In an NPN transistor 1010 electrons enter the emitter in 10−6s. 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×(1.6×10−19)10−6=1.6mA
The base current (IB) is given by,
IB=2100×1010×(1.6×10−19)10−6=0.032mA
In a transistor, IE=IB+Ic
⇒Ic=IE−IB=1.6−0.032=1.568mA
The current transfer ratio=IcIE=1.5681.6=0.98=98100