Capacitance: How to get 50 microF w/ 80.0 microF

[Z COOL]

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!

 

[ranger]

Well you obviously solved for the unknown incorrectly. Why not try to solve the simpler equation (assuming series capacitors):

$$C_{eq} = \frac{C1 \cdot C2}{C1+C2}$$

 

[astrokreng]

Hello there! It seems like you're on the right track with using the equation Ceq = (1/C1 +1/C2)^-1 to find the equivalent capacitance. However, since you already have an 80.0 microF capacitor, you can rearrange the equation to solve for C2 (the additional capacitor you need) instead of C1. This would give you C2 = (1/Ceq - 1/C1)^-1, where Ceq is the desired capacitance of 50.0 microF and C1 is the capacitance of the 80.0 microF capacitor you already have. Solving for C2 gives you a value of 126.3 microF, which is the additional capacitance you need to get a total capacitance of 50.0 microF. I hope this helps and good luck with your Mastering physics question!