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PhysicsMediumMCQAIIMS · 1989
In an $NPN$ transistor circuit,the collector current is $10 \,mA$. If $90\%$ of the electrons emitted reach the collector,the emitter current $(i_E)$ and base current $(i_B)$ are given by
A
$i_E = -1 \,mA, i_B = 9 \,mA$
B
$i_E = 9 \,mA, i_B = -1 \,mA$
C
$i_E = 1 \,mA, i_B = 11 \,mA$
D
$i_E = 11 \,mA, i_B = 1 \,mA$
Solution
(D) Given that the collector current $i_C = 10 \,mA$.
Since $90\%$ of the electrons emitted from the emitter reach the collector,we have the relation $i_C = 0.90 \times i_E$.
Substituting the value of $i_C$,we get $10 = 0.90 \times i_E$.
Therefore,$i_E = \frac{10}{0.9} \approx 11.11 \,mA$. Rounding to the nearest integer as per the options,$i_E = 11 \,mA$.
Using the Kirchhoff's current law for a transistor,$i_E = i_B + i_C$.
Substituting the values,$11 = i_B + 10$.
Thus,$i_B = 11 - 10 = 1 \,mA$.
Since $90\%$ of the electrons emitted from the emitter reach the collector,we have the relation $i_C = 0.90 \times i_E$.
Substituting the value of $i_C$,we get $10 = 0.90 \times i_E$.
Therefore,$i_E = \frac{10}{0.9} \approx 11.11 \,mA$. Rounding to the nearest integer as per the options,$i_E = 11 \,mA$.
Using the Kirchhoff's current law for a transistor,$i_E = i_B + i_C$.
Substituting the values,$11 = i_B + 10$.
Thus,$i_B = 11 - 10 = 1 \,mA$.
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