In the circuit shown in figure, when the input voltage of the base resistance is 10 ~V , V_ b e is zero and V_ c e is also zero. Value of is

Step-by-step solution

Given V_i=10 ~V , R_b= 00 k =400 10^3 aligned & R_c=3 k =3 10^3 , V_ b e =0 & V_ c e =0, V_ C C =10 ~V & As V_i-V_ b e =R_b I_b aligned aligned & 10-0= (400 10^3 ) I_b & I_b= 10 400 10^3 =25 10⁻⁶ ~A =25 A & and V_ C C -V_ c e =I_c R_c aligned aligned & I_c= 10 3 10^3 =3.33 10⁻³=3.33 ~mA & = I_c I_b = 3.33 10⁻³ 25 10⁻⁶ =133 aligned