Q145 / 200 · NEET 2012 (Screening) · Physics

In a CE transistor amplifier, the audio signal voltage across the collector resistance of $2 \mathrm{k} \Omega$ is $2 \mathrm{~V}$. If the base resistance is $1 \mathrm{k} \Omega$ and the current amplification of the transistor is 100 , the input signal voltage is

  • A
    $0.1 \mathrm{~V}$
  • B
    $1.0 \mathrm{~V}$
  • C
    $1 \mathrm{mV}$
  • Correct answer
    D
    $10 \mathrm{mV}$

Solution

Current amplification factor $\beta=\frac{\Delta I_C}{\Delta I_B}$ Collector resistance $\Delta I_C=\frac{2 \mathrm{~V}}{2 \times 10^3 \Omega}=1 \times 10^{-3} \mathrm{~A}$ Base current $\Delta I_B=\frac{V_B}{R_B}=\frac{V_B}{1 \times 10^3}=V_B \times 10^{-3}$ Given, $\beta=100$ Now, $100=\frac{10^{-3}}{V_B \times 10^{-3}}$ $V_B=\frac{1}{100} \mathrm{~V}=10 \mathrm{mV}$

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

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