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Question: In an \( n-p-n \) transistor circuit, the collector current is \( 10\;mA \) . If \( 96\;% \) of the ...

In an n−p−nn-p-n transistor circuit, the collector current is 10  mA10\;mA . If 96  96\;% of the electrons emitted reach the collector:
(A) The base current will be 1  mA1\;mA
(B) The base current will be −1  mA-1\;mA
(C) The emitter current will be 9  mA9\;mA
(D) The emitter current will be 15  mA15\;mA

Explanation

Solution

Hint : In an n−p−nn-p-n circuit, the base is sandwiched between the emitter region and the collector region. Hence, the current is released by the emitter and is distributed between the base and the collector

Complete Step By Step Answer:
As per the given question, the electrons are always emitted by the emitter. Hence, suppose the emitter emits x  mAx\;mA current.
In the question, it is given 96  96\;% of the electrons emitted reach the collector.
We know that the current can be calculated as
I=netI=\dfrac{ne}{t} , where ee is the charge of an electron, tt is the time of travel, and nn is the number of electrons.
As the charge of an electron is a constant value, and here the time of travel is not considered, the current depends on the number of electrons.
Hence, we can conclude that 96  96\;% of the electrons reaching the collector signifies 96  96\;% of the current emitted by the emitter reaching the collector.
As the emitter current is x  mAx\;mA , 96  96\;% of emitter current is calculated as
Ic=96×x100{{I}_{c}}=\dfrac{96\times x}{100}
∴Ic=0.96x\therefore {{I}_{c}}=0.96x
Here, we are given that the collector current is 10  mA10\;mA . Thus substituting the value,
∴10mA=0.96x\therefore 10mA=0.96x
∴x=10mA0.96\therefore x=\dfrac{10mA}{0.96}
Writing the numerator and denominator without units in powers of   10\;10
∴x=10×10−39.6×10−1\therefore x=\dfrac{10\times {{10}^{-3}}}{9.6\times {{10}^{-1}}}
Shifting the power term in the numerator,
∴x=109.6×10−2\therefore x=\dfrac{10}{9.6}\times {{10}^{-2}}
∴x=1.08×10−2A\therefore x=1.08\times {{10}^{-2}}A
Rearranging the decimal point to get a smaller power,
∴x=10.8×10−3A\therefore x=10.8\times {{10}^{-3}}A
∴x≈11mA\therefore x\approx 11mA
Hence, the emitter current is 11  mA11\;mA which is distributed in the collector and the base,
∴IE=IB+IC\therefore {{I}_{E}}={{I}_{B}}+{{I}_{C}}
Substituting the given and the obtained values,
∴11mA=IB+10mA\therefore 11mA={{I}_{B}}+10mA
∴IB=1mA\therefore {{I}_{B}}=1mA
Hence, the correct answer is Option (A)(A) .

Note :
We know that as the current emitted by the emitter is distributed between the base and the collector, the current reaching the collector is always less than the current emitted by the emitter. However, as the base is very thin, it absorbs very little current. Hence, the difference between the currents of the emitter and collector is very less.