The total capacitance of the circuit above is

JAMB 1980

The total capacitance of the circuit above is?

A. 0.25mF

B. 0.50mF

C. 0.75mF

D. 1.25mF

E. 1.50mF

Correct Answer: Option E

Explanation

 CTotal in parallel = C1 + C2 + C3

 CTotal = 2 + 2 + 2 → 6

1/CT = 1 / C1 + 1 / C2 

Where CT is the Total Capacitance

1 / CT = 1/2 + 1/6

1 / CT = 4/6

cross multiply

CT =  3/2 or 1.5mF

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