JAMB - Physics (1980 - No. 43)
The total capacitance of the circuit above is?
0.25mF
0.50mF
0.75mF
1.25mF
1.50mF
Explicació
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