The current amplification factor of a transistor is 50. The input resistance when used in common emitter mode is 1 k Ω . The peak value of the a.c. output current for an a.c. input voltage of 0.01 V p

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The current amplification factor of a transistor is 50. The input resistance when used in common emitter mode is 1 k Ω . The peak value of the a.c. output current for an a.c. input voltage of 0.01 V peak is

TestHub · Accepted Answer

Given β = 50 R in ​ = 1 k Ω = 1000Ω V in ​ = 0.01 V The peak a.c. input current is given by: I b ​ = R in ​ V in ​ ​ = 1000 0.01 ​ = 1 0 − 5 A The peak a.c. output current is: I c ​ = β I b ​ = 50 × 1 0 − 5 A = 500 × 1 0 − 6 A = 500 μ A Answer: 500 μ A

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