Question:
If an emitter current is changed by $4 \mathrm{~mA}$, the collector current changes by $3.5 \mathrm{~mA}$. The value of $\beta$ will be :
Correct Option: 1
Solution:
$\mathrm{I}_{\varepsilon}=\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{B}}$
$\Rightarrow \Delta \mathrm{I}_{\varepsilon}=\Delta \mathrm{I}_{\mathrm{C}}+\Delta \mathrm{I}_{\mathrm{B}}$
$4 \mathrm{~mA}=3.5 \mathrm{~mA}+\Delta \mathrm{I}_{\mathrm{B}}$
$\Rightarrow \Delta \mathrm{I}_{\mathrm{B}}=0.5 \mathrm{~mA}$
$\Rightarrow \beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}$
$\beta=\frac{3.5}{0.5}$
$\Rightarrow \beta=7$