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JAMB Physics 2019 Paper

Question : 64

Total: 100

The resultant capacitance in the figure above is

Solution:

Concept:

Capacitors connected in series have an equivalent capacitance

C_T

given by the reciprocal relation:

\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_4}.

Explanation:

From the figure, take the individual capacitances as

2\mu\text{F}

3\mu\text{F}

6\mu\text{F}

and

4\mu\text{F}

Apply the series capacitance formula:

\frac{1}{C_T} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \frac{1}{4}.

Use a common denominator of

12

\frac{1}{C_T} = \frac{6}{12} + \frac{4}{12} + \frac{2}{12} + \frac{3}{12} = \frac{15}{12}.

Take the reciprocal to find the resultant capacitance:

C_T = \frac{12}{15} = 0.8\mu\text{F}.

This corresponds to option B.

Answer:

0.8\mu\text{F}

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