Question

A capacitor loses half of its voltage every 1 seconds. If the initial energy it holds is E, how much energy will it have after 3 seconds?
$\begin{array}{l} {\rm{A}}{\rm{. }}\dfrac{E}{3}\\\ {\rm{B}}{\rm{. }}\dfrac{E}{8}\\\ {\rm{C}}{\rm{. }}\dfrac{E}{9}\\\ {\rm{D}}{\rm{. }}\dfrac{E}{{64}} \end{array}$

Answer 1

Hint : We need the relation between the energy stored in a capacitor and the voltage on the capacitor. The energy is directly proportional to the square of the voltage on the capacitor. Now by using given information on the reduction in voltage with time, we can find the required answer.
Formula used:
The energy stored in a capacitor is given as
$E = \dfrac{1}{2}C{V^2}$
Here E represents the energy that is stored in the given capacitor; C is the capacitance of the given capacitor while V represents the voltage of the capacitor.

Complete step by step solution :
The capacitor is an electrical device which is used to store electrical energy in it. The capacitance of a capacitor tells how much charge the capacitor can store per unit voltage. The energy stored in a capacitor is given in terms of the capacitance of a capacitor and the voltage on the capacitor by the following formula:
$E = \dfrac{1}{2}C{V^2}$
We are given that a capacitor loses half of its voltage every 1 second. We know from the above relation that energy stored in a capacitor is directly proportional to the square of the voltage.
Now after 1 second, the voltage on the capacitor is given as
$V' = \dfrac{V}{2}$
After two seconds, the voltage has been further reduced by half and is given as
$V'' = \dfrac{{V'}}{2} = \dfrac{V}{4}$
After three seconds, the voltage has reduced further by half and is given as
$V''' = \dfrac{{V''}}{2} = \dfrac{V}{8}$
Now the energy stored in the capacitor after three seconds can be written as follows:
$E''' = \dfrac{1}{2}CV''{'^2} = \dfrac{1}{2}C{\left( {\dfrac{V}{8}} \right)^2} = \dfrac{1}{2}C\left( {\dfrac{{{V^2}}}{{64}}} \right) = \dfrac{E}{{64}}$
Here we have used the initial value of the energy stored in the capacitor. The energy stored after 3 seconds is $\dfrac{1}{{64}}$ times the initial energy in the capacitor. Hence, the correct answer is option D.

Note : 1. The capacitance of the capacitor remains unchanged, only the voltage on the capacitor is decreasing.
2. The student should not go about solving the question by simply relating voltage directly to the energy. If we simply go on reducing energy in the same way as voltage then we get option B as answer but that method is wrong.