In the circuit shown in the figure, the input voltage `V_i = +5 V, V_(BE) = + 0.8 V` and `V_(CE) = +0.12 V`. Find the values of `I_B I_C` and `beta`.

Explanation

`I_B=V_i/R_B=(5V)/(80kOmega)`
`=5/(80xx10^3)=5/8xx10^(-4)=0.625xx10^(-4)=62.5xx10^(-6)`
`I_B=62.5muA`
`I_C=V_"CC"/R_C =(12V)/(5kOmega)`
`=12/(5xx10^3)=2.4xx10^(-3)`
`I_C`=2.4 mA
`beta=I_C/I_B=(2.4xx10^(-3))/(62.5xx10^(-6))=0.0384xx10^3 , beta=38`