Question

In the following common emitter circuit if $\beta =100,V_{CE}=7V,V_{BE}$ = Negligible $R_C=2k\Omega$ then $I_B$ = ?

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

Answer 1

The correct option is B

0.04 mA

$V=V_{CE}+I_C{R}_L$

$\Rightarrow 15=7+I_C\times 2\times {10}^3{i}_C=4mA$

$\because \beta =\frac{i_C}{i_B}\Rightarrow i_B=\frac{4}{100}=0.04mA$