The current amplification factor of a transistor is 50. The input resistance when used in common emitter mode is 1 kΩ. The peak value for an a.c. input voltage of 0.01 V peak is (a) 100 (b) 0.01 mA (c) 0.25 mA (d) 500

Answer: (d). Peak base current IB = (Vi)/(Ri) = (0.01)/(1000) = 10 μ A. Peak collector current IC = β IB = 50 × 10 = 500 μ A. So the answer is (d) 500 μ A.

The current amplification factor of a transistor is 50. The input… | MHT CET 2024 (9 May, Shift 1)

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Junction Transistor

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

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Answer: (d)

Peak base current $I_B = \dfrac{V_i}{R_i} = \dfrac{0.01}{1000} = 10\ \mu$ A.
Peak collector current $I_C = \beta I_B = 50 \times 10 = 500\ \mu$ A.
So the answer is (d) $500\ \mu$ A.

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