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A transistor is operated in common-emitter configuration at $V_c=2$ volt 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
For a transistor
$$
\begin{aligned}
& \text { where } \quad \begin{array}{l}
\mathrm{I}_{\mathrm{E}}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}} \\
\mathrm{I}_{\mathrm{E}}=\text { emitter current } \\
\mathrm{I}_{\mathrm{B}}=\text { base current } \\
\mathrm{I}_{\mathrm{C}}=\text { collector current } \\
\text { and current gain } \beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}} \\
\Delta \mathrm{I}_{\mathrm{C}}=5 \times 10^{-3} \mathrm{~A} \\
\Delta \mathrm{I}_{\mathrm{B}}=100 \times 10^{-6} \mathrm{~A} \\
\beta=\frac{5}{100} \times 1000=50
\end{array}
\end{aligned}
$$
$$
\begin{aligned}
& \text { where } \quad \begin{array}{l}
\mathrm{I}_{\mathrm{E}}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}} \\
\mathrm{I}_{\mathrm{E}}=\text { emitter current } \\
\mathrm{I}_{\mathrm{B}}=\text { base current } \\
\mathrm{I}_{\mathrm{C}}=\text { collector current } \\
\text { and current gain } \beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}} \\
\Delta \mathrm{I}_{\mathrm{C}}=5 \times 10^{-3} \mathrm{~A} \\
\Delta \mathrm{I}_{\mathrm{B}}=100 \times 10^{-6} \mathrm{~A} \\
\beta=\frac{5}{100} \times 1000=50
\end{array}
\end{aligned}
$$
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