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JAMB Physics 2019 Paper

Question : 24

Total: 100

Solution:

Concept:

For capacitors in series, the total capacitance is less than any individual capacitance, and the charge on each capacitor is the same.

Explanation:

For two capacitors in series, use the product-over-sum formula:

C_T = \frac{C_1 \times C_2}{C_1 + C_2}

Substitute

C_1 = 3\,\mu\mathrm{F}

and

C_2 = 6\,\mu\mathrm{F}

C_T = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\,\mu\mathrm{F}

Convert to farads:

C_T = 2 \times 10^{-6}\,\mathrm{F}

The charge stored is given by:

Q = C_T V

Q = 2 \times 10^{-6} \times 400

Q = 800 \times 10^{-6} = 8 \times 10^{-4}\,\mathrm{C}

Answer:

The charge is

8 \times 10^{-4}\,\mathrm{C}

, which is option A.

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