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For an n-p-n transistorpercentage of electronsemitted reaching the collector is 80% if th ecollectcurrentis 1 mA |
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Answer» the EMITTER CURRENT is 1.5 mA `rArrI_(e) = (10)/(8)I_(c )` Given `I_(e) = 1 mA` `I_(b) = I _(e)- I_(e)` `=(1.25 -1) mA = 0 .25 mA` Henceoption (c ) and (d) are correct. |
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