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  • In an n-p-n transistor electrons enter the emitter in <img alt="10^{-6}s." src="https://entrancecorner.oncodecogs.com/gif.latex?10%5E%7B-6%7Ds."

In an n-p-n transistor 10^10 electrons enter the emitter in 10^{-6}s. 2% of the electrons are lost in the base. The current amplification factor is:

Option: 1

19


Option: 2

29


Option: 3

39


Option: 4

49


Answers (1)

best_answer

\mathrm{\mathrm{IE}=\frac{\mathrm{N}_{\mathrm{e}}}{\mathrm{t}}=\frac{10^{10} \times\left(1.6 \times 10^{-19}\right)}{10^{-6}}=1.6 \mathrm{~mA}}

The base current I_B is given by

\mathrm{\mathrm{I}_{\mathrm{B}}=\frac{2}{100} \times 1.6=0.032 \mathrm{~mA}}

In a transistor

\mathrm{\begin{aligned} & I_E=I_B+I_C \\ & I_C=I_E-I_B=1.6-0.032=1.568 \mathrm{~mA} \end{aligned}}

Current amplification factor 

\mathrm{=\frac{I_c}{l_k}=49 .}

Posted by

Devendra Khairwa

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