Q140 / 180 · JIPMER 2016 · Physics

The transfer ratio $\beta$ of a transistor is 50 . The input resistance of the transistor when used in the common emitter configuration is $2 \mathrm{k} \Omega$. The peak value of the collector AC current for an AC input voltage of 0.02 V peak is

  • A
    $200 \mu \mathrm{A}$
  • B
    0.01 mA
  • C
    0.25 mA
  • Correct answer
    D
    $500 \mu \mathrm{A}$

Solution

Given, $\beta=50, R=2 \mathrm{k} \Omega=2 \times 10^3 \Omega$, $\begin{aligned} & \Delta V_b=0.02 \mathrm{~V} \\ & i_c=?\end{aligned}$ Input current, $i_b=\frac{\Delta V_b}{R_b}=\frac{0.02}{2000}=10^{-5} \mathrm{~A}$ Also, current gain, $\beta=\frac{i_c}{i_b}$ $\begin{aligned} \therefore \quad i_c & =\beta i_b=50 \times 10^{-5} \mathrm{~A} \\ & =500 \times 10^{-6} \mathrm{~A}=500 \mu \mathrm{A}\end{aligned}$

Hence, the correct answer is option: 'D'.

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