Question

In a transistor connected in common emitter mode, $R_{C} = 4k\Omega$ $R_{I} = 1k\Omega,I_{C} = 1$ mA and $I_{B} = 20\mu A$. The voltage gain is.

• A. 100
• B. 200
• C. 300
• D. 400

Answer 1

Answer: (B)

200

Answer 2

: Voltage gain, $A_{V} = \frac{\beta R_{C}}{R_{I}} = \frac{I_{C}}{I_{B}} \times \frac{R_{C}}{R_{I}}$

$= \frac{1 \times 10^{- 3} \times 4 \times 10^{3}}{20 \times 10^{- 6} \times 1 \times 10^{3}} = 200$