Q102 / 125 · BITSAT 2015 · Physics
In CE transistor amplifier, the audio signal voltage across the collector resistance of $2 \mathrm{k} \Omega$ is $2 \mathrm{~V}$. If the base resistance is $1 \mathrm{k} \Omega$ and the current amplification of the transistor is $100,$ the input signal voltage is
- A$2 \mathrm{mV}$
- B$3 \mathrm{mV}$
- Correct answerC$10 \mathrm{mV}$
- D$0.1 \mathrm{mV}$
Solution
Given : Voltage across the collector $\mathrm{V}_{0}=2$ $\mathrm{V} ;$ collector resistance, $\mathrm{R}_{\mathrm{c}}=2 \times 10^{3} \Omega ;$ Base resistance $\mathrm{R}_{\mathrm{B}}=1 \times 10^{3} \Omega ;$ Input signal voltage, $\mathrm{V}_{\mathrm{i}}=?$
$$ \begin{array}{l} \mathrm{V}_{0}=\mathrm{I}_{\mathrm{C}} \mathrm{R}_{\mathrm{C}}=2 \\ \Rightarrow \quad \mathrm{I}_{\mathrm{C}}=\frac{2}{2 \times 10^{3}}=10^{-3} \mathrm{~A} \\ \text { Current gain } \alpha=\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}=100 \\ \Rightarrow \quad \mathrm{I}_{\mathrm{B}}=\frac{\mathrm{I}_{\mathrm{C}}}{100}=\frac{10^{-3}}{100}=10^{-5} \mathrm{~A} \\ \quad \mathrm{~V}_{\mathrm{i}}=\mathrm{R}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}} \Rightarrow \mathrm{V}_{\mathrm{i}}=1 \times 10^{3} \times 10^{-5} \\ \Rightarrow \quad \mathrm{V}_{\mathrm{i}}=10^{-2} \mathrm{~V} \Rightarrow \mathrm{V}_{\mathrm{i}}=10 \mathrm{mV} \end{array} $$
$$ \begin{array}{l} \mathrm{V}_{0}=\mathrm{I}_{\mathrm{C}} \mathrm{R}_{\mathrm{C}}=2 \\ \Rightarrow \quad \mathrm{I}_{\mathrm{C}}=\frac{2}{2 \times 10^{3}}=10^{-3} \mathrm{~A} \\ \text { Current gain } \alpha=\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}=100 \\ \Rightarrow \quad \mathrm{I}_{\mathrm{B}}=\frac{\mathrm{I}_{\mathrm{C}}}{100}=\frac{10^{-3}}{100}=10^{-5} \mathrm{~A} \\ \quad \mathrm{~V}_{\mathrm{i}}=\mathrm{R}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}} \Rightarrow \mathrm{V}_{\mathrm{i}}=1 \times 10^{3} \times 10^{-5} \\ \Rightarrow \quad \mathrm{V}_{\mathrm{i}}=10^{-2} \mathrm{~V} \Rightarrow \mathrm{V}_{\mathrm{i}}=10 \mathrm{mV} \end{array} $$
Hence, the correct answer is option: 'C'.