1. Use the Zener regulator condition

When the Zener diode is in breakdown, it maintains the output voltage at $V_Z = 6\,\text{V}.$

So the load resistance $R_L = 4\,\text{k}\Omega$ also has $6\,\text{V}$ across it.

2. Find the load current

$I_L = \frac{V_Z}{R_L} = \frac{6}{4\times 10^3} = 1.5\times 10^{-3}\,\text{A} = 1.5\,\text{mA}.$

3. Find the current through the series resistor

The series resistor is $R_i = 1\,\text{k}\Omega$.

Current through it is $I = \frac{V_B - V_Z}{R_i}.$

This current splits into load current and Zener current: $I = I_L + I_Z.$

Hence, $I_Z = I - I_L = \frac{V_B - 6}{1\,\text{k}\Omega} - 1.5\,\text{mA}.$

4. Minimum Zener current

Minimum $I_Z$ occurs when battery voltage is minimum, i.e. $V_B = 8\,\text{V}$.

Then series current is $I = \frac{8-6}{1\times 10^3} = \frac{2}{1000} = 2\,\text{mA}.$

So, $I_Z = 2 - 1.5 = 0.5\,\text{mA}.$

5. Maximum Zener current

Maximum $I_Z$ occurs when battery voltage is maximum, i.e. $V_B = 16\,\text{V}$.

Then series current is $I = \frac{16-6}{1\times 10^3} = \frac{10}{1000} = 10\,\text{mA}.$

So, $I_Z = 10 - 1.5 = 8.5\,\text{mA}.$

6. Match with options

Thus,

• Minimum Zener current $= 0.5\,\text{mA}$
• Maximum Zener current $= 8.5\,\text{mA}$

So the correct option is A.