$C_{\textrm{series}} = 3.08 \textrm{ }\mu\textrm{F}$

$C_{\textrm{parallel}} = 13.0 \textrm{ }\mu\textrm{F}$

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This is College Physics Answers with Shaun Dychko. We have capacitance one is 5 microfarads and capacitance two is eight microfarads. And the question is, what equivalent capacitances can you make of that two together? And so, there are two ways to connect them together. One is in series, and then the other is in parallel. And, we'll get different answers for each type of connection. So in series, we have one and then the other in this circuit. And, to find the equivalent capacitance here, we have to take the reciprocal of each capacitance. And then, add those reciprocals together, and then take the reciprocal of that result. So, we have one over five microfarads plus one over eight microfarads. And then, take the reciprocal of that sum and we get 3.08 microfarads. Maybe I should write that down. So, one over five microfarads plus one over eight microfarads. Take that sum and then raise that result to negative one and we end up with 3.08. And then in parallel, we have the wire going to one side of capacitor one, and then also going to one side of capacitor two, and then the other side of these two capacitors are connected together, this is better. And so, the equivalent capacitance will be the arithmetic sum of the individual capacitances. And so, we have five microfarads plus eight, giving us 13.0 microfarads.