### Video Transcript

The current in three wires of the
circuit shown are known. The currents 𝐼 one and 𝐼 two are
unknown. Find 𝐼 one. Find 𝐼 two.

So taking a look at this circuit,
we see five different currents labeled. There are three that are known —
this one here, 1.5 amps; this one here, 2.0 amps; and then this current, 2.5
amps. And along with this, there are two
unknown currents, 𝐼 one and 𝐼 two.

We want to solve for both of these
unknown currents. And to do this, we can recall
Kirchhoff’s current law, also called Kirchhoff’s first law. We can state this law in words like
this: the current entering a junction equals the current leaving. This means that at any junction in
an electrical circuit, if we add up all the currents moving into that junction
point, that sum will equal the sum of all the currents leaving that point.

We can apply this law to certain
junction points in this circuit over here to answer the question of what is 𝐼 one
and what is 𝐼 two. Let’s first solve for the current
𝐼 one. To do that, we’ll pick a junction
point in this circuit, where 𝐼 one is either entering or leaving the junction. The junction we’ll pick is right
here. At this location, we have these two
currents entering and then current 𝐼 one leaving.

If we apply Kirchhoff’s current law
to this junction, we can say that the sum of currents entering it, 1.5 amps added to
2.5 amps, is equal to the sum of currents leaving this junction. But we see that only one is
leaving, 𝐼 one. So then, 𝐼 one is the only current
on the right side of our equation. And adding together 1.5 amps and
2.5 amps, we get 4.0 amps. This is the value of the current 𝐼
one.

Now let’s use the same law to solve
for current 𝐼 two. This time, we’ll pick a different
junction point in our circuit. A good one to choose will be this
junction right here. We can see that there are two
currents, 𝐼 two, and then the current of 2.0 amps coming into this junction point
and one current of 2.5 amperes leaving it.

Once again, applying Kirchhoff’s
current law, we can say that the sum of currents entering this junction is 𝐼 two
added to 2.0 amps and that that equals the total current leaving, which we can see
is 2.5 amps. Using this equation to solve for 𝐼
two, if we subtract 2.0 amps from both sides, we see then that 𝐼 two is equal to
0.5 amps. So 𝐼 one is 4.0 amps and 𝐼 two is
0.5 amps.