Question

An $\text{npn}$ transistor circuit has $\alpha =\mathrm{0.985.}$ If $I_C=2\text{mA}$, then the value of $I_B$ is

• A. $0.03\text{mA}$
• B. $0.003\text{mA}$
• C. $0.66\text{mA}$
• D. $0.015\text{mA}$

Answer 1

The correct option is A $0.03\text{mA}$

Given,
$\alpha =0.985;I_C=2\text{mA}$

As we know that current gain for $\text{npn}-$ transistor is,

$\alpha =\frac{I_C}{I_E}=\frac{I_C}{I_C+I_B}=0.985$

$I_C=0.985(I_C+I_B)$

$\Rightarrow 0.985I_B=I_C-0.985I_C$

$\Rightarrow I_B=\frac{0.015\times 2\times {10}^{-3}}{0.985}$

$\Rightarrow I_B\approx 0.03\text{mA}$

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