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NEET 2016 (Phase 1) · Junction Transistor

Junction Transistor question from NEET 2016 (Phase 1)

Q98NEET 2016 · PHASE 1Difficult

Junction Transistor

A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800 Ω\Omega is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 Ω\Omega, the voltage gain and the power gain of the amplifier will respectively be

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Answer: (c)

  1. Taking the given current amplification factor as β=0.96\beta = 0.96 (as in the official solution).
  2. Voltage gain: Av=βRLRi=0.96×800192=4A_v = \beta \dfrac{R_L}{R_i} = 0.96 \times \dfrac{800}{192} = 4
  3. Power gain: Ap=βAv=0.96×4=3.84A_p = \beta A_v = 0.96 \times 4 = 3.84
  4. So the answer is (c) 4, 3.84.
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