NEET-XII-Physics

pp year:2018

with Solutions -
  • #6
    In the circuit shown in the figure, the input
    voltage``V_i`` is 20 V, ``V_{BE}`` = 0 and
    ``V_{CE}`` = 0. The values of ``I_B``, ``I_C``
    and ``\beta`` are given by :-
    p6.png
    (1) ``I_B``= 40 ``\mu A``, ``I_C``= 10 mA, ``\beta``= 250
    (2) ``I_B``= 25 ``\mu A``, ``I_C``= 5 mA, ``\beta``= 200
    (3) ``I_B``= 20 ``\mu A``, ``I_C``= 5 mA, ``\beta``= 250
    (4) ``I_B``= 40 ``\mu A``, ``I_C``= 5 mA, ``\beta``= 125
    digAnsr:   4
    Ans : (4)
    Sol. Vi = IBRB + VBE
    20 = IB × (500 × 10
    3) + 0
    = =

    B 3
    20
    I 40µA
    500 10
    = +CC C C CEV I R V
    20 = IC × (4 × 10
    3) + 0
    IC = 5 × 10
    -3 = 5 mA



     = = =

    3
    C
    6
    B
    I 5 10
    125
    I 40 10