In an NPN transistor circuit, the collector current is 10rnA.If 90% of the electrons emitted reach the collector, the emitter current (iE) and base current (iB) are given by
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a
iE=1 mA, iB=9 mA
b
iE=9 mA, iB=-1 mA
c
iE=1 mA, iB=11 mA
d
iE=11 mA, iB=1 mA
answer is D.
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Detailed Solution
iC=90100×iE⇒10=0.9×iE⇒iE=11 mA Also, iE=iB+iC⇒11-10=1 mA