Question

A capacitor discharges through a resistance. The stored energy U0 in one capacitive time constant falls to:
A) $\dfrac{{{U_0}}}{{{e^2}}}$
B) $e{U_0}$
C) $\dfrac{{{U_0}}}{\begin{array}{l} e\\\\\ \end{array}}$
D) none of these

Answer 1

First we have to write RC circuit q and U0 values with the help of that we find stored energy at any instant. According to the question we have to finalize the answer by calculating one capacitive time constant stored energy. We have to show the value of stored energy after one capacitive constant fall.

Complete answer:
Initial stored energy of capacitance is
${U_0} = \dfrac{{{q_0}^2}}{{2C}}$ ------(1)
Where q0 charge on capacitance
C is the capacitance.
While discharging through the circuit it gives.
$q = {q_0}.{e^{\dfrac{{ - t}}{\tau }}}$
After one time constant above equation become
$\Rightarrow q = {q_0}{e^{ - \dfrac{\tau }{\tau }}}$ is cancel above and below we get
$\begin{array}{l} \therefore q = \dfrac{{{q_0}}}{e}\\\ \end{array}$
Therefore we get the charge in one capacitive time constant
Then total stored energy become
$U = \dfrac{{{q^2}}}{{2C}}$
Substitute the q value in the above equation
$\begin{array}{l} \Rightarrow U = \dfrac{{\dfrac{{{q_0}^2}}{{{e^2}}}}}{{2C}}\\\ \end{array}$
We can write this in another form
$\begin{array}{l} \Rightarrow U = \dfrac{1}{{{e^2}}}\dfrac{{{q_0}^2}}{{2C}}\\\ \end{array}$
From the equation one we have
$\therefore U = \dfrac{{{U_0}}}{{{e^2}}}$

Hence, option A is the correct answer.

Additional information:
Capacitors are nothing but it provides temporary storage of energy in circuits and can be made to release it when required. By the property of a capacitor we can say that capacitance like that characterises its ability to store energy is called its capacitance.
The stored energy in a capacitor can be associated with the electric field. In other words energy can be associated with the existence of an electric field. The study of capacitors and capacitance leads us to an important aspect of electric fields,so we can say the energy of an electric field.

Note:
If we go on giving charge to a conductor its potential keeps on rising, so we can say the charge is directly proportional to the potential. This is a very basic question but we have to put some effort to remember the steps of the derivation.