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For a transistor amplifier in common emitter configuration having load impedance of 1 k ( h _ fe .=50 and h_ o e =25 ) the current gain is

Options

  1. A-5.2
  2. B-15.7
  3. C-24.8
  4. D-48.78

Correct answer

D. -48.78

Step-by-step solution

In CE configuration, A_ i = - h _ fe 1+ h _ 0 e R_ L = -50 1+25 10⁻⁶ 1 10^3 =-48.78

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