Answer
The potential at point a is $~~9~V$
The potential at point b is $~~1~V$
Work Step by Step
We can use the loop rule to find the current in the circuit:
$9~V-(2~\Omega)~I-6~V - (1~\Omega)~I = 0$
$(3~\Omega)~I = 3~V$
$I = \frac{3~V}{3~\Omega}$
$I = 1~A$
We can find the potential difference across the $2~\Omega$ resistor:
$\Delta V = I~R = (1~A)(2~\Omega) = 2~V$
From ground to point a, there is a potential increase of 9 V. Therefore, the potential at point a is $~~9~V$
From point a across the $2~\Omega$ resistor, the potential decreases by 2 V. Then the potential decreases another 6 V before reaching point b. Therefore, the potential at point b is $9~V - 2~V-6~V$ which is $1~V$
