Capacitance: How to get 50 microF w/ 80.0 microF

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SUMMARY

The discussion centers on calculating the additional capacitance needed to achieve a total of 50 microfarads (µF) using an existing 80.0 µF capacitor. The correct approach involves using the formula for capacitors in series, specifically Ceq = (C1 * C2) / (C1 + C2). The user initially attempted to use the parallel capacitor formula incorrectly, leading to confusion. The solution requires determining the appropriate value of an additional capacitor to achieve the desired capacitance.

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Z COOL
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Hey gang,
I am doing a Mastering physics question, and it's got me stumped. ><

Q: You need a capacitance of 50.0 microF, but you don't happen to have a 50.0 microfarads capacitor. You do have a 80.0 microF capacitor.

What additional capacitor is needed to get a capacitance of 50 microF?

I was thinking it has something to do w/ C = Q / deltaV. But that would mean that I would need to have some sort of knowledge of Q. I then tried

Ceq = (1/C1 +1/C2)^-1
and solved for 1/C1 to get 49.9875 microF. But that doesn't seem logical (+ I got it wrong ><)

Any suggestions / help? Greatly appreciated!
 
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Well you obviously solved for the unknown incorrectly. Why not try to solve the simpler equation (assuming series capacitors):

[tex]C_{eq} = \frac{C1 \cdot C2}{C1+C2}[/tex]
 

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