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The transfer ratio $β$ of a transistor is 50. The input resistance of the transistor when used in the common emitter configuration is 1 k $Ω$ . The peak value of the collector alternating current input peak voltage of 0.01 V is

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$100 μA$

$0.01 μA$

$500 μA$

$0.25 μA$

answer is D.

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Detailed Solution

$β = 50, R_{i} = 1000 Ω, V_{i} = 0.01 V β = \frac{i_{c}}{i_{b}} and i_{b} = \frac{V_{i}}{R_{i}} = \frac{0.01}{10^{3}} = 10^{- 5} A Hence, i_{c} = 50 \times 10^{- 5} A = 500 μA$

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