1. Given data

   • Zener breakdown voltage: $V_Z = 8\,\text{V}$
   • Power rating: $P_Z = 1.6\,\text{W}$
   • Input voltage fluctuates up to $10\,\text{V}$
   • We need series resistance $R_s$ for safe operation.

2. Maximum safe zener current

   The zener power must not exceed its rating: $P_Z = V_Z I_{Z,\max}$ So, $I_{Z,\max} = \frac{P_Z}{V_Z} = \frac{1.6}{8} = 0.2\,\text{A}$

3. Condition for safe operation

   The zener current is maximum when the supply voltage is maximum.

   At maximum input voltage $V_{in}=10\,\text{V}$, the voltage across the series resistor is: $V_R = V_{in} - V_Z = 10 - 8 = 2\,\text{V}$

   Hence, $I_Z = \frac{V_R}{R_s} = \frac{2}{R_s}$

   For safe operation, $I_Z \le I_{Z,\max} = 0.2\,\text{A}$

   Therefore, $\frac{2}{R_s} \le 0.2$ $R_s \ge \frac{2}{0.2} = 10\,\Omega$

4. Required value

   The minimum safe resistance is: $R_s = 10\,\Omega$

5. Check options

   • A: $13\,\Omega$ → safe, but not the minimum required value
   • B: $13.3\,\Omega$ → safe, but not the minimum required value
   • C: $10\,\Omega$ → exactly the minimum safe value
   • D: $12\,\Omega$ → safe, but not the minimum required value

Therefore, the correct option is C.