In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 k Ω is 2 V. If the base resistance is 1 k Ω and the current amplification of the transistor is 100, the input signal voltage is (a) 0.1 V (b) 1.0 V (c) 1 mV (d) 10 mV

Answer: (d). IC = (Vo)/(RC) = 2/2000 = 1 mA. IB = (IC)/(β) = (1 mA)/(100) = 10 μ A. Vi = IB RB = 10 × 10⁻⁶ × 1000 = 10 mV So the answer is (d) 10 mV.

In a CE transistor amplifier, the audio signal voltage across the… | NEET 2012 (AIPMT, Prelims)

Q107NEET 2012 · AIPMT · PRELIMSMedium

Junction Transistor

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

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Answer: (d)

$I_C = \dfrac{V_o}{R_C} = \dfrac{2}{2000} = 1$ mA.
$I_B = \dfrac{I_C}{\beta} = \dfrac{1\text{ mA}}{100} = 10\ \mu$ A.
$V_i = I_B R_B = 10 \times 10^{-6} \times 1000 = 10$ mV
So the answer is (d) 10 mV.

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