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JAMB Physics 2017 Paper

Question : 20

Total: 49

The net capacitance in the circuit above is

Solution:

For capacitance i n parallel, 2µF and 2µF are in parallel,

their equivalence is 2µF and 2µF = 4µF

The 4µF generated is now in series with the remaining 4µF.

The net capacitance for series connection is

\frac{1}{C}

\frac{1}{4}

\frac{1}{4}

= 1 +

\frac{1}{4}

\frac{2}{4}

C =

\frac{4}{2}

= 2µF

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