

C_{\text{p}} = C_1 + C_2 = 12.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 18.0\text{ }\mu\text{F}

Capacitors in parallel add directly.

Calculate total equivalent capacitance of the network

C_{\text{eq}} = \frac{C_{\text{p}} \times C_3}{C_{\text{p}} + C_3} = \frac{18.0 \times 9.0}{18.0 + 9.0} = 6.0\text{ }\mu\text{F}

The parallel combination

C_{\text{p}}

is in series with

C_3

, using the reciprocal formula for series capacitors.

Calculate total charge supplied

Q = C_{\text{eq}} V = 6.0 \times 10^{-6}\text{ F} \times 45.0\text{ V} = 270\text{ }\mu\text{C}

Total charge from the power source depends on total equivalent capacitance and total supply voltage.

Key Concept

Equivalent capacitance of mixed series-parallel networks

Estimated Time:1m 30s

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