Solve this:

Dear Student, 

Resistors connected between A and D, D and C, C and B are in series, 
$$\begin{gathered} R\text{'}=R+R+R=3R \\ Now, 3R\parallel R \\ {R}_{effective}=\frac{3R\times R}{3R+R}=\frac{3}{4}R \\ R=\frac{16}{4}=4 \mathrm{ohm} \\ \Rightarrow R_{\mathrm{effective}}=\frac{3}{4}\times 4=3 \mathrm{ohm} \\ \mathrm{So}, \mathrm{current} \mathrm{drawn}, \mathrm{I}=\frac{V}{R_{effective}}=\frac{9}{3}=3 \mathrm{A} \end{gathered}$$
Total voltage will equally divide among Resistors connected between A and D, D and C, C and B.
So, potential difference between ends C and D, 
$V_{CD}=\frac{9}{3}=3 \mathrm{V}$
Here is a link also given for the final part of the question .
https://www.meritnation.com/ask-answer/question/solve-this-problem-without-providing-links/electrostatic-potential-and-capacitance/11600905


Regards