In the circuit shown in Fig. when the input voltage of the base resistance is `10V, V_(be)` is zero and `V_(ce)` is also zero. Find the values of `I_, I_` and `beta`.
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Here, `V_(i)=10V, V_(BE)=0, V_(CE)=0, V_(C C)=10V, R_B=400kOmega=400xx10^(3)Omega, R_C=3kOmega=3xx10^(3)Omega` Now `V_(i)-V_(BE)=R_b I_b` `:. 10-0=(400xx10^(3))I_b` or `I_b=10/(400xx10^(3))=25xx10^(-6)A=25muA` Also `V_(C C)-V_(CE)=I_c R_c` or `10-0=I_cxx(3xx10^(3))` or `I_c=10/(3xx10^(3))=3.33xx10^(-3) A=3.33mA` `beta=I_c/I_b=(3.33xx10^(-3))/(25xx10^(-6))=133`
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