00:01
To solve this question, let us first draw the circuit diagram.
00:05
So, this is vth connected with a register that is denoted by rth and a connection is made in this direction where this value is given as 1k with a connection of voltage of 12v followed by 1k and the direction of the loop is in this direction denoted by the arrow.
00:41
As in this question, both the given stages are identical.
00:53
So by considering the concept of dc equivalent, we can calculate the value for rth and this is equal to 10k in a parallel connection with 5k which is equal to 3 .33 kω and the value of vth is given by 12 multiplied with 5k divided by 5k plus 10k that is equal to 4v.
01:38
Now the next step, we can solve by applying kvl in the loop of the circuit diagram that we have drawn above.
01:51
The equation can be written as vth minus ibrth minus 0 .7 minus ie multiplied with 1k and the overall value is equal to 0.
02:14
From here, the value of ie can be written as beta plus 1 multiplied with ib which is equal to 101 ib.
02:31
Now substituting the values, we get ie equals to 4 minus 0 .7 divided by 1k plus 3 .33k divided by 101.
02:52
Upon solving, the value of ie will be equal to 3 .1945 ma and by using this value of ie, the value of ic can be determined which is given by 100 divided by 101 into ie.
03:25
Now using the value of ie, we get the value of ic equals to 3 .16 ma.
03:35
So r pi 1 is further equal to r pi 2 and this is given by the formula beta vt divided by ic.
03:52
Upon substituting the values, that is 100 multiplied with 26 divided by the value of ic that we have derived above that is 3 .16...