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franci2005

  • 3 years ago

You have a 1.0 μF, 2.0 μF and 3.0 μF capacitors. What capacitances can you get by connecting them?

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  1. Carl_Pham
    • 3 years ago
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    \[(1.0 + 2.0 + 3.0) \mu{\rm F}\] \[\left(\frac{1.0 + 2.0}{2} + 3.0\right)\mu{\rm F}\] \[\left(\frac{1.0 + 3.0}{2} + 2.0\right)\mu{\rm F}\] \[\left(\frac{2.0 + 3.0}{2} + 1.0\right)\mu{\rm F}\] \[\left(\frac{1.0 + 2.0 + 3.0}{3} \right)\mu{\rm F}\]

  2. NoelGreco
    • 3 years ago
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    Not quite Carl. The first one is correct for all three caps in parallel. The correct formula for capacitors in series is:\[\frac{ 1 }{ C _{Eq} }= \frac{ 1 }{ C _{1} }+\frac{ 1 }{ C _{2} }+\frac{ 1 }{ C _{3} }...\] which reduces to: \[C _{Eq}=\frac{ C _{1}C _{2} }{C _{1}+C _{2} }\] for two capacitors in series.

  3. Carl_Pham
    • 3 years ago
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    RIght you are. For some reason I had a braino and thought we were dealing with resisters. Substitute the inverses for the numbers in all the equations, and invert at the end.

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