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Question 13.PE.NAQ.3: Find the value of Ab for the case discussed in Question 2....

Find the value of A_b for the case discussed in Question 2.

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We know that

h_{ fb }=-\frac{h_{ fe }}{1+h_{ fe }}

Therefore,

A_{ b }=h_{ fb }=\frac{100}{1+100}=-0.99 \text { or }-1

Ans. (−0.99, −1)

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