00:01
Hi, this is a circuit diagram for the given question.
00:03
So, we have to find a norton current in the terminal ab.
00:06
Now, so first in the first part we have to find our ab is short circuited.
00:13
So, short circuited.
00:15
So, applying the nodal analysis.
00:17
So, va by 5.
00:20
So, at point a.
00:22
So, applying the nodal analysis va by 5 plus va plus 13 divided by 9 is equal to 8.
00:28
So, 9 va plus 5 va plus 65 is equal to 45 into 8 which is equal to.
00:37
So, we have va will be.
00:40
So, substituting for va we have 295 by 14.
00:44
So, this is equal to 21 .07 volt.
00:51
So, this is the voltage across point a that is at point a.
00:55
So, point a is about here.
00:59
So, this is the point of va.
01:01
Now, so considering the point o this is point o.
01:06
So, here we have to apply kirchhoff current law.
01:08
So, by applying kirchhoff current law we have i is equal to here at point o the current flowing through this system is i which is equal to i1 plus another current is norton current.
01:22
So, this is i1 plus in.
01:25
So, in is equal to i minus i1 which is equal to.
01:30
So, current iin is equal to.
01:36
So, this is i minus i1 is va plus 13 by 9 which is equal to.
01:43
So, voltage va is flowing across the circuit.
01:47
So, plus 13 divided by 9 which is equal to.
01:52
So, this value is 3 .785.
01:56
Then so this is the total voltage of v1 plus va this is divided by 9.
02:03
So, applying va from the previous equation we have in as 3 .785 amperes...