Kirchhoff's Laws exercise
There is an ideal current source of 3 amperes, which means it will arrange the things to make 3 amperes flow through it in the direction of the arrow, no matter what happens.

We are provided that voltage across the 3 ohms resistor is 6 V, which leads a 2 amperes of current through it from c to d. So, the remaining 1 ampere must be flowing through 2 ohms resistor. Making node c higher in potential than node a by 2 V.

Here I have used Kirchoff's Current Law, which states, the current entering to any node will be equal to the current leaving that node.


Start from node a, when you pass to node c, you are now at 2 volts higher with respect to your starting point, node a.
Move towards node d, you are lowered by 6 volts, so you are at 2 - 6 = - 4 volts now.
Pass to node e, the current can be calculated as 2 amperes by applying KCL. You again went to a lower potential by 12 Volts(6 ohms * 2 amperes). You are now at 2 - 6 - 12 = -16 Volts.
Finally, pass to node b, you have gained 12 Volts, which -at the end- sums up to 2 - 6 - 12 + 12 = - 4 volts.


So the change in the voltage while going from node a to node b is -4 Volts. Which means node a is higher in potential than node b by 4 Volts.