i) Produce the linear portion CAB of the graph further to meet Voltage axis

CALCULATIONS :-

i) Consider the graph for INPUT CHARACTERISTICS

r_i = [delta(V_EB) / delta(I_E)] at V_CB = constant

a) r_i = [delta(V_EB) / delta(I_E)] at V_CB = 0 Volts
= (V_EB corresponding to point E - V_EB corresponding to point F) / (I_E corresponding to point E - I_E corresponding to point F)
= (0.65 - 0.6) Volts / (20 - 2.5) mA
= 0.05 / 17.5 x 10^-3
= 2.857 ohms.

b) r_i = [delta(V_EB) / delta(I_E)] at V_CB = 4 Volts
= (V_EB corresponding to point C - V_EB corresponding to point D) / (I_E corresponding to point C - I_E corresponding to point D)
= (0.6 - 0.55) Volts / (11 - 1.75) mA
= 0.05 / 9.25 x 10^-3
= 5.405 ohms.

c) r_i = [delta(V_EB) / delta(I_E)] at V_CB = 8 Volts
= (V_EB corresponding to point A - V_EB corresponding to point B) / (I_E corresponding to point A - I_E corresponding to point B)
= (0.6 - 0.55) Volts / (20 - 5) mA
= 0.05 / 15 x 10^-3
= 3.333 ohms.

So, mean value of r_i = 3.865 ohms.

ii) Consider the graph for OUTPUT CHARACTERISTICS

r_o = [delta(V_CB) / delta(I_C)] at I_E = constant

Here we see that for any change in V_CB, the corresponding change in I_C is zero.
So, r_o = Finite number / zero = infinity
So, r_o = infinite.

iii) & iv) Consider the graph for TRANSFER CHARACTERISTICS

a) alpha_DC = (I_C) / (I_E)

Choose point B on the graph
So, alpha_DC = (I_C corresponding to point B) / (I_E corresponding to point B)
= 10 mA / 10 mA = 1

b) alpha_AC = [delta(I_C) / delta(I_E)] at V_CB = constant

Choose points A and C on the graph
Then, alpha_AC = (I_C corresponding to point A - I_C corresponding to point C) / (I_E corresponding to point A - I_E corresponding to point C)] at V_CB = 50 mV
= (15 - 10) mA / (15 - 10) mA = 1