# Dissipated energy in RC circuit

Hi there.
I'm trying to evaluate the energy dissipated due to Joule effect on a resistor within an RC circuit
(R, C, and battery in series), with the capacitor initially uncharged.
Till now I just came up with the formula for the current flowing in the circuit
$i(t) = e^{-\frac{t}{RC}}$, resulting from a differential equation.
The energy dissipated by the resistance should be the power over R integrated from 0 to ∞:
$\displaystyle E = \int_0^\infty P(t)dt = \int_0^\infty e^{-\frac{t^2}{(RC)^2}}dt$
which appears to be something I just can't calculate.
I'm quite sure there's a simpler way to do this.. do yo have any hint?

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ehild
Homework Helper
Hi there.
I'm trying to evaluate the energy dissipated due to Joule effect on a resistor within an RC circuit
(R, C, and battery in series), with the capacitor initially uncharged.
Till now I just came up with the formula for the current flowing in the circuit
$i(t) = e^{-\frac{t}{RC}}$, resulting from a differential equation.
The energy dissipated by the resistance should be the power over R integrated from 0 to ∞:
$\displaystyle E = \int_0^\infty P(t)dt = \int_0^\infty e^{-\frac{t^2}{(RC)^2}}dt$
which appears to be something I just can't calculate.
I'm quite sure there's a simpler way to do this.. do yo have any hint?
What is the formula for the power on a resistor? And what do you mean with the integrand? It is not the power.

ehild

Okay, thank you, the power dissipated by the resistor is IV = I^2R, then the integral had to be
$R\int_0^\infty e^{-\frac{2t}{RC}}dt$ which gives $-\frac{R^2C}{2}[e^{-\frac{2t}{RC}}]^\infty_0 = \frac{R^2C}{2} [\frac{Vs}{A}]$ which is... dimensionally wrong?
Edit: ok, by checking again everything I found out the expression for I was dimensionally wrong (e^kt is a pure number). The correct one is $i(t) = I_0e^{-\frac{t}{RC}}$, which make everything else correct.
Thanks a lot!

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ehild
Homework Helper
There is nothing to be thanked, you did it alone Well done!

ehild

Well, I still have some doubts.
In the end, the formula for the energy dissipated during the process of discharging a capacitor is equal to the potential energy stored in the capacitor $\frac{1}{2}Q_0V_0$, which was actually quite obvious, since there's nothing else providing energy. My first intuitive (and wrong) thought was that some extra energy would have been spent producing heat. Evidences show that $R$ only concerns the time needed to complete the discharge.
Now, what would happen if I completely charge a capacitor, and then I connect its plates each other with a PEC with no resistance? Would the discharge be instantaneous, since R is 0? And what about energy? We shouldn't be able to detect any energy change due to joule effect, yet the capacitor no longer possess energy. Where did it go?

ehild
Homework Helper
What is PEC????

Unless it is a superconductor, anything you connect the plates with have some resistance. If it is very small, the current will be very high and heat is produced.

Moreover, if you connect the capacitor plate with something of very small resistance, the current increases to a high value in a very short time. It can be also a spark. The current produces magnetic field. The change of magnetic field produces electric field. That electromagnetic disturbance travels in form an electromagnetic wave and spreads the energy of the capacitor to the environment. You can even hear some noise from your radio if there is a spark nearby: Some of the energy of the capacitor reached the radio.

ehild

NascentOxygen
Staff Emeritus