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Question

In an NPN transistor 1010 electrons enter the emitter in 106 s. If 2% of the electrons are lost in the base. The current transfer ratio of the transistor 98n. The value of n is .(integer only)

A
100.0
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B
100.00
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C
100
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Solution

The emitter current (IE) is given by,

IE=Net=1010×(1.6×1019)106=1.6 mA

The base current (IB) is given by,

IB=2100×1010×(1.6×1019)106=0.032 mA

In a transistor, IE=IB+Ic

Ic=IEIB=1.60.032=1.568 mA

The current transfer ratio=IcIE=1.5681.6=0.98=98100

n=100

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