You have already seen that electric charge is conserved: charge cannot be created or destroyed. First, check the face of that law you met in series circuits:
Which statement is true of the charge flowing around a series circuit?
The last lesson's rule, that branch currents add up to the supply current, follows from that law. Here is how.
Here is the two-lamp circuit from the last lesson, with its junction. Each second, 0.5 C of charge arrives at that junction.
So the 0.5 C arriving each second must all leave each second. It splits: 0.3 C goes into one branch, and 0.2 C goes into the other. 0.3 C + 0.2 C = 0.5 C. Nothing is lost, and nothing piles up at the junction.
Charge per second is current. So the branch currents, 0.3 A and 0.2 A, must add up to the supply current, 0.5 A.
The next video shows the same ledger in motion: one stream of charge around a series loop, then a stream splitting at one junction and recombining at the far junction.
Each second, 0.9 C of charge flows into a junction. How much charge flows out of it each second, in C?
The same holds at the far junction, where the branches rejoin: 0.3 C + 0.2 C arrives each second, so 0.5 C leaves each second, back toward the battery.
Why do the branch currents of a parallel circuit add up to the supply current?
Each second, 1.2 C flows into a junction. The first branch carries away 0.8 C each second. Why must the second branch carry away exactly 0.4 C each second?
Charge is conserved: all the charge that flows into the junction each second must flow out of it.
The charge leaving through the two branches must therefore total 1.2 C each second.
0.8 C + 0.4 C = 1.2 C
The ledger balances only if the second branch carries 0.4 C each second. Charge cannot be created, destroyed, or used up at the junction.
The two branches of the same circuit deliver 0.8 C and 0.4 C to the far junction each second. How much charge flows back toward the battery each second?
Charge is conserved at the far junction too: all the charge flowing in each second must flow out.
Add the charge the branches deliver:
0.8 C + 0.4 C = 1.2 C
1.2 C flows back toward the battery each second, a current of 1.2 A.
Conservation fixes only the total: the charge out must equal the charge in. It does not make the split equal.
Theo says a junction uses up some charge, so the branch currents add up to less than the supply current. Which fact shows the claim is wrong?
Each second, 4.0 C arrives at a junction and 1.5 C leaves into the first branch. How much charge enters the second branch each second, in C?
The two branch currents of a parallel circuit recombine at the far junction. What is the current leaving that junction?
The branches of a parallel circuit carry 0.9 A and 0.4 A. Using conservation of charge, what is the supply current, in A?