Question

In an n-p-n transistor 10^10 electrons enters the emitter in 10^-6 seconds, 2% electrons are lost in the base then current amplification is

Answer 1

Answer:

0.98, 49

Explanation:

$l _{e} = \frac{n _{e})(e) }{t}$

where he is number of electrons entees into the emitter,e is

change on election and t is time

$i _{e} = \times \frac{10 {}^{10} }{10 {}^{ - 6} } (1.6 \times {10 }^{ - 19} )$

$i _{e} = {10}^{3} = 1.6 ma$

2% of the electrical are electrons are listed in the base hence , the base current is