Question

Consider the circuit shown in the figure below:

All the resistors are identical. The charge stored in the capacitor once it is fully charged is
A. 0
B. $\dfrac{5}{13}CV$
C. $\dfrac{2}{3}CV$
D.$\dfrac{5}{8}CV$

Answer 1

Hint: As a very first step, one could find the equivalent resistance of the given circuit by considering the series and parallel combination into account. Thereby you could find the equivalent current in the circuit. Then you could find the voltage drop across the capacitor and hence the charge stored.

Complete answer:
In the question, we are given a circuit in which a number of resistors are connected in various combinations and we have a capacitor and also a battery. Let us now recall the fact that when the capacitor gets charged fully, the voltage across the capacitor is considered equivalent to voltage of the cell. Also, note that all the connected resistors are identical.
We could conclude that the current through the capacitor zero for the fully charged capacitor. On finding the equivalent resistance of the circuit we see the value to be,
${{R}_{eq}}=\dfrac{8R}{11}$
Now, for the current that is flowing through the first loop, we have the ohms law given by,
$I=\dfrac{V}{R}$
Now we could simply substitute the equivalent resistance into this expression to get,
${{I}_{eq}}=\dfrac{11V}{8R}$
What we have to understand here is that the current is being distributed between the $\dfrac{8R}{3}$and $R$resistances that are connected in parallel. So, the current through the $\dfrac{8R}{3}$ resistance would be $\dfrac{3V}{8R}$
Now the voltage drop could be given by,
$V-\dfrac{3V}{8R}R=\dfrac{5V}{8}$
Now the charge stored in the capacitor would be,
Q=CV
Substituting the voltage drop found above here, we get,
$Q=\dfrac{5CV}{8}$

So, we found the correct answer to be D.

Note:
You may have noted that we have directly substituted the equivalent resistance of the given combination as this is a pretty basic stuff. For the resistors connected in series equivalent resistance is given by the sum and for the resistors connected in parallel, it would be,
${{R}_{EQ}}=\dfrac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}$
Using these two basic steps, we could easily find the answer.