00:02
Hello students, let's solve the first part of the question.
00:06
Here we know that r1 is equal to vcc minus vb divided by i.
00:14
Substituting the values we can write 15 minus 3 divided by 0 .1 into 10 raised to minus 3.
00:24
Hence, the value for r1 is equal to 120 kω.
00:30
Now, ib1 is equal to ie1 divided by 1 plus β which is equal to 1 ma divided by 101 and the value is 0 .0099 ma.
00:50
Next one is ir2 is equal to i minus ib1 which is equal to 0 .1 minus 0 .0099.
01:02
Hence, the value is 0 .0901 ma.
01:09
Now, let's find out the resistance r2 and the equation is vb1 divided by ir2.
01:17
It will be equal to 3 divided by 0 .0901 into 10 raised to minus 3.
01:25
Hence, we will get the value for r2 as 33 .3 ω.
01:36
Now, vc1 is equal to vbe2 plus ve2.
01:45
Adding the values that is 0 .7 plus 4 is equal to 4 .7 v.
01:53
Let's find out the ic1 which is equal to β into ib1.
02:00
Let's substitute the values that is 100 into 0 .009 is equal to 0 .99 ma...