Semiconductors

JEE Main level, single correct, medium

4 of 209

Question 4 of 209

In a common emitter NPN transistor amplifier, input resistance is 1 kΩ1\,\text{k}\Omega and load resistance is 10 kΩ10\,\text{k}\Omega. If β=50\beta=50, voltage gain is:

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Answer: (B) 500500

Av=βRLRin=50×100001000=500A_v = \beta \frac{R_L}{R_{in}} = 50 \times \frac{10000}{1000} = 500.

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