Electronic Devices Question 3
Question: For a transistor the parameter b = 99. The value of the parameter a is
[Pb CET 1998]
Options:
A) 0.9
B) 0.99
C) 1
D) 9
Show Answer
Answer:
Correct Answer: B
Solution:
$ \alpha =\frac{\beta }{1+\beta }=\frac{99}{1+99}=0.99 $ .