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JAMB Physics 2010 Paper

Question : 40

Total: 47

Solution:

Concept:

In a parallel connection, the total capacitance is the sum of the individual capacitances. The energy stored is given by

E = \frac{1}{2}CV^2

Explanation:

For capacitors in parallel,

C_{\text{eq}} = C_1 + C_2 + C_3

C_{\text{eq}} = 2\,\mu\mathrm{F} + 4\,\mu\mathrm{F} + 8\,\mu\mathrm{F} = 14\,\mu\mathrm{F}

The voltage

V

across each capacitor is

6\,\mathrm{V}

Substitute into the energy formula:

E = \frac{1}{2} \times 14 \times 10^{-6}\,\mathrm{F} \times (6\,\mathrm{V})^2

E = \frac{1}{2} \times 14 \times 10^{-6} \times 36

E = 2.52 \times 10^{-4}\,\mathrm{J}

Therefore, the total energy stored is

2.52 \times 10^{-4}\,\mathrm{J}

Answer:

Option C:

2.52 \times 10^{-4}\,\mathrm{J}

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