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A transistor is operated in common-emitter configuration at $\mathrm{V}_{\mathrm{C}}=2 \mathrm{~V}$ such that a change in the base current from $100 \mu \mathrm{A}$ to $200 \mu \mathrm{A}$ produces a change in the collector current from $5 \mathrm{~mA}$ to $10 \mathrm{~mA}$. The current gain is :
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The correct answer is:
50
$$
\begin{aligned}
\Delta \mathrm{I}_{\mathrm{B}} & =100 \mu \mathrm{A} \\
\Delta \mathrm{I}_{\mathrm{C}} & =5 \mathrm{~mA} \\
\beta & =\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{5 \times 10^{-3}}{100 \times 10^{-6}}=50
\end{aligned}
$$
\begin{aligned}
\Delta \mathrm{I}_{\mathrm{B}} & =100 \mu \mathrm{A} \\
\Delta \mathrm{I}_{\mathrm{C}} & =5 \mathrm{~mA} \\
\beta & =\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{5 \times 10^{-3}}{100 \times 10^{-6}}=50
\end{aligned}
$$
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