00:01
Hello students, this question is related to ce transistor.
00:04
Here rb is given as 1 kω which is equal to 1000 ω.
00:09
Then re is equal to 2 .2 kω which is 2200 ω.
00:14
Then vee is equal to 10 v.
00:16
Ve is equal to 0 .7 v.
00:19
Vcc is equal to 10 v.
00:21
Rc is equal to 1 kω which is equal to 1000 ω.
00:26
Then the βdc value it is given as 200.
00:29
Here we have to find out ib, ic, ie, then vc, ce and vb.
00:51
So first let's find out ib.
00:53
For that as in the figure let's apply the quick chop rule, loop rule, loop rule in the bottom most part that is we have minus ib rb minus ib rb minus ie re is equal to 0.
01:32
Substituting the values we have implies minus 1000 ib minus 0 .7 is equal to 0 implies ib is equal to 0 .7, sorry 0 .7 minus 0 .7 divided by minus 1000 implies ib is equal to 0 .007 a.
02:24
Further let's find out ic...