Answer
$V_B=5V, V_C=-10V$
Work Step by Step
We can determine the potential at point B and C as follows:
$V_A-V_C=(11\Omega) I$.....eq(1)
The equivalent resistance of the right-most branches is
$R_{eq}=6.2\Omega+12\Omega$
$R_{eq}=18.2\Omega$
The equivalent resistance of the two parallel current paths between A and B is
$\frac{1}{\frac{1}{18.2\Omega}+\frac{1}{7.2\Omega}}=5.3\Omega$
The equivalent resistance for the entire network is
$5.3\Omega+11\Omega=16.3\Omega$
Now $I=\frac{15V}{16.3\Omega}$...eq(2)
Using eq(1) and eq(2), we obtain:
$V_A-V_C=0-V_C=11\Omega \times 0.92A$
$\implies V_{C}=-10V$
Similarly, the potential at B is given as
$V_B=5V$
