A 16ohm resistance wire is bent to form square. A source of emf 9V is connected across one of its sides as shown. Calculate the current drawn from the source. Find the potential difference between the ends C and D.

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, 
$V_{CD}=\frac{9}{3}=3 \mathrm{V}$