Question

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

Answer 1

Hint :In an $ n-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\;mA $ current.
In the question, it is given $ 96\;% $ of the electrons emitted reach the collector.
We know that the current can be calculated as
 $ I=\dfrac{ne}{t} $ , where $ e $ is the charge of an electron, $ t $ is the time of travel, and $ n $ 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\;% $ of the electrons reaching the collector signifies $ 96\;% $ of the current emitted by the emitter reaching the collector.
As the emitter current is $ x\;mA $ , $ 96\;% $ of emitter current is calculated as
 $ {{I}_{c}}=\dfrac{96\times x}{100} $
 $ \therefore {{I}_{c}}=0.96x $
Here, we are given that the collector current is $ 10\;mA $ . Thus substituting the value,
 $ \therefore 10mA=0.96x $
 $ \therefore x=\dfrac{10mA}{0.96} $
Writing the numerator and denominator without units in powers of $ \;10 $
 $ \therefore x=\dfrac{10\times {{10}^{-3}}}{9.6\times {{10}^{-1}}} $
Shifting the power term in the numerator,
 $ \therefore x=\dfrac{10}{9.6}\times {{10}^{-2}} $
 $ \therefore x=1.08\times {{10}^{-2}}A $
Rearranging the decimal point to get a smaller power,
 $ \therefore x=10.8\times {{10}^{-3}}A $
 $ \therefore x\approx 11mA $
Hence, the emitter current is $ 11\;mA $ which is distributed in the collector and the base,
 $ \therefore {{I}_{E}}={{I}_{B}}+{{I}_{C}} $
Substituting the given and the obtained values,
 $ \therefore 11mA={{I}_{B}}+10mA $
 $ \therefore {{I}_{B}}=1mA $
Hence, the correct answer is Option $ (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.