Question

Let C be the capacitance of a capacitor discharging through a resistor R. Suppose ${{t}_{1}}$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and ${{t}_{2}}$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $\dfrac{{{t}_{1}}}{{{t}_{2}}}$ will be
$\begin{align}
  & (A)1 \\
 & (B)\dfrac{1}{2} \\
 & (C)\dfrac{1}{4} \\
 & (D)2 \\
\end{align}$

Answer 1

Hint: First find the initial and final energy of the capacitor. Then by applying the concept that the discharging curve always exponentially decreases find ${{q}_{2}}$. Taking natural logarithm on both sides we get a new equation. Then substituting and taking ratio we will get the value of $\dfrac{{{t}_{2}}}{{{t}_{1}}}$ .
Formula used:
${{q}_{2}}={{q}_{1}}{{e}^{\dfrac{-t}{CR}}}$
where, ${{q}_{1}}\And {{q}_{2}}$ are the initial and final charges.
 C is the capacitance.
R is the resistance.
t is the time taken.

Complete answer:
Given C is the capacitance, R is the resistor and t is the time taken.
Initial energy of the capacitor is given by,
$E_1$=$\dfrac{1}{2}\dfrac{{{q}^{2}}}{c}$
Final energy of the capacitor,
$E_2$=$\dfrac{1}{2}{{E}_{1}}=\dfrac{{{q}_{1}}^{2}}{4c}$
${{t}_{_{2}}}$ is the time taken to reduce one fourth of initial value.
${{q}_{2}}={{q}_{1}}{{e}^{\dfrac{-t}{CR}}}$
$\Rightarrow \dfrac{{{q}_{2}}}{{{q}_{1}}}={{e}^{\dfrac{-t}{CR}}}$
Taking natural logarithm on both sides,
$\Rightarrow \ln \left( \dfrac{{{q}_{2}}}{{{q}_{1}}} \right)=\ln \left( {{e}^{\dfrac{-t}{CR}}} \right)$
$\Rightarrow \ln \dfrac{1}{\sqrt{2}}=-\dfrac{t}{CR}$
Therefore \[\ln \left( \dfrac{1}{\sqrt{2}} \right)=-\dfrac{{{t}_{1}}}{CR}\]
Similarly,
$\ln \dfrac{1}{4}=-\dfrac{{{t}_{2}}}{CR}$
Thus,
$\dfrac{{{t}_{1}}}{{{t}_{2}}}=\dfrac{\ln (\dfrac{1}{\sqrt{2}})}{\ln \left( \dfrac{1}{4} \right)}$ = $\dfrac{1}{4}$

Hence option(C) is correct.

Additional information:
As charge flows from one plate to another through the resistor the charge is neutralized and so the current falls and rate of decrease of potential difference also falls. Eventually the charge on the plate is zero and the current and the potential difference are also zero. That is, the capacitor is fully discharged.

Note:
The charging curve always increases exponentially starting from zero and the discharging curve exponentially decreases. And at a particular point charging and discharging voltages across the capacitor appears the same. The point of that time is ${{T}_{\dfrac{1}{2}}}$. Then at ${{T}_{\dfrac{1}{2}}}$charging and discharging voltages are the same. As charge flows from one plate to another through the resistor the charge is neutralized and so the current falls and rate of decrease of potential difference also falls. Eventually the charge on the plate is zero and the current and the potential difference are also zero. That is, the capacitor is fully discharged.