00:01
Hello students, in this question we have provided an npn transistor in common emitter configuration as in the figure.
00:08
So we have to calculate the q point.
00:10
First we can look at the vcc.
00:12
So given vcc is equal to 15 volt, hence it is equal to vce.
00:19
And also the maximum ic we can find using the equation vce divided by rc plus re.
00:29
So in this equation we have vce 15 volt divided by rc is 1 and re is 1 kilo ohm.
00:38
So we have 2 kilo ohm that is equal to 7 .5 milliampere.
00:46
Now we can look at the case of the voltage across q.
00:50
So to find the q point we can look at the case.
00:54
So we have r1 is equal to 10 kilo ohm and r2 is equal to 5 kilo ohm and the voltage across them is 15 volt.
01:06
Hence voltage across r2.
01:08
So v2 that is voltage across r2 will be equal to 15 5 volt...