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MHT CET 2024 (9 May, Shift 1) · Junction Transistor

Junction Transistor question from MHT CET 2024 (9 May, Shift 1)

Q40MHT CET 2024 · 9 MAY · SHIFT 1Medium

Junction Transistor

The current amplification factor of a transistor is 50. The input resistance when used in common emitter mode is 1 kΩ1\text{ k}\Omega. The peak value for an a.c. input voltage of 0.01 V peak is

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Answer: (d)

  1. Peak base current IB=ViRi=0.011000=10 μI_B = \dfrac{V_i}{R_i} = \dfrac{0.01}{1000} = 10\ \muA.
  2. Peak collector current IC=βIB=50×10=500 μI_C = \beta I_B = 50 \times 10 = 500\ \muA.
  3. So the answer is (d) 500 μ500\ \muA.
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