In $N P N$ transistor, $10^{10}$ electrons enters in emitter region in $10^{- 6}$ sec. If $2$ electrons are lost in base region then collector current and current amplification factor $(β)$ respectively are

1.57 mA, 49

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1.92 mA, 70

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2 mA, 25

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2.25 mA, 100

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Solution

The correct option is A 1.57 mA, 49
$I_{e} = 10^{10} \times$
$1.6 \times 10^{- 19} \times \frac{1}{10^{- 6}} = 1.6 m A (I = \frac{Q}{t})$ Since 2 percent electrons are absorbed by base, hence 98
percent electrons reaches the collectori.e. a=0.98 $I c = α I_{e} = 0.98 \times 1.6 = 1.568 m A \approx 1.57 m A A l s o c u r r e n t a m p l i f i c a t i o n f a c t o r β = α 1 - α = 0.980.02 = 49$

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In NPN transistor, 1010 electrons enters in emitter region in 10−6 sec. If 2 electrons are lost in base region then collector current and current amplification factor (β) respectively are

The correct option is A 1.57 mA, 49Ie=1010×1.6×10−19×110−6=1.6mA(I=Qt) Since 2 percent electrons are absorbed by base, hence 98percent electrons reaches the collectori.e. a=0.98 Ic=αIe=0.98×1.6=1.568mA≈1.57mAAlsocurrentamplificationfactorβ=α1−α=0.980.02=49

In $N P N$ transistor, $10^{10}$ electrons enters in emitter region in $10^{- 6}$ sec. If $2$ electrons are lost in base region then collector current and current amplification factor $(β)$ respectively are