Explanation

(a) When diode is forward biased, then there is a small voltage drop across it
$i = \frac{V - V^{'}}{R}$
Given, $V = 8\ volt,\ V^{'} = 0.5\ volt,$
$R = 2.2\ k\ \Omega = 2.2 \times 10^{3}\Omega$
$\therefore = \ \frac{8 - 0.5}{2.2 \times 10^{3}} = \frac{7.5 \times 10^{3}}{2.2} = 3.4 \times 10^{- \ 3}$ A
$i = 3.4\ mA.$