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A change of 8.0 mA in the emitter current brings a change of 7.9 mA in the collector current. The value of $\alpha$ will be
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Verified Answer
The correct answer is:
0.99
We have,
$\begin{aligned} \quad I_E & =I_B+I_C \\ \Delta I_E & =\Delta I_B+\Delta I_C\end{aligned}$
According to question,
$\text { Thus, } \quad \begin{aligned}
\Delta I_E & =8.0 \mathrm{~mA} \\
\Delta I_C & =7.9 \mathrm{~mA} \\
\Delta I_B & =8.0 \mathrm{~mA}-7.9 \mathrm{~mA} \\
& =0.1 \mathrm{~mA}
\end{aligned}$
Thus, $\quad \Delta I_B=8.0 \mathrm{~mA}-7.9 \mathrm{~mA}$
Now, $\alpha=\frac{I_C}{I_E}=\frac{\Delta I_C}{\Delta I_E}=\frac{7.9 \mathrm{~mA}}{8.0 \mathrm{~mA}} \cong 0.99$
$\begin{aligned} \quad I_E & =I_B+I_C \\ \Delta I_E & =\Delta I_B+\Delta I_C\end{aligned}$
According to question,
$\text { Thus, } \quad \begin{aligned}
\Delta I_E & =8.0 \mathrm{~mA} \\
\Delta I_C & =7.9 \mathrm{~mA} \\
\Delta I_B & =8.0 \mathrm{~mA}-7.9 \mathrm{~mA} \\
& =0.1 \mathrm{~mA}
\end{aligned}$
Thus, $\quad \Delta I_B=8.0 \mathrm{~mA}-7.9 \mathrm{~mA}$
Now, $\alpha=\frac{I_C}{I_E}=\frac{\Delta I_C}{\Delta I_E}=\frac{7.9 \mathrm{~mA}}{8.0 \mathrm{~mA}} \cong 0.99$
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