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The input resistance of a common emitter amplifier is $330 \Omega$ and the load resistance is 5 $\mathrm{k} \Omega$. A change of base current is $15 \mu \mathrm{A}$ results in the change of collector current by $1 \mathrm{~mA}$. The voltage gain of amplifier is
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1010
Given: $\Delta \mathrm{I}_{\mathrm{C}}=1 \mathrm{~mA}=10^{-3} \mathrm{~A}$ $\Delta \mathrm{I}_{\mathrm{b}}=15 \mu \mathrm{A}=15 \times 10^{-6} \mathrm{~A}$ $\mathrm{R}_{\mathrm{L}}=5 \mathrm{k} \Omega=5 \times 10^{3} \Omega$ $\mathrm{Ri}=330 \Omega$ The voltage gain of an amnl
The voltage gain of an amplifier
$\begin{array}{l}
\mathrm{A}_{\mathrm{r}}=\frac{\Delta \mathrm{I}_{\mathrm{C}} \times \mathrm{R}_{\mathrm{L}}}{\Delta \mathrm{I}_{\mathrm{b}} \times \mathrm{R}_{\mathrm{i}}} \\
=\frac{10^{-3} \times 5 \times 10^{3}}{15 \times 10^{-6} \times 330} \approx 1010
\end{array}$
The voltage gain of an amplifier
$\begin{array}{l}
\mathrm{A}_{\mathrm{r}}=\frac{\Delta \mathrm{I}_{\mathrm{C}} \times \mathrm{R}_{\mathrm{L}}}{\Delta \mathrm{I}_{\mathrm{b}} \times \mathrm{R}_{\mathrm{i}}} \\
=\frac{10^{-3} \times 5 \times 10^{3}}{15 \times 10^{-6} \times 330} \approx 1010
\end{array}$
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