Question

In a $\text{NPN}$ transistor circuit, the collector current is $10\text{mA}.$ If $90\%$ of the electrons emitted reaches the collector, the emitter current $\left(I_E\right)$ and base current $\left(I_B\right)$ are given by-

• A. $I_E=1\text{mA};I_B=11\text{mA}$
• B. $I_E=11\text{mA};I_B=1\text{mA}$
• C. $I_E=-1\text{mA};I_B=9\text{mA}$
• D. $I_E=9\text{mA};I_B=-1\text{mA}$

Answer 1

The correct option is B $I_E=11\text{mA};I_B=1\text{mA}$

Given,

$I_C=10\text{mA}\text{also}{I}_C=90\%I_E$

As we know, $I_E=I_B+I_C$

$I_C=0.9I_E$

$\Rightarrow I_E=\frac{10}{0.9}\approx 11\text{mA}$

$\Rightarrow I_B=I_E-I_C=11\text{mA}-10\text{mA}=1\text{mA}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.