00:01
Here to solve this problem, first we simplify this circuit.
00:05
Here 12 om and 20 om are connected in parallel.
00:08
So their parallel combination is given by 12 into 22 over 12 plus 22.
00:16
This is equal to 7 .7647 now in the circuit we can replace these two registers by an equivalent 7 .76.
00:31
4 -7 -oom register.
00:34
Now suppose current flowing through this branch is equal to i1 and current flowing through this branch is equal to i2.
00:43
Let's say that the voltage of this node is equal to v1 and voltage of this node is equal to v2.
00:51
Here we ground node number 2, that means voltage of node number 2 is equal to 0.
00:58
Or here we take node 2 as a reference node.
01:01
Now in this circuit we can say v1 minus v2 is equal to 6 here v2 is equal to 0 so from this we obtain v1 is equal to 6 volt.
01:16
Now by using the arm's law we can say v1 minus v2 is equal to 14 into i2.
01:25
V1 minus v2 is equal to 14 into i2.
01:30
V2 is equal to 0, v1 is equal to 6, 14, 14.
01:35
Into i2 from this we obtain i 2 is equal to 3 over 7 ampere 3 over 7 is approximately equal 0 .42857 ampere now again by omns law we can say v1 minus v2 is equal to i 1 into 7 .7647 plus 4 .5.
02:12
V2 is equal to 0, v1 is equal to 6 into i1 7 .7647 plus 4 .5.
02:22
By simplifying this, we obtain i1 is equal to 0 .48 9 to 1 ampere.
02:28
Now suppose this current is equal to i...