00:01
Alright, so let's find the thévanines and norton's equivalent circuits for the following circuit.
00:07
So let's find the open circuit voltage voc is an open circuit voltage and the short circuit current.
00:14
Voc is the voltage across the terminals a and b when under open circuit conditions.
00:22
Alright, so under open circuit conditions, what happens is the current will flow only through this path.
00:31
So that means these two resistors will be in series right so i can i can convert this circuit something like this so this is a 15 old source and this is the 5 oms source and this is the dependent source so from here 0 .5 ix of the current is going and this is ix and then the 10 oms and 10 oms i can combine to 20 oms and this is our terminals a and b right so obviously when you apply kitchchof's voltage law, so let's see how much current will, i mean how much voltage across kb gets dropped.
01:14
Now since ix is coming from 5 oms and 0 .5 ix went to this, so obviously 0 .5 ix will be distributed along this resistor and let's apply kcho's voltage law to the bigger loop.
01:27
Since we can't apply kvl for the dependent loop containing the dependent sources, so we should apply the kchox voltage law along this.
01:33
So if you apply it will be 15 minus 5 into ix minus 0 .5 x into 20 is equal to 0.
01:43
So 15 is equal to 5 ix plus 0 .5 into 20 that is 10 x.
01:52
So when you add 5 x and 10 x, you'll be getting 15 is equal to 15 x...