The charges stored in each capacitor ${C_1}$ and ${C_2}$ in the circuit shown below are.

$A.6\mu C,6\mu C$
$B.6\mu C,3\mu C$
$C.3\mu C,6\mu C$
$D.3\mu C,3\mu C$

Hint: Find out the current flowing through the circuit which is given by the ratio of voltage to the equivalent resistance $I = \dfrac{V}{R}$ . Voltage is the product of current and resistance which need to be calculated across the capacitors. The capacitors are connected in series so the effective capacitance will be evaluated from $C = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}$ . Now, charge is the product of capacitance and voltage which can be found by substituting the data. The charge of both the capacitors will be the same due to their series connection.

Complete step-by-step answer:

Consider the capacitors as fully charged, then current is not drawn from the cell.
The equivalent resistance of the circuit is \[6 + 3 + 3 = 12\]
Current of the circuit is given by,
$I = \dfrac{V}{R}$
Given that, $V = 12V$
$I = \dfrac{{12}}{{12}}$
$I = 1A$
Hence the current in the circuit is 1A.
Now the potential across the capacitors is
$V = IR$
$V = 6I + 3I + 3I$
$V = 9V$
The net capacitance between them is
$C = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}$
$C = \dfrac{2}{{2 + 1}}$
$C = \dfrac{2}{3}\mu F$
Now the equivalent charge is given by
$Q = CV$
$Q = \dfrac{2}{3} \times 9$
$Q = 6\mu C$
The capacitors are in series so the charge will be the same on both capacitors that is $Q = 6\mu C$.

The correct option is A.

Note: the effective capacitance in parallel grouping is given by,
$C = {C_1} + {C_2} + {C_3}$
For n identical capacitors in series, ${C_e} = \dfrac{C}{n}$
In parallel,${C_e} = nC$ .