I'm just confused in general about what happens with charge, voltage, etc in parallel and series circuits with capacitors.
Anyways, I'm trying to find the total energy stored in $2$ equivalent capacitors in series vs in parallel, vs 1 capacitor alone. They're charged by a battery that has a constant voltage and current.
If I use $\rm P=\frac{1}{2}CV^2$ , I know using the equivalent capacitance for $2$ capacitors in series would be : $$\rm C_{eq}=\frac{1}{\frac{1}{C}+\frac{1}{C}}=\frac{1}{2}C$$
And for parallel it's just : $$\rm C_{eq}=2C$$
The equivalent capacitance for a single capacitor is simply $\rm C$.
My question(s) :
(a) What I do about the voltage? Is it the same for both in all $3$ cases ($2$ capacitors in series, $2$ in parallel, and 1 lone capacitor)?
(b) What if I wanted to instead use $P=\frac{1}{2}\frac{Q^2}{C}$? What happens with the charge in the $3$ cases?
Thanks for the help folks
Best Answer
I'm assuming the battery charged the series, parallel, and single capacitors to the same battery voltage $V$.
For series capacitors the charge on each capacitor is the same regardless of the value of each capacitance, while the voltage on each depends on the capacitance. For two equally sized capacitors, the charge and voltage are the same and the voltage across each is one half the battery voltage. Therefore the total stored energy is
$$E_{series}=\frac{1}{2}C(V/2)^2+\frac{1}{2}C(V/2)^2=\frac{CV^2}{4}$$
Or, in terms of the single equivalent capacitance of $C/2$
$$E_{equiv}=\frac{1}{2}\frac{C}{2}V^{2}=\frac {CV^2}{4}$$
The capacitors in parallel have the same voltage across them and the charge depends on the capacitance. So the total stored energy for two equal parallel capacitors is
$$E_{parallel}=\frac{1}{2}CV^2+\frac{1}{2}CV^2=CV^2$$
Or, in terms of the single equivalent parallel capacitance of $2C$
$$E_{equiv}=\frac{1}{2}(2C)V^{2}=CV^2$$
Finally, for the single capacitor
$$E=\frac{1}{2}CV^2$$
Just plug in $\frac{Q}{C}$ for $V$ in each of the above equations.
Hope this helps.