Solution 3:

(a) Input low current,

I(IL) = (Vcc - Vbe(SAT) - Vce(SAT)) / Rb

        = (5 - 0.8 - 0.2) / 40K

= 0.1 mA

(b) Power Dissipated,

Pdd(avg) = (ILH + IHL + IHH + ILL) * Vcc / 4

When both the inputs are low,

Irba = Irbb = (5 - 0.8 - 0.2) / 40 = 0.1 mA

ILL = Irba + Irbb = 0.1 + 0.1 = 0.2 mA

When both the inputs are high,

Irba = Irbb = (5 - 0.7 - 0.8 - 0.8) / 40 = 0.0675 mA

Irc = (5 - 0.2 - 0.8) / 20 = 0.2 mA

IHH = 0.0675 * 2 + 0.2 = 0.335 mA

When Vina is high and Vinb is low,

Irba = 0.0675 mA

Irbb = 0.1 mA

Irc = 0.2 mA

IHL = 0.1 + 0.2 + 0.0675 = 0.3675 mA

When Vina is low and Vinb is high,

Irba = 0.1 mA

Irb = 0.0675 mA

Irc = 0.2 mA

ILH = 0.1 + 0.2 + 0.0675 = 0.3675 mA

Pdd = (0.3675 * 2 + 0.2 + 0.335) * 5 / 4 = 1.5875 mW