In the circuit shown below, the diode has $20 \Omega$ in forward bias resistance. When $V_i$ increases from 8 V to 12 V, then change in the current is found to be $x \mathrm{~mA}$. Find the value of $x$.

Diode will be in forward bias when $V_i$ is greater than 10 V and it conducts till $V_i$ is 12 V . Total circuit resistance $=200+20=220 \Omega$ When $V_i$ is less than 10 V , the diode will not conduct due to its reverse biasing. Thus, initial current, $i_1=0$. When $V_i=12 \mathrm{~V}$, then diode will be in forward bias with effective voltage, $V^{\prime}=12-10=2 \mathrm{~V}$ $\therefore$ Maximum current, $I_2=\frac{V^{\prime}}{R}=\frac{2}{220}=0.009 \mathrm{~A}=9 \mathrm{~mA}$ $\therefore$ Change in current, $\Delta I=I_2-I_1=9 \mathrm{~mA}$