The total capacitance of the circuit above is?
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
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