[SOLVED] calculating maximum energy stored in capacitor

I am working through a problem in Edminister's Electric Circuits, 6th edition (I notice it has been recommended once or twice in a different forum).

The problem goes like this:

A capacitance of C farads has a current i = (V_m/R)e^-(t/RC) (A). Show that the maximum stored energy is (1/2)CV_m².

I have taken the basic equation i = Cdv/dt and integrated to get v. Then I multiply the two to get power. Unfortunately this looks as if it is not going anywhere so I have not attempted to take the time integral of this to get energy.

The only assumption I can make is that the maximum energy will be stored at t = infinity, i.e. when it is fully charged.

The equation suggests a resistor in series but no hint of this is given in the question so I have not assumed there is one. I have worked solely from the facts given along with my assumption above.


Have I tackled this the wrong way? I don't mind working through the math if someone can point me in the right direction.

Thanks for any help with this one.

As you said, the maximum energy is stored only at t=infinity. But the point is, the equation derived (1/2cv^2) gives the energy stored in the capacitor when it is charged to v volts. It is not required to know how much time 't' it has taken to charge to v volts. Hence you dont encounter either t or the resistance in the charging path while deriving the equation.

The total work done when moving charge Q starting at V=0 equals:
W=1/2QV
since Q=CV
W=1/2(CV)V=1/2CV²

Compute the current needed to discharge 50%Vc in 1 second
Ic=C*50%Vc then the work done = 1/2CV²

jmvalks said:

......
A capacitance of C farads has a current i = (V_m/R)e^-(t/RC) (A). Show that the maximum stored energy is (1/2)CV_m².

Click to expand...

Is the discharge of a capacitor through a resistor.

--> Instant power: P = i²R

--> Then W = ∫(i²R)dt between 0 to ∞ (at infiniy is fully discharged)

integrating... W = ½ C V_m²

I have taken the basic equation i = Cdv/dt

Click to expand...

Good alternative

and integrated to get v. Then I multiply the two to get power.

Click to expand...

:shock: No. Multiply first to get power, then integrate to get energy.


Regardless de above expression for the current, if you have at t=0 --> v=0, and at t=T --> v=V_m

Then
\[W = \displaystyle\int_{0}^{T}{v(C\frac{dv}{dt})dt} = C\int_{v(t=0)}^{v(t=T)}{v\;dv} = C\int_{0}^{V_m}{v\;dv} = \frac{1}{2}C\;V_m^2\]

yes I agree with all above.
the equation you give is the stock equation of a working electrical engineer, for the voltage on a cap that's charging from a voltage v via a resistor r.

Heres a sheet of the maths that's generally useful in general hard switched smps work.
Generally theres rarely a need to do integration, unless you are doing research style work.

I had a junior engineer telling me that laplace transforms would be needed to design a 5w boost converter......but not so.

I think its good if people know what integration is but don't worry if you cant remember standard integrals. as you know, Integration is just the act of finding the total of some quantity by splitting it into a large number of equal pieces and then summing those up.......rarely is it used outside of research as theres standard forms etc.

Thank you all for your help. They are all instructive in different ways. The calculus solutions are the ones that did it for me.I think that's what the authors were really looking for.