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A change of $8.0 \mathrm{~mA}$ in the emitter current brings a change of $7.9 \mathrm{~mA}$ in the collector current. The values of $\alpha$ and $\beta$ are
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Verified Answer
The correct answer is:
$0.99,79$
Given that, change in emitter current, $\Delta I_{E}=8 \mathrm{~mA}$
and change in collector current, $\Delta I_{C}=7.9 \mathrm{~mA}$ We know that,
$$
\alpha=\frac{\Delta I_{C}}{\Delta I_{E}} \Rightarrow \alpha=\frac{7.9}{8} \Rightarrow \alpha \simeq 0.99
$$
Also we know that
$$
\begin{array}{c}
\quad \beta=\frac{\alpha}{1-\alpha} \\
\Rightarrow \quad \beta=\frac{\frac{7.9}{8}}{1-\frac{7.9}{8}}=\frac{7.9}{8-7.9}
\end{array}
$$
or $\quad \beta=\frac{7.9}{0.1}=79$
Hence, the required answer is $\alpha=0.99$ and $\beta=79$
and change in collector current, $\Delta I_{C}=7.9 \mathrm{~mA}$ We know that,
$$
\alpha=\frac{\Delta I_{C}}{\Delta I_{E}} \Rightarrow \alpha=\frac{7.9}{8} \Rightarrow \alpha \simeq 0.99
$$
Also we know that
$$
\begin{array}{c}
\quad \beta=\frac{\alpha}{1-\alpha} \\
\Rightarrow \quad \beta=\frac{\frac{7.9}{8}}{1-\frac{7.9}{8}}=\frac{7.9}{8-7.9}
\end{array}
$$
or $\quad \beta=\frac{7.9}{0.1}=79$
Hence, the required answer is $\alpha=0.99$ and $\beta=79$
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