Question

In a NPN transistor, 108 electrons enter the emitter in ${10}^{-8}\text{sec.}$ If $1\%$ electrons are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively

• A. $0.8\text{and}49$
• B. $0.9\text{and}90$
• C. $0.7\text{and}50$
• D. $0.99\text{and}99$

Answer 1

The correct option is D $0.99\text{and}99$

As $1\%$ of electrons are lost in the base,
$\therefore I_B=\frac{I_E}{100}=0.01I_E$

And, the remaining $99\%$ electrons enter the collector region, so that, $I_C=0.99I_E$

$\Rightarrow \frac{I_C}{I_E}=0.99$

Now, current amplification factor is,
$\beta =\frac{I_C}{I_B}=\frac{0.99I_E}{0.01I_E}=99$

Hence, $\left(D\right)$ is the correct answer.