MHTCET 14 May 2023 Shift 2 Solved Paper

In an n-p-n transistor, the collector current (Ic) is primarily due to the flow of electrons from the emitter to the collector. However, not all the electrons emitted by the emitter reach the collector; a small percentage of them recombine in the base. This percentage is given by the common base current gain $(\alpha)$, which in this case is $80\%$ or 0.80 .The base current $(I_b)$ can be calculated by knowing that the emitter current $(I_e)$ is the sum of the collector current $(I_c)$ and the base current $(I_b)$. The relationship is given by the equation:$I_e=I_c+I_b$ Given that $\alpha$, which is the common base current gain, is the ratio of the collector current to the emitter current:$\alpha = \frac{I_c}{I_e}$Let's solve for $I_e$ :$I_e = \frac{I_c}{\alpha} = \frac{28\,\mathrm{mA}}{0.80} = 35\,\mathrm{mA}$ Now, we can find the base current using the initial relation between $I_e, I_c$, and $I_b$ :$I_b = I_e - I_c = 35\,\mathrm{mA} - 28\,\mathrm{mA}$$I_b = 7\,\mathrm{mA}$Hence, the base current $I_b$ is $7\,\mathrm{mA}$, which corresponds to Option D.