For a transistor amplifier in common emitter configuration having load impedance of 1 kΩ ( h_fe = 50 and h_oe = 25 × 10^-6 A/V ), the current gain is: - Physics MCQ AIEEE 2004 2026 | ExamTest.live

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For a transistor amplifier in common emitter configuration having load impedance of 1 k Ω (h_{f e} = 50 and h_{oe} = 25 \times 1 0^{- 6} A/V), the current gain is:

Solution & Step-by-step Explanation

In a common emitter configuration, the current gain A_{i} is given by: A_{i} = \frac{- h _{f e}}{1 + h _{oe} R _{L}} Given: h_{f e} = 50 h_{oe} = 25 \times 1 0^{- 6} S (or A/V) R_{L} = 1 k Ω = 1000 Ω $ A_{i} = \frac{- 50}{1 + ( 25 \times 1 0 ^{- 6} \times 1000 )} $ A_{i} = \frac{- 50}{1 + 0.025} = \frac{- 50}{1.025} \approx - 48.78

Try it yourself before checking the explanation above.

For a transistor amplifier in common emitter configuration having load impedance of 1 k Ω (h_{f e} = 50 and h_{oe} = 25 \times 1 0^{- 6} A/V), the current gain is:

- 5.2

- 15.7

- 24.8

- 48.78

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