Question

It takes a capacitor 18 ms for the potential difference across it to fall to half the original value, Vo / 2. How long does it take for the stored energy to have fallen to one-16th of its original value?

(a) 18 ms

(b) 144 ms

(c) 36 ms

(d) 72 ms

Answer 1

Let's first find the time constant from the capacitor's discharging equation:

$$V(t)=V_0e^{-\dfrac{t}{\tau}},$$ $$0.5V_0=V_0e^{-\dfrac{t}{\tau}},$$ $$ln(0.5)=ln(e^{-\dfrac{t}{\tau}}),$$ $$\tau=-\dfrac{t}{ln(0.5)}=-\dfrac{18\cdot10^{-3}\ s}{ln(0.5)}=0.026\ s=26\ ms.$$

The energy stored in the capacitor can be written as follows:

$$E=\dfrac{1}{2}CV^2.$$

As we know, from the condition of the question, the energy stored in the capacitor falls to one-16th of its original value, that means that the potential difference across the capacitor falls to $V_0/4$ of its original value:

$$E_{new}=\dfrac{1}{2}C(\dfrac{V}{4})^2=\dfrac{1}{16}\cdot\dfrac{1}{2}CV^2=\dfrac{1}{16}E.$$

Therefore, we can find the time that takes for the stored energy to have fallen to one-16th of its original value from the same capacitor's discharging equation:

$$0.25V_0=V_0e^{-\dfrac{t}{\tau}},$$ $$ln(0.25)=ln(e^{-\dfrac{t}{\tau}}),$$ $$t=-\tau ln(0.25)=-0.026\cdot ln(0.25)=0.036\ s=36\ ms.$$

Therefore, the correct answer is (c).

Answer:

(c) 36 ms