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In a transistor `10^(8)` electrons enter at the emitter in `10^(-4)` s, out of which `2%` electron go to the base. The current transfer ratio in common base configuration isA. 98B. 2C. 0.98D. 0.2 mA |
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Answer» Correct Answer - C `I_(b)=2% ` of `I_(e), I_(c)=98% ` of `I_(e), alpha=?` `alpha=(I_(c))/(I_(e))=(98% " of " I_(e))/(I_(e))=98% =0.98` |
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