The total capacitance of the circuit above is

JAMB 1980

The total capacitance of the circuit above is

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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