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Energy lost charging a capacitor - can someone check my calculations.

  • Thread starter MalachiK
  • Start date
  • #1
137
4

Homework Statement


I'm trying to show that not all (half) of the energy supplied by the power supply ends up in the capacitor when the capacitor is charged through a resistor. I've looked at some of the other threads on these sorts of topics and I'd thought I'd have a go at working through it myself. The treatments of this problem that I've seen seem to consider the work done moving each charge element through a potnetial differnece but I thought I could do this another way. Anyway, it's not going well and I'd appreciate any advice.


Homework Equations



I've included these in the working below...


The Attempt at a Solution


It seems to me that you can start by considering the power transfered to C and to R over a given time interval and then integrate to get the energy transfered in this time.

At first, all of the p.d. is dropped across R. Let's call this V0. We can also call the initial current I0. None of the p.d. is dropped across the cap at first.

After time t, the p.d. across R is V0e-t/RC and the p.d. across the cap is V0(1-e-t/RC). The current is I0e-t/RC.

At ant time t, the power delivered to the resistor is V(t)I(t). So after some time T, the energy lost as heat in the resistor should be V0I0[tex]\int^T_0[/tex]e-2t/RC dt

Setting Vo Io as Po and with some calculating I get...

ER=(2Po/RC)(1 - e-2T/RC)

It also occured to me that the total energy delivered to the circuit would be [tex]\int^T_0[/tex]Vo Io e-t/RC dt [the total p.d. across the circuit doesn't change, only the way it is shared out changes. The current falls with time.]


Evaluating this gives me
ET = (Po/RC)(1 - e-T/RC)


So far so good (unless I've made some stupid mistake!) Now for the energy to the capacitor. Following the same reasoning as before I get that...

EC= Po [tex]\int^T_0[/tex] e-t/RC - e-2t/RC dt

Evaluating this gives me

EC = (Po/RC)(2e-2T/RC -e-T/RC - 1)


Now I'm doubtful of this result because it looks like it's got too many terms in it, but on the other hand, when I calculate EC = ET - ER I get the same result so at least if I'm wrong I'm internally consistent!

At this point my algebra fails me and I can't see how I can compare my expressions for each of the energies to show something like EC = 0.5 ET. Is this because there's something seriously wrong with my thinking?
 
Last edited:

Answers and Replies

  • #2
1,953
248
[itex]\int e^{at} dt [/itex] is not equal to [itex] a e^{at} [/itex]


You want the energy for time from 0 to infinity, so you have to take the limit for t->inf,
wich should be easy as all the exponentials become 0.
 
  • #3
gneill
Mentor
20,793
2,773
For the capacitor, I get

[tex] P_c = \int_0^T I_o e^{-\frac{t}{\tau}} V_o \left(1 - e^{-\frac{t}{\tau}}\right) dt [/tex]

[tex] P_c = -\frac{1}{2} \tau I_o V_o \left(2 e^{-\frac{T}{\tau}} - e^{-\frac{2T}{\tau}} - 1 \right) [/tex]

[tex] P_c = \frac{1}{2} \tau I_o V_o \left(1 - e^{-\frac{2T}{\tau}} + e^{-\frac{T}{\tau}} \right) [/tex]

Now, τ = RC, so

[tex] P_c = \frac{1}{2} R C I_o V_o \left(1 - e^{-\frac{2T}{\tau}} + e^{-\frac{T}{\tau}} \right) [/tex]

But R Io = Vo, so

[tex] P_c = \frac{1}{2} C V_o^2 \left(1 - e^{-\frac{2T}{\tau}} + e^{-\frac{T}{\tau}} \right) [/tex]

In the limit as T → ∞ this becomes

[tex] P_c = \frac{1}{2} C V_o^2 [/tex]
 
Last edited:
  • #4
137
4
Doh! I am a moron.

Thanks for the help.
 

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