In a `n-p-n` transistor `10^(10)` electrons enter the emitter in `10^(-6)s`. `2%` of the elecrons are lost in the base. The current transfer ratio and the current amplification factor will beA. `0.98,49`B. `98,0.49`C. `0.49,98`D. `49,0.49`

In a `n-p-n` transistor `10^(10)` electrons enter the emitter in `10^(

1.	In a `n-p-n` transistor `10^(10)` electrons enter the emitter in `10^(-6)s`. `2%` of the elecrons are lost in the base. The current transfer ratio and the current amplification factor will beA. `0.98,49`B. `98,0.49`C. `0.49,98`D. `49,0.49`
Answer» Correct Answer - A `alpha=(I_(C ))/(I_(B))=(0.98)` `beta=(I_(C ))/(I_(B))=(0.98)/(0.02)=49`

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In a `n-p-n` transistor `10^(10)` electrons enter the emitter in `10^(-6)s`. `2%` of the elecrons are lost in the base. The current transfer ratio and the current amplification factor will beA. `0.98,49`B. `98,0.49`C. `0.49,98`D. `49,0.49`

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