Apply Kirchhoff's law Kirchhoff's law says that when you sum up all the currents (positive for the currents going to a junction, and negative for current leaving a junction), you will always get as result 0.

Look at the following diagram:



Using Kirchhoff's law, you can see that i1 + i4 - i2 - i3 = 0, so i1 + i4 = i2 + i3.

Given two lists, one with all the currents entering the junction and one with all the currents leaving the junction except one, output the last one.

Testcases:

[1, 2, 3], [1, 2] = 3
[4, 5, 6], [7, 8] = 0
[5, 7, 3, 4, 5, 2], [8, 4, 5, 2, 1] = 6


The second list always has one item less than the first list. The output cannot be negative. Smallest program wins.
Jelly, 2 bytes

_S


Try it here!

Takes the entering currents in the first argument, and the leaving currents in the second argument. _ subtracts them pairwise, leaving the single element from the longer list as-is, and S sums the result.