Consider three capacitors, of capacitance 3600 pF, 7200 pF, and 0.011 mu F. Part C What...

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Consider three capacitors, of capacitance 3600 pF,

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Answer 1

Answer #1

C1 = 3600 pF

C2 = 7200 pF

C3 = 11000 pF

capacitance is minimum when all are connected in series

1/Cmin = 1/C1 + 1/C2 + 1/C3

1/Cmin = 1/3600 + 1/7200 + 1/11000

Cmin = 1970.15 pF <<<<<-------answer

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Answer 2

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