In the figure below, what is the potential difference between the point A and B and between B and C respectively in steady state?

  $$\begin{gathered} C_{equivalent}=\frac{(3+3)\times (1+1)}{(3+3)+(1+1)}+1 \\ =\frac{(6\times 2)}{6+2}+1=\frac{12}{8}+1=\frac{5}{2}\mu F \\ \therefore Q=C\times V=\frac{5}{2}\times 100=250\mu C \end{gathered}$$

So the charge of $250\mu C$ will divide as $100\mu C in the 1\mu F$,$150\mu C in the 2\mu F and 6\mu F branch$

$\therefore V_{AB}=\frac{150}{6}=25Vand{V}_{BC}=100-V_{AB}=75V$