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

[MalachiK]

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 transferred to C and to R over a given time interval and then integrate to get the energy transferred 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$$\int^T_0$$e-^2t/RC dt

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

E_R=(2Po/RC)(1 - e^-2T/RC)

It also occurred to me that the total energy delivered to the circuit would be $$\int^T_0$$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
E_T = (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...

E_C= Po $$\int^T_0$$ e^-t/RC - e^-2t/RC dt

Evaluating this gives me

E_C = (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 E_C = E_T - E_R 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 E_C = 0.5 E_T. Is this because there's something seriously wrong with my thinking?

 

[willem2]

$\int e^{at} dt$ is not equal to $a e^{at}$


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.

 

[gneill]

For the capacitor, I get

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

$$P_c = -\frac{1}{2} \tau I_o V_o \left(2 e^{-\frac{T}{\tau}} - e^{-\frac{2T}{\tau}} - 1 \right)$$

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

Now, τ = RC, so

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

But R I_o = V_o, so

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

In the limit as T → ∞ this becomes

$$P_c = \frac{1}{2} C V_o^2$$

 

[MalachiK]

Doh! I am a moron.

Thanks for the help.

 

[rwisz]

First of all, it's great that you're trying to work through this problem on your own. It shows a strong understanding and dedication to learning. However, I do see some errors in your calculations.

The power delivered to the capacitor is not just the potential difference (V) multiplied by the current (I), but also the time derivative of the charge on the capacitor (Q). So the correct expression for the power delivered to the capacitor would be V(t)I(t) + dQ/dt. This is because the capacitor is continuously charging and the current is changing over time.

Also, when calculating the energy delivered to the capacitor, you need to take into account the initial energy stored in the capacitor (0.5*C*V0^2). So the correct expression would be 0.5*C*V0^2 + the integral you calculated.

After making these corrections, you should be able to show that EC = 0.5 ET.

Additionally, I would recommend checking your units throughout your calculations to ensure that they are consistent. This can often help identify errors.

Overall, your approach is correct and you just need to make a few adjustments to your calculations. Keep up the good work!