The current amplification factor of a transistor is 50. The input resistance when used in common emitter…

The current amplification factor of a transistor is 50. The input resistance when used in common emitter mode is $1 \mathrm{k} \Omega$. The peak value for an a.c. input voltage of 0.01 V peak is
  1. $\quad 100 \mu \mathrm{~A}$
  2. 0.01 mA
  3. 0.25 mA
  4. $\quad 500 \mu \mathrm{~A}$

Solution

$\begin{aligned} & \text { Given: } \beta=50, R_i=1 \mathrm{k} \Omega=10^3 \Omega, \\ & V_i=0.01 \mathrm{~V} \\ & \beta=\frac{I_C}{I_B}=50 \Rightarrow I_C=50 \times I_B \\ & V_i=I_B \times R_i \\ \therefore \quad & I_B=\frac{V_i}{R_i}=\frac{0.01}{10^3}=10^{-5} \\ \therefore \quad & I_C=50 \times 10^{-5}=500 \times 10^{-6} \mathrm{~A}=500 \mu \mathrm{~A}\end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 1)

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