A diode has Zener voltage of 10 V and maximum power dissipation of...

Given Zener voltage $V_Z = 10$ V and maximum power dissipation P_\max = 0.5 W.

The maximum safe current through the Zener diode is given by I_\max = \dfrac{P_\max}{V_Z} = \dfrac{0.5}{10} = 0.05 A.

The voltage drop across the series resistor is $V_R = V_S - V_Z = 25 - 10 = 15$ V.

For safety, the current through the circuit should not exceed I_\max. Thus, the minimum resistance $R$ is R = \dfrac{V_R}{I_\max} = \dfrac{15}{0.05} = 300 $\Omega$.

Answer: $300$