Question

If an emitter current is changed by $4mA,$ the collector current changes by $3.5mA$. The value of $\beta$ will be

• A. 7
• B. 0.5
• C. 0.875
• D. 3.5

Answer 1

Answer: (A)

$I_{\epsilon }=I_C+I_B$
$\Rightarrow \Delta I_{\epsilon }=\Delta I_C+\Delta I_B$
$4mA=3.5mA+\Delta I_B$
$\Rightarrow \Delta I_B=0.5mA$
$\Rightarrow \beta =\frac{\Delta I_C}{\Delta I_B}$
$\beta =\frac{3.5}{0.5}$
$\Rightarrow \beta =7$