- #1

mhrob24

- 53

- 9

- Homework Statement
- Three capacitors

C1 = 10.0 µF, C2 = 16.0 µF, and C3 = 29.3 µF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum potential difference across the series combination.

- Relevant Equations
- Capacitors in series circuit: C(eq) = 1/C(1) +1/C(2) +1/C(3) .......

in parallel circuit: C(eq) = C(1) + C(2) + C(3) ........

C = Q/V

I'm not really sure what I need to find exactly. From what I'm seeing, I could give C1 the max potential difference of 125V because it has the lowest capacitance, and because V = Q/C, this means the capacitor with the highest potential difference across its plates will be the one with the lowest capacitance. Then I can just solve for Q, and once I have Q, I'd have everything I need in order to solve for the voltages between C2 and C3...but I'm not sure where to go from there. I usually see questions worded as find the potential differences "across each capacitor in the combination", not "across the series combination". If they were asking for the maximum potential DROP across the series I think I could do that by just finding the lowest voltage in the series (should be C3 since C3 has the highest capacitance out of the 3 capacitors), and then subtracting that from 125 which would give me to total drop in voltage across the series.