Question

In an n-p-n transistor, the collector current is 15
mA. If 95% of the electrons reach the collector,
then the emitter current will be
(1) 19 mA
(2) 300 mA
(3)19/300mA
(4)300/19mA​

Answer 1

Explanation:

(3) must be that .

That is correct

Relativeness of time.

Answer 2

$\underline{\blacksquare\:\:\:\footnotesize{\red{Figure}}}$

$\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\linethickness{5 mm}\put(0,0){\line(1,0){40}}\put(0,0){\line(0,1){40}}\put(0,40){\line(1,0){40}}\put(40,40){\line(0,-1){40}}\put(11,36){ \underline{n-p-n \:\:transistor} }\linethickness{0.1 mm}\put(4,20){\line(1,0){15}}\put(4,20.095){\vector(1,0){12}}\put(19,16){\line(0,1){8}}\put(23,24.8){\line(0,1){6}}\put(23,15){\line(0,-1){6}}\put(23,30.85){\vector(0,-1){4}}\put(23,15){\vector(0,-1){4}}\put(12,22){ I_B }\put(24,28){ I_C }\put(24,11){ I_E }\thicklines\put(19,22){\line(4,3){4}}\put(19,18){\line(4,-3){4}}\end{picture}$

$\underline{\blacksquare\:\:\:\footnotesize{\red{SolutioN}}}$

$\footnotesize{\text{collector current }\:,\:I_C\:=\:15mA}$

$\footnotesize{\text{According to the given condition , }\:I_C=95\%\:of \:I_E}$

$\footnotesize{\implies\:I_C=95\%\:of \:I_E}$

$\footnotesize{\implies\:I_C=\dfrac{95}{100}\times I_E}$

$\footnotesize{\implies\:I_E=\dfrac{100\times I_C}{95}}$

$\footnotesize{\implies\:I_E=\dfrac{100\times 15mA}{95}}$

$\footnotesize{\implies\:I_E=\dfrac{1500}{95}\:mA}$

$\footnotesize{\implies\:\boxed{\red{I_E=\dfrac{300}{19}\:mA}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{2 mm}\put(1,1){\line(1,0){6.5}}\end{picture}$

$\footnotesize{Now\:,\:\:\implies\:I_B=I_E-I_C}$

$\footnotesize{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\implies\:I_B=(\dfrac{300}{19}-10)\:mA}$

$\footnotesize{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\implies\:I_B=\dfrac{300-190}{17}\:mA}$

$\footnotesize{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\implies\:I_B=\dfrac{110}{19}\:mA}$

$\footnotesize{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\implies\:\boxed{\red{I_B=\dfrac{110}{19}\:mA}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{4 mm}\put(1,1){\line(1,0){6.5}}\end{picture}$