Maybe a low ohmic resistor 0.01ohm 5W is suitable in the application.

The hall effect sensor normally is very expensive in my country.

Let say the total amp flows is 15A,:

Voltage dropped = 15 X 0.01 = 0.15V
Wattage dissipated = 15 X 0.15 = 2.25W

I have been using this with the volt meter to measure the voltage up to 22A, and it is working fine. The heat will not be a issue for the resistor. But it is a bit expensive.