# Capacitors connected in series.

by ness9660
Tags: capacitors, connected, series
 P: 400 what do you mean treid all sorts of ways using the series capacitors and Q=VC? did you use $$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$$? did you get all the three equations? (so you could calculate the forth) write them here, lets see whats wrong with them.
 P: 36 Well you would have $$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$$ and $$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_3}$$ and finally $$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$ but how could you find the final charge without using a voltage?
 P: 400 Capacitors connected in series. right, so you got $$30.8*10^{-6}=C_1V$$ and $$23.1*10^{-6}=C_{eq_1}V$$ and $$25.2*10^{-6}=C_{eq_2}V$$ you should be able to find $$C_1$$, $$C_2$$, and $$C_3$$ with these equations (they all depend on the voltage, but because the fourth eq. should use the same voltage, it cancels out...)