Physics - Semiconductors Question with Solution | TestHub

We have, $\begin{array}{rl}{I}_E& =I_B+I_C\\ \Delta I_E& =\Delta I_B+\Delta I_C\end{array}$ According to question, $\text{ Thus, }\begin{array}{rl}\Delta I_E& =8.0\mathrm{mA}\\ \Delta I_C& =7.9\mathrm{mA}\\ \Delta I_B& =8.0\mathrm{mA}-7.9\mathrm{mA}\\ & =0.1\mathrm{mA}\end{array}$ Thus, $\Delta I_B=8.0\mathrm{mA}-7.9\mathrm{mA}$ Now, $\alpha =\frac{I_C}{I_E}=\frac{\Delta I_C}{\Delta I_E}=\frac{7.9\mathrm{mA}}{8.0\mathrm{mA}}\cong 0.99$