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MHT CET 2024 (2 May, Shift 1) · Junction Transistor

Junction Transistor question from MHT CET 2024 (2 May, Shift 1)

Q39MHT CET 2024 · 2 MAY · SHIFT 1Medium

Junction Transistor

In an NPN transistor 101010^{10} electrons enter the emitter in 10−6 s10^{-6}\text{ s} and 2%2\% electrons recombine with holes in base. The current ratios ' α\alpha ' and ' β\beta ' of a transistor are respectively (nearly)

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Answer: (a)

  1. 2%2\% of electrons recombine in the base, so 98%98\% reach the collector: α=ICIE=0.98\alpha = \dfrac{I_C}{I_E} = 0.98.
  2. β=α1−α=0.980.02=49\beta = \dfrac{\alpha}{1-\alpha} = \dfrac{0.98}{0.02} = 49
  3. So the answer is (a) 0.98,490.98, 49.
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