A transistor is used in common-emitter configuration. Given its $\alpha = 0.9$, calculate the change in collector current when the base current changes by $2\mu \mathrm{A}$.}
Show Hint
- $1\mu \mathrm{A}$
- $0.9\mu \mathrm{A}$
- $30\mu \mathrm{A}$
- $18\mu \mathrm{A}$
The Correct Option is D
Solution and Explanation
$\beta = \frac{\alpha}{1-\alpha}$ and $\beta = \frac{\Delta I_C}{\Delta I_B}$.
Step 2: Detailed Explanation:
$\beta = \frac{0.9}{0.1} = 9$. $\beta = \frac{\Delta I_C}{\Delta I_B}$, so $\Delta I_C = \beta \times \Delta I_B = 9 \times 2 = 18 \mu \mathrm{A}$.
Step 3: Final Answer:
The change in collector current is $18\mu \mathrm{A}$.
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