The discussion revolves around a circuit with two capacitors of 0.40 µF and 0.50 µF connected in series to a 9V battery. Participants are exploring the charge distribution on the capacitors and the relationship between charge, capacitance, and voltage in series circuits.
Some participants have provided guidance on how to approach the problem by suggesting that the voltages across the capacitors must add up to the total voltage of 9V. There is an ongoing exploration of the implications of the charge being the same for capacitors in series, but no consensus has been reached regarding the specific charge values on each capacitor.
Participants are working under the assumption that the charge on capacitors in series is the same, while also grappling with the calculations needed to determine the voltage across each capacitor. There is mention of potential rounding errors affecting the understanding of the charge distribution.
slammer said:Homework Statement
A circuit composed of 2 capacitors of .40MF and .50MF are connected in series to a 9V battery. Calculate the charge on each
I just have a general question. I know that capacitors in parallel have the same charge.
I found Q to be 1.8MC and I was wondering if 1.8MC is the charge on both caps or is it .9MC charge on both caps.
slammer said:Oops sorry my bad i mean capacitors in series have same charge. But I can't solve for V w/o finding out the capacitance on each capacitor. So what I am trying to say is does the both of the capacitors get 1.8mC of charge or do they each get .9mC?
LowlyPion said:What voltage do you get across each capacitor for both of those possibilities?
Which possibility adds up to 9 v adding the voltages of both capacitors? Because that's the one that's your answer.
Btw, where did the 1.8 mC come from? Didn't you calculate the equivalent capacitance and multiply by the voltage?
slammer said:round errors and I figured it out. 1.8mC gets dumped on both of the caps.