SATHEE: Electronic Devices Question 3

Electronic Devices Question 3

Question: For $\alpha$ transistor the parameter $\beta$ = 99. The value of the parameter $\alpha$ 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 $ .

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Electronic Devices Question 3

Question: For $\alpha$ transistor the parameter $\beta$ = 99. The value of the parameter $\alpha$ is

Options:

Answer:

Solution:

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