00:01
Now as shown in the figure that is circuit diagram here, we need to find the currents through the three registers, r1, r2, and r3.
00:12
So now to start with the solution, let us consider the voltage at node a, b, va, and the voltage at node b, va, and the voltage at node b, va, and the voltage at node b, b, vb, and its value is 0 volt because that is the ground node.
00:50
Now, let's start with the step one here.
00:58
Consider all the currents, that means i1, i2, and i3 are moving out of the node a.
01:16
And applying kcl, that is kirchop's current law at node a.
01:25
And just to mention the kcl is that sum of the currents entering the node a is equal to some of the currents leaving the node a.
01:53
This is what kcl states.
01:57
So therefore what we are going to make use of, now no current is a actually entering the node a if you consider here because we have assumed that all the currents are moving out of the node a.
02:11
So this will be zero and some of the currents which are leaving node a that is going to be i1 plus i3 plus i2.
02:23
So this is what we get as an expression out of the kcl...