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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