Question

In the given circuit, the charge on $4\mu F$ capacitor will be:

${\text{A}}{\text{. 5}}{\text{.4}}\mu {\text{C}} \\\ {\text{B}}{\text{. 24}}\mu {\text{C}} \\\ {\text{C}}{\text{. 13}}{\text{.4}}\mu {\text{C}} \\\ {\text{D}}{\text{. 9}}{\text{.6}}\mu {\text{C}} \\\$

Answer 1

First, we need to calculate the equivalent capacitance of the given combination of capacitors and find out the total charge stored in the circuit. Then we need to use the fact that voltage remains same in parallel and charge gets distributed in parallel and vice-versa to find the amount of charge stored in the $4\mu F$ capacitor.

Formula used:
For a capacitor, we have the following relation for the charge stored in it.
$Q = CV$

Complete step by step answer:
For the given circuit, let us first find out the equivalent capacitance. For the two capacitors connected in parallel to each other, if C’ is their equivalent capacitance then it is given as
$C' = 5\mu F + 1\mu F = 6\mu F$
Now this equivalent capacitance is connected in series with the $4\mu F$ capacitor, if C’’ is their equivalent capacitance then it is given as
$C'' = \dfrac{{6\mu F \times 4\mu F}}{{6\mu F + 4\mu F}} = \dfrac{{6\mu F \times 4\mu F}}{{10\mu F}} = 2.4\mu F$
Now this equivalent capacitance is connected in parallel with the $3\mu F$ capacitor. Therefore, the equivalent capacitance of the whole circuit is given as
$C = C'' + 3\mu F = 2.4\mu F + 3\mu F = 5.4\mu F$
Now the voltage supplied by the battery is given as
$V = 10V$
Now we can find out the amount of charge in the circuit by using the following formula.
$Q = CV = 5.4\mu F \times 10V = 54\mu C$
Now we know that voltage remains the same in parallel combination while the charge gets distributed in the parallel combination. So, the voltage will be 10V through the $3\mu F$ capacitor, and hence, the charge through this capacitor will be $10V \times 3\mu F = 30\mu C$. Now the amount of charge stored in the upper branch of the capacitor is $54\mu C - 30\mu C = 24\mu C$. Since, the amount of charge will remain the same in series combination of capacitors, therefore, the charge through $4\mu F$ capacitor is $24\mu C$.
This is the required answer and hence, the correct answer is option B.

Note:
For calculating the equivalent capacitance for a given combination of capacitors, we need to remember that the rules for capacitors are opposite to that used for calculating equivalent resistance. When connected in series, the equivalent capacitance uses the formula used for calculating equivalent resistance in parallel combination while when connected in parallel, the equivalent capacitance is equal to the sum of the capacitances.