Q138 / 160 · WBJEE 2010 · Physics
In a common emitter configuration, a transistor has $\beta=50$ and input resistance $1 \mathrm{k} \Omega$. If the peak value of a.c. input is $0.01 \mathrm{~V}$ then the peak value of collector current is
- A$0.01 \mu \mathrm{A}$
- B$0.25 \mu \mathrm{A}$
- C$100 \mu \mathrm{A}$
- Correct answerD$500 \mu \mathrm{A}$
Solution
$$
\text { Hints: } \beta=50 \Rightarrow \beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}} \Rightarrow \Delta \mathrm{I}_{\mathrm{C}}=\beta \times \Delta \mathrm{I}_{\mathrm{B}}
$$
$$
\begin{aligned}
& \Delta \mathrm{I}_{\mathrm{B}}=\frac{0.01}{10^3}=10^{-2} \times 10^{-3}=10^{-5} \\
& \Delta \mathrm{I}_{\mathrm{C}}=50 \times 10^{-5}=500 \times 10^{-6}=500 \mu \mathrm{A}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.