I RC charging - intuition behind equal dissipated energies?

coquelicot

Summary
Resistor R charging a capacitance C. The energy dissipated by R is equal to the energy stored by C.
Assume that a resistor R charges a capacitor C, whose other terminal is connected to the ground.
The charge at time t = 0 is assumed to be null and the supply voltage is equal to V.
We have, as is well known, $i = \frac{V}{R} e^{-\frac{t}{RC}}$. Integrating $\frac{i^2}{R}$ between t = 0 and infinity, I found that the energy dissipated by R is $\frac{1}{2} CV^2$. That is, the same as the potential energy stored in the capacitor after it has completed its charging.

Is there an intuitive way to understand this equality ?

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ZapperZ

Staff Emeritus
2018 Award
Summary: Resistor R charging a capacitance C. The energy dissipated by R is equal to the energy stored by C.

Assume that a resistor R charges a capacitor C, whose other terminal is connected to the ground.
The charge at time t = 0 is assumed to be null and the supply voltage is equal to V.
We have, as is well known, $i = \frac{V}{R} e^{-\frac{t}{RC}}$. Integrating $\frac{i^2}{R}$ between t = 0 and infinity, I found that the energy dissipated by R is $\frac{1}{2} CV^2$. That is, the same as the potential energy stored in the capacitor after it has completed its charging.

Is there an intuitive way to understand this equality ?
Conservation of energy?

BTW, the resistor does not "charge" the capacitor. It can't physically do that. A power or voltage source can.

Zz.

coquelicot

Yes you are right regarding your BTW. But give me some freedom.

NB: I don't see where there is a point to the conservation of energy here. Some energy is wasted in the air, and another energy is stored as a potential energy by the capacitor.

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2018 Award

ZapperZ

Staff Emeritus
2018 Award
Yes you are right regarding your BTW. But give me some freedom.

NB: I don't see where there is a point to the conservation of energy here. Some energy is wasted in the air, and another energy is stored as a potential energy by the capacitor.
Energy wasted in the air? What is this?

Zz.

coquelicot

ZapperZ, the energy wasted in the air is called "heat".

ZapperZ

Staff Emeritus
2018 Award
ZapperZ, the energy wasted in the air is called "heat".
But that's ohmic heating due to the resistor, which is what you were asking about. What do you think is the nature of the energy lost in the resistor? There is no "heat" being lost in the ideal capacitor.

Zz.

coquelicot

Lord Jestocost, thx for the link. It does not give the answer to the question though (at least according to my understanding).

coquelicot

ZapperZ , I think everything is clear. I'm not posting at this level.

vanhees71

Gold Member
Let's do the energy balance. You are right, the energy you calculated,
$$\int_0^{\infty} \mathrm{d} t i^2(t)/R=\frac{1}{2} C V^2$$
is the energy dissipated into heat.

But that's not all the battery must do for you. It also must build up the electric field between the plates of the capacitor. This energy is the energy stored in the field. You can evaluate it by thinking about the work need to carry the little charge portion $\mathrm{d} Q'=\mathrm{d} t i(t)$ from the one plate to the other against the field already built up at time $t$, which is $\mathrm{d} W=\mathrm{d} Q' V_C(t)=\mathrm{d} Q' Q'/C$. Now with your initial condition you get
$$W=\int_{Q}^{Q} \mathrm{d} Q' Q'/C=\frac{Q^2}{2C}=\frac{C V^2}{2},$$
i.e., together the battery has to provide the energy to carry the charges from one plate to the other, "working" against the already built-up field, and the energy to heat up the resistor, i.e.,
$$E=W+W_{\text{therm}}=C V^2.$$
The energy stored in the capacitor is stored in the field between the plates and just as large as the heat, which is dissipated in the resistor.

Another way to calculate the energy stored in the capacitor is to think about discharging it through the resistor. So let the capacitor be charged at voltage $V_0$ at $t=0$ and now connect it to a resistor. The circuit equation gives
$$i=-\mathrm{d} Q/\mathrm{d} t=V(t)/R$$
Since $Q= C V$,
$$\dot{V}=-\frac{1}{R C} V \; \Rightarrow \; V(t)=V_0 \exp[-t/(RC)].$$
The total heat dissipated in the resistor thus is
$$W_{\text{heat}}=\int_0^{\infty} \mathrm{d} t V^2(t)/R= \frac{V_0^2}{R} \int_0^{\infty} \mathrm{d} t \exp[-2t/(RC)]=\frac{C}{2} V_0^2.$$
Since the capacitor is completely discharged for $t \rightarrow \infty$ this heat must be the total electric-field energy stored in the capacitor before discharging it.

coquelicot

In the first part of your answer, you say what I said with more details (right?)

The second part is more interesting, with your idea to see the reverse process: discharging the capacitor through the resistor. This needs no calculus though: it is evident that after the cap has discharged, all the potential energy that was stored in it has been dissipated as heat by the resistor.

This may be the key to understand intuitively the equality $E_{cap} = E_{therm}$, since at the level of the infinitesimals, the process of charging and discharging is the same up to a sign change. Thank you for your answer.

coquelicot

I just want to build up upon the idea of Vanhees71 and detail my remark above.

Dealing with elementary calculus, it is easy to show that the current flowing through an RC circuit whenever the capacitor is charged at constant voltage V via the resistor R is given by
$i = i_0 e^{-t\over RC}$, with $i_0 = i(t = 0) = V/R$.

Now, after a very long time (assumed to be infinite) the capacitor is charged up to the voltage V and the current stop. Then one may connect the terminal of R that was previously fed with the supply voltage V to the ground (that is, it is now connected to the other side of C), in order to discharge the capacitor. Again, dealing with elementary calculus, you find that the current flowing through the circuit is
$i' = i'_0 e^{-t\over RC}$, with $i'_0 = i'(t=0) = -V/R$.
So, we have $i' = -i$.
But after the discharge, all the potential energy $E_{cap}$ of the cap has obviously been dissipated as heat by R, that is, the dissipated energy is $E_{therm} = E_{cap}$ for the discharge.
Since we have $i= -i'$ for all $t$, it follows that the energy dissipated by the resistor as the capacitor was charging is also equal to $E_{therm}$, that is, to $E_{cap}$.

"RC charging - intuition behind equal dissipated energies?"

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