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In an `NPN` transistor the collector current is `10 mA`. If `90%` of electrons reach collector, the emitter current `(i_(E))` and base current `(i_(B))` are given byA. `I_(E) = 1mA , I_(B) = 11 mA`B. `I_(E) = 11 mA l I_(B) = 1mA`C. `I_(E) = -1mA , I_(B) = 9mA`D. `I_(E) = 9mA , I_(B) = -1 mA` |
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Answer» Correct Answer - B `I_(C)=90%I_(E)=(90)/(100)I_(E)rArr I_(E)=(100)/(90)xxI_(C)=11mA` `I_(B)=I_(E)-I_(C)=11mA-10mA=1mA`. |
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