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NEET 2012 (AIPMT, Mains) · Junction Transistor

Junction Transistor question from NEET 2012 (AIPMT, Mains)

Q106NEET 2012 · AIPMT · MAINSMedium

Junction Transistor

The input resistance of a silicon transistor is 100 $\Omega$ . Base current is changed by 40 $\mu$ A which results in a change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load resistance of 4 k $\Omega$ . The voltage gain of the amplifier is

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Answer: (a)

$\beta = \dfrac{\Delta I_C}{\Delta I_B} = \dfrac{2 \times 10^{-3}}{40 \times 10^{-6}} = 50$
$A_V = \beta \dfrac{R_L}{R_i} = 50 \times \dfrac{4000}{100} = 2000$
So the answer is (a) 2000.

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