Given data

• Zener voltage: $V_Z = 8\,\text{V}$
• Supply voltage: $V_S = 20\,\text{V}$
• Maximum Zener current: $I_{Z,\max} = 25\,\text{mA} = 0.025\,\text{A}$

Voltage across the series resistor Since the Zener diode maintains $8\,\text{V}$ across itself in breakdown, the remaining voltage appears across $R$: $V_R = V_S - V_Z = 20 - 8 = 12\,\text{V}$

Condition for minimum resistance The current through the resistor is $I = \frac{V_R}{R}$ To ensure the Zener current does not exceed its maximum value, the largest allowed current is $I_{\max} = I_{Z,\max} = 25\,\text{mA}$ (This gives the minimum possible value of $R$.)

Calculate $R_{\min}$ Using Ohm’s law, $R_{\min} = \frac{V_R}{I_{\max}} = \frac{12}{0.025} = 480\,\Omega$

Final answer $\boxed{480\,\Omega}$

Comparison with stored answer The derived answer is $480\,\Omega$, which matches the stored correct answer.