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MHT CET 2021 (22 Sep, Shift 1) · Junction Transistor

Junction Transistor question from MHT CET 2021 (22 Sep, Shift 1)

Q74MHT CET 2021 · 22 SEP · SHIFT 1Difficult

Junction Transistor

In an n-p-n transistor 200 electrons enter the emitter in 10 −8^{-8} second. If 1% electrons are lost in the base, then the current that enters the emitter and the current amplification factor are respectively [e = 1.6 ×\times 10 −19^{-19} C]

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Answer: (b)

  1. Emitter current: IE=net=200×1.6×10−1910−8=3.2×10−9 AI_E = \dfrac{ne}{t} = \dfrac{200 \times 1.6 \times 10^{-19}}{10^{-8}} = 3.2 \times 10^{-9}\ \text{A}
  2. 1% lost in base, so α=ICIE=0.99\alpha = \dfrac{I_C}{I_E} = 0.99
  3. β=α1−α=0.990.01=99\beta = \dfrac{\alpha}{1 - \alpha} = \dfrac{0.99}{0.01} = 99
  4. So the answer is (b) 3.2×10−93.2 \times 10^{-9} A and 99.
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