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

In summary, the author is trying to solve a problem where energy is lost in a circuit. He first considers the power transferred to and from the capacitor and then the energy lost in the resistor. He gets that the energy lost in the resistor is V0I0e-2t/RC dt and the energy lost in the capacitor is EER=(2Po/RC)(1 - e-2T/RC). He then calculates P_c as Po/RC*(-\frac{1}{2}*tau*I_o*V_o^2) and realizes that he is a moron because he forgot to integrate.
  • #1
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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[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 occurred 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?
 
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  • #2
[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
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]
 
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  • #4
Doh! I am a moron.

Thanks for the help.
 
  • #5


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!
 

1. How is energy lost when charging a capacitor?

Energy is lost when charging a capacitor due to the resistance in the circuit. This resistance causes the capacitor to heat up, and some of the electrical energy is converted into thermal energy.

2. Can you explain the formula for calculating energy lost when charging a capacitor?

The formula for calculating energy lost when charging a capacitor is E = 0.5 x C x V2, where E is the energy lost in joules, C is the capacitance in farads, and V is the voltage across the capacitor.

3. How do I know the resistance value to use in the energy loss calculation?

The resistance value to use in the energy loss calculation can be determined by measuring the total resistance in the circuit using a multimeter. Alternatively, if the resistance value is not known, it can be estimated based on the type of material used in the circuit and the length and thickness of the wires.

4. Is there a way to reduce energy loss when charging a capacitor?

There are several ways to reduce energy loss when charging a capacitor. One way is to use a circuit with lower resistance. Additionally, using capacitors with higher voltage ratings can also reduce energy loss. Properly sizing and selecting the components in the circuit can also help minimize energy loss.

5. Can someone check my calculations for energy loss when charging a capacitor?

Yes, someone can check your calculations for energy loss when charging a capacitor. It is always a good idea to have a second person review your work to ensure accuracy. Additionally, double-checking your calculations and using multiple methods for verification can also help validate your results.

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