In the circuit shown in figure, when the input voltage of the base resistance is 10 V, `V_(be)` is zero and `V_(ce)` is also zero. Then

Options

A. `beta=110`
B. `I_b=25 muA`
C. `I_c=3.33 mA`
D. both (A) and (C )

Answer

D

Explanation

Given `V_i`=10 V, `R_b=400 kOmega=400xx10^3 Omega`
`R_c=3 kOmega=3xx10^3 Omega, V_(be)=0,V_(ce)=0,V_("CC")=10 V`
As `V_i-V_(be)=R_b I_b`
`therefore 10-0=(400xx10^3)I_b`
`l_b=10/(400xx10^3)`
`=25xx10^(-6) A = 25 muA`
`V_("CC")=V_(ce)=I_c R_c`
`I_c=10/(3xx10^3)=3.33xx10^(-3)=3.33 mA`
`beta=I_c/I_b=(3.33xx10^(-3))/(25xx10^(-6))=133`