Again, we're using Kirchhoff's Current Law to determine I1, I4 and
I6 so we treat each one of those complex circuits as a node.
And so it's kind of a node that is encompassing part of the circuit.
And flowing out of that node we have the three currents at the top,
we have I1, I or 60 milliamps and 20 milliamps.
So, we're going to start with the top and write Kirchhoff's current law for the top
and again, before we do that, let's state what Kirchhoff's current law is.
And that's, that the algebraic sum of all the currents entering any node at any
instant in time is 0, so looking at the top node,
the top node being the complex circuit element at the top, elements at the top,
we have a current I1 flowing into it, or summing currents into that node.
I1 is flowing in, it's a positive flowing in.
We have 60 flowing out so it's a minus 60 milliamps and
we also have 20 flowing out so it's a minus 20 milliamps and
that's all the currents associated with the top complex circuit node, and
we determine from that that I1 is equal to 80 milliamps.