Four capacitors and a battery are connected as shown in figure. If the potential difference across the 7 $\mathrm{\mu F}$ capacitor is 6 V then which of the following statement(s) is/are correct?

$C_{\mathrm{eq}}^{\mathrm{\prime }}=3.9+{\left(\frac{1}{3}+\frac{1}{7}\right)}^{-1}=3.9+2.1=6\mu \mathrm{F}$

$C_{\mathrm{eq}}=3.9+{\left(\frac{1}{6}+\frac{1}{12}\right)}^{-1}=4\mu \mathrm{F}\text{ so }\mathrm{q}=4\mathrm{E}\mu \mathrm{C}$

$q_1+q_2=4E\text{ and }\frac{q_1}{2.1}=\frac{q_2}{3.9}\Rightarrow q_1=\frac{2.1}{6}\times 4E$

$=0.7\times 2\mathrm{E}=1.4\mathrm{E}$

$\text{ so }\mathrm{\Delta V}\text{ across }7\mu \mathrm{F}\Rightarrow \mathrm{ql}/7\mu \mathrm{F}=\frac{0.7\times 2\mathrm{E}\mu \mathrm{C}}{7\mu \mathrm{F}\times 10}=6$

$\Rightarrow E=30V$

$\Rightarrow q_1=0.7\times 2\times 30=42\mu C$

$\mathrm{\Delta V}\text{ across }3\mu \mathrm{F}=\frac{\mathrm{q}}{3\mu \mathrm{F}}=14\mathrm{V}$

$\mathrm{\Delta V}\text{ across }12\mu \mathrm{F}=4\times \frac{30}{12}=10\mathrm{V}$