MCQ
If an emitter current is changed by $4\, mA ,$ the collector current changes by $3.5\, mA$. The value of $\beta$ will be
- ✓$7$
- B$0.5$
- C$0.875$
- D$3.5$
$\Rightarrow \Delta I _{\varepsilon}=\Delta I _{ C }+\Delta I _{ B }$
$4 mA =3.5 mA +\Delta I _{ B }$
$\Rightarrow \Delta I _{ B }=0.5 mA$
$\Rightarrow \beta=\frac{\Delta I _{ C }}{\Delta I _{ B }}$
$\beta=\frac{3.5}{0.5}$
$\Rightarrow \beta=7$
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$(A)$ the particle will hit $T$ if projected at an angle $45^{\circ}$ from the horizontal
$(B)$ the particle will hit $T$ if projected either at an angle $30^{\circ}$ or $60^{\circ}$ from the horizontal
$(C)$ time taken by the particle to hit $T$ could be $\sqrt{\frac{5}{6}} \mu s$ as well as $\sqrt{\frac{5}{2}} \mu s$
$(D)$ time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} \mu s$
