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NEET 2016 (Phase 2) · Junction Transistor

Junction Transistor question from NEET 2016 (Phase 2)

Q96NEET 2016 · PHASE 2Medium

Junction Transistor

For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 k $\Omega$ is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 k $\Omega$ , then the input signal voltage is

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Answer: (b)

$I_C = \dfrac{V_o}{R_C} = \dfrac{4}{2000} = 2$ mA.
$I_B = \dfrac{I_C}{\beta} = \dfrac{2\text{ mA}}{100} = 20\ \mu$ A.
$V_i = I_B R_B = 20 \times 10^{-6} \times 1000 = 20$ mV
So the answer is (b) 20 mV.

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