Question regarding Energy in Circuits with Capacitors

[Phy_TR]

Hey everyone!

I have a doubt that relates to energy stored in different capacitors (when they're in a circuit). I understand how in capacitors connected in parallel, the capacitor with the biggest capacitance stores the most energy because:

E = 1/2 CV²

So naturally the bigger the capacitance the bigger the stored energy. However, I don't understand how this is reversed with two capacitors connected in series. Any help would be greatly appreciated.

Thank you in advance and I hope I posted this in the correct forum!

 

[sophiecentaur]

Hey everyone!

I have a doubt that relates to energy stored in different capacitors (when they're in a circuit). I understand how in capacitors connected in parallel, the capacitor with the biggest capacitance stores the most energy because:

E = 1/2 CV²

So naturally the bigger the capacitance the bigger the stored energy. However, I don't understand how this is reversed with two capacitors connected in series. Any help would be greatly appreciated.

Thank you in advance and I hope I posted this in the correct forum!

The linking formula is Q = CV
Both series capacitors have the same charge so the PD across each one is inversely proportional to the capacitance value. Fit that into your Energy equation and you have your answer; it's in the V² term, which 'beats' the C term

 

[Wee-Lamm]

Consider the result of connecting 2 capacitors is the opposite to connecting 2 resistors, as capacitors will allow current to pass through and will oppose a change in voltage.

With 2 equal capacitors connected in parallel, they will equally share the voltage and accordingly the current. The capacitors will store equal amounts of energy during the upswing phase of the voltage supply and release equal amounts during the downward sweep, noting that voltage will lag the current by 90 degrees in phase. Using capacitors of different values will shift the amount of energy through each one but as they share a common input and output line, the output will be reflective of the input save for the voltage being 90 degrees behind the current.

With any 2 capacitors connected in series; the second capacitor can only be charged by the energy released by the first, as such they will all have the same charge. Even if the second capacitor is considerably larger in value than the first, it is still limited in how much energy is released to it. As such, the voltage is divided across all of the capacitors while the current remains constant. This in turn declares that total capacitance will be less than the value of the smallest capacitor in the chain by virtue of Kirchoff's Voltage Law.