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_BE) / delta(I_B)] at V_CE = constant

r_i = [delta(V_BE) / delta(I_B)] at V_CE = 10 Volts constant
Choosing points A & C on the graph for V_CE = 10 Volts constant.
r_i = (V_BE corresponding to point A - V_BE corresponding to point C) / (I_B corresponding to point A - I_B corresponding to point C)
= (0.35 - 0.15) Volts / (220 - 72.5) microA
= 0.2 V/ 147.5 x 10^-6micro A
= 1.356 k ohms.

For other graphs at V_CE at 0 volts constant and 5 volts constant, the slope seems to be nearly same.
So, for these graphs also r_i values will be nearly equal to 1.356 k ohms.

So, mean value of r_i = 1.356 k ohms.

ii) Consider the graph for OUTPUT CHARACTERISTICS

r_o = [delta(V_CE) / delta(I_C)] at I_B = constant

Choosing points A & C on the graph. These points are common for all three graphs for I_B constant at 50 micro A, 150 micro A & 250 micro A.

So, r_o = (V_CE corresponding to point A - V_CE corresponding to point C) / (I_C corresponding to point A - I_C corresponding to point C)
= (4 - 2) V / (8.5 - 4) mA = 2 V / 4.5 x 10^-3 A
= 444.44 ohms.

So, mean value of r_o = 444.44 ohms.

iii) & iv) Consider the graph for TRANSFER CHARACTERISTICS

a) beta_DC = (I_C) / (I_B) at constant V_CE

Choosing point A on the graph
beta_DC = (I_C corresponding to point A) / (I_B corresponding to point A)
= 6.25 mA / 25 microA
= 6.25 x 10^-3 A / 25 x 10^-6 A
= 250

b) beta_AC = [delta(I_C) / delta(I_B)] at V_CB = constant

Choose points A and B on the graph
Then, beta_AC = (I_C corresponding to point A - I_C corresponding to point B) / (I_B corresponding to point A - I_B corresponding to point B)] at V_CE = 5 V
= (25 - 10) mA / (6.25 - 2.4) microA
= 10 x 10^-3 A / 3.85 x 10^-6 A
= 259.74