Question

If for the following common emitter circuit, $\beta =100,V_{CE}=7V,V_{BE}$ is negligible and $R_C=2k\Omega$4, then $i_B=$?

• A. $0.04mA$
• B. $0.03mA$
• C. $0.02mA$
• D. $0.01mA$

Answer 1

The correct option is A $0.04mA$

$V=V_{CE}+i_C{R}_C$
$15=7+i_C\times 2\times {10}^3$
$i_C=\frac{8}{2\times {10}^3}=4mA$
$\therefore \beta =\frac{i_C}{i_B}$
or $i_B=\frac{i_C}{\beta }=\frac{4}{100}-0.04mA$.