Q 12-14-102JEE MainJEE Main 2023 (8 Apr, Shift 2)Medium
For a given transistor amplifier circuit in CE configuration $V_{CC}=1$ V, $R_C=1\ \text{k}\Omega$, $R_b=100\ \text{k}\Omega$ and $\beta=100$. Value of base current $I_b$ is
Answer: (D) $I_b=10\ \mu\text{A}$
Taking the collector–emitter voltage as negligible (transistor driven to saturation), $I_C=\dfrac{V_{CC}}{R_C}=\dfrac{1}{1000}=1$ mA.
$$I_b=\frac{I_C}{\beta}=\frac{1\ \text{mA}}{100}=10\ \mu\text{A}$$
Solution by Sreeraj P, M.Sc Physics