00:01
First, let us draw the circuit to understand in a more clear way.
00:04
So here we have the 5 -oam resistance.
00:07
So this is here, it is v1, and here we have the positive and negative charge of 8 volt, and here it is v2.
00:16
So here have the 10 -oam resistance, which is attached with v3 node.
00:23
This is here v -3.
00:24
So from v -1, we have the 2 -oom resistance.
00:28
So here v5 note and here the charge is of plus a negative which is 4 ix and this is attached with v2.
00:43
So this v2 is here we have.
00:45
In this we have the 4 ome resistance which the knot v4 and here to a current which is here have the earthing.
01:00
In this it is earth -ate.
01:02
Now this is from here v4 we have voltage of 16 volt.
01:10
So this is attached with this.
01:14
So here the current flow is i -x like this and here have the 8 -oom resistance.
01:22
So basically a super node exists when an idle voltage source appears between any 2 -node of an electric circuit.
01:30
So let us say that from here to here this is node 1, node 1, and from here to here it is node.
01:48
So at super node 1, super node 1, v1 minus v2 will be equal to 8.
02:02
So from this, v1 will be equal to v2 plus 8.
02:06
Let be equation number 1 now at super node super node 2 v4 minus v3 will be equal to 16 let be equation number 2 and v5 and this v5 is equal to 4 i x so from this i x will be equal to v5 divided by 4 let the equation number 3 so u -s -sign g the kcel at v3 so ix will be equal to v1 minus v3 divided by 5 which will be added v2 minus v3 divided by 10 so from this v5 divided by 4 will be equal to v1 minus v3 divided by 5 plus v2 minus v3 divided by 10 this is from equation number 3 as i x is equal to v5 divided by 4 so we have replaced i x with v5 divided by fro so from this the 10 v5 will be equal to 10 v5 will be equal to 8 v1 plus 4 v2 minus 12 v3 so this u -s -sign g the this is u -sign g the equation the equation will be that will be 1 1 2 v2 minus 12 v 3 minus 10 v 5 will be equal to minus 64 let the equation number so from this the kcel at super node 1 super node super node super node 1 it will be v1 minus v5 divided by 2 plus v1 minus v3 divided by 5 plus v2 minus v4 divided by 4 plus v2 minus v3 divided by 4 will be equal to this will be equal to 0 1 4 v1 now from this v1 plus v2 that is 7 v2 minus 6 v3 minus 5 v4 minus 10 v5 will be equal to 0 u sine g now equation that is this is double 1 0 0 0 0 0 0 0 0 this is 1 114 114 v2 plus 8 plus 7 v2 minus 6 v3 minus 5 v4 minus 10 v5 will be equal to 0 to 1 so from this v2 minus 11 v3 minus 10 v5 will be equal to minus 32 now let us say that equation number it is now kcl adds super node super node 2 it will be v4 minus v2 divided by 4 plus v3 minus v2 divided by 10 plus v3 minus v1 divided by 5 plus v3 divided by 8 will be equal to...