Question

A capacitor loses half of its voltage every 2 seconds. If the initial voltage is V, what will the voltage be after 8 seconds?

& \text{A}\text{. }\dfrac{V}{2} \\\ & \text{B}\text{. }\dfrac{V}{8} \\\ & \text{C}\text{. }\dfrac{V}{16} \\\ & \text{D}\text{. }\dfrac{V}{32} \\\ \end{aligned}$$

Answer 1

After each half-life capacitor loses half of its charge or voltage across it. You can calculate the voltage across the capacitor after every 2 seconds by taking half of its previous value. Another way is to use the generalized expression to calculate voltage after a given number of half-lives.

Formula used:
Final voltage after n half-lives,
${V}'=\dfrac{V}{{{2}^{n}}}$
where,
V is the initial voltage across the capacitor

Complete step by step answer:
The time after which a capacitor loses its half of the charge or voltage of its initial value is called the half-life.
Given that the initial voltage is V, therefore, we assume at $t=0s$the voltage across the capacitor is V.
Also, given that the capacitor loses half of its voltage every 2 seconds, therefore,
After 2s voltage across the capacitor will be $\dfrac{V}{2}$
After 4s voltage across the capacitor will be $\dfrac{V}{4}$
After 6s voltage across the capacitor will be $\dfrac{V}{8}$
After 8s voltage across the capacitor will be $\dfrac{V}{16}$
Sometimes elapsed time is very large so it will be a very lengthy and complicated process to calculate this way.
We can find the remaining voltage as follows:
If t is the time of observation and ${{T}_{1/2}}$ is the half-life then the number of half-lives in a given time of observation is
$n=\dfrac{t}{{{T}_{1/2}}}$
Given that, $t=8s$ and ${{T}_{1/2}}=2s$
Therefore,
$n=\dfrac{8}{2}=4s$
If V is the initial voltage then the voltage after n half-lives is given as
${V}'=\dfrac{V}{{{2}^{n}}}$
Thus, after 4 half-lives, the final voltage will be
${V}'=\dfrac{V}{{{2}^{4}}}=\dfrac{V}{16}$
Therefore, the voltage after 8 seconds will be equal to $\dfrac{V}{16}$.

Answer - C. $\dfrac{V}{16}$

Note:
In RC circuit, during the discharging of the capacitor through a resistance R, the voltage across capacitor having capacitance C at any time t is given as $V={{V}_{0}}{{e}^{-t/RC}}$, where ${{V}_{0}}$ is the initial voltage across the capacitor and RC is known time constant denoted by $\tau$. Discharging of a capacitor is an exponential decay. Therefore all the terms used in capacitor discharge also apply for other exponential decays like radioactive decay.