From Kirchoff’s current law, the current through the 1kΩ resistor is 5mA – 2mA = 3mA. This 3mA current flows through the 1kΩ resistor and creates a voltage of −3V at node B. The 2mA current flowing through the 2kΩ resistor creates a drop across the resistor of 4V. The voltage at A is therefore −3V + 4V = +1V. The 5mA flowing through the 3kΩ resistor creates a drop across the resistor of 15V. The voltage at C is therefore −3V −15V = −18V. The steps are labeled on the following diagram.