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WAEC Physics 2011 Paper

Question : 48

Total: 49

Solution:

Concept:

Total charge stored is found by using the equivalent capacitance and the supply voltage:

Q = C_{\text{eq}} V

Explanation:

The

2\ \mu\text{F}

and

6\ \mu\text{F}

capacitors are in parallel.

Their combined capacitance is

C_p = 2 + 6 = 8\ \mu\text{F}

This combination is in series with the other

2\ \mu\text{F}

capacitor.

So,

\frac{1}{C_{\text{eq}}} = \frac{1}{8} + \frac{1}{2} = \frac{5}{8}

Therefore,

C_{\text{eq}} = \frac{8}{5}\ \mu\text{F} = 1.6\ \mu\text{F}

The supply voltage is

10\ \text{V}

Hence,

Q = C_{\text{eq}}V = \frac{8}{5} \times 10 = 16\ \mu\text{C}

Answer:

16\ \mu\text{C}

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