JAMB - Physics (1980 - No. 43)

The total capacitance of the circuit above is?

0.25mF
0.50mF
0.75mF
1.25mF
1.50mF

Explanation

 CTotal in parralel = C1 + C2 + C3
 CTotal = 2 + 2 + 2 → 6

C Total in series = 1 / C\(_1\) + 1 / C\(_2\) 

1 / C\(_T\) = 1/2 + 1/6

1 / C\(_T\) = \(\frac{4}{6}\)

cross multiply

C\(_T\) =  \(\frac{3}{2}\) or 1.5mF

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