Question

In the following common emitter circuit, if $\beta =100$ , $V_{CE}=7V$ , $R_{C}=2k\Omega$ and $V_{BE}$ is negligible, then $I_{B}$ is

• A. $0.01 \, \, \text{mA}$
• B. $0.04 \, \, \text{mA}$
• C. $0.02 \, \, \text{mA}$
• D. $0.03 \, \, \text{mA}$

Answer 1

Answer: (B)

$V=V_{C E}+I_{C}R_{C}$
$\Rightarrow 15=7+I_{C}\times 2\times 10^{3}\Rightarrow I_{C}=4mA$
$\because \beta =\frac{I_{C}}{I_{B}}\Rightarrow I_{B}=\frac{4}{100}=0.04 \, mA$