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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His Name is Tiamiyu Abdulbazeet Olawale. He is a student at the University of Ilorin studying Electrical and Electronics Engineering. He likes to give the latest updates about Nigeria and international institutions as he aims at helping people to lay their hands on the right information.