Question

In a P-N-P transistor, the collector current is $10\text{mA}$. If $90\%$ of the holes reach the collector, then emitter current will be,

• A. $13\text{mA}$
• B. $12\text{mA}$
• C. $11\text{mA}$
• D. $10\text{mA}$

Answer 1

The correct option is C $11\text{mA}$

Here, $I_C=10\text{mA}$

As $90$% of the holes reach the collector,
$\Rightarrow \frac{I_C}{I_E}=\frac{90}{100}$

So, $I_E=\frac{10}{9}\times I_C=\frac{10}{9}\times 10=\frac{100}{9}$

$I_E\approx 11\text{mA}$

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