Question

In a transistor connected in CE configuration, the emitter current increases from 10 mA to 20 mA when the base current increases from 0.12 mA to 0.24 mA. Then the current gain is

• A. 41.3
• B. 82.3
• C. 80.2
• D. 96.4

Hint:

Use $\beta = \dfrac{\Delta I_C}{\Delta I_B}$ for current gain in CE configuration.

Answer 1

The Correct Option is B

In CE (common emitter) configuration: $I_E = I_B + I_C \Rightarrow I_C = I_E - I_B$
Initial: $I_E = 10$ mA, $I_B = 0.12$ mA ⇒ $I_C = 9.88$ mA
Final: $I_E = 20$ mA, $I_B = 0.24$ mA ⇒ $I_C = 19.76$ mA
Current gain $\beta = \dfrac{\Delta I_C}{\Delta I_B} = \dfrac{19.76 - 9.88}{0.24 - 0.12} = \dfrac{9.88}{0.12} \approx 82.3$