A voltage regulator circuit uses a Zener diode of breakdown voltage $6\ \text{V}$ connected across a load resistance $R_L = 4\ \text{k}\Omega$. A series resistance $R_i = 1\ \text{k}\Omega$ connects this combination to a battery of voltage $V_B$. If $V_B$ varies from $8\ \text{V}$ to $16\ \text{V}$, what are the minimum and maximum values of the current through the Zener diode?
Answer: (C) $0.5\ \text{mA};\ 8.5\ \text{mA}$
Load current is fixed by the Zener: $I_L = \dfrac{6}{4\ \text{k}\Omega} = 1.5\ \text{mA}$.
$V_B = 8\ \text{V}$: $I = \dfrac{8-6}{1\ \text{k}\Omega} = 2\ \text{mA}$, so $I_Z = 0.5\ \text{mA}$.
$V_B = 16\ \text{V}$: $I = \dfrac{16-6}{1\ \text{k}\Omega} = 10\ \text{mA}$, so $I_Z = 8.5\ \text{mA}$.
Solution by Sreeraj P, M.Sc Physics