A diode has Zener voltage of 10 V and maximum power dissipation of 0.5 W, then the minimum resistance to be used in series with this diode for safety when it is connected to a 25 V power supply is ______ $$\Omega$$.

The Zener diode will operate in its breakdown (regulating) region when the voltage across it is $$V_Z = 10\text{ V}$$.

Its maximum power dissipation is $$P_{\text{max}} = 0.5\text{ W}$$.
Using the power formula $$P = VI$$, the maximum permissible current through the Zener is

$$I_{\text{Z(max)}} = \frac{P_{\text{max}}}{V_Z} = \frac{0.5\text{ W}}{10\text{ V}} = 0.05\text{ A} = 50\text{ mA}.$$

The supply voltage is $$V_S = 25\text{ V}$$. A series resistor $$R_S$$ is placed so that, at the worst case (no external load, all current through the Zener), the current does not exceed $$I_{\text{Z(max)}}$$.

Applying Kirchhoff’s Voltage Law:

$$V_S = I_{\text{Z}}\,R_S + V_Z.$$

At the limiting condition, $$I_{\text{Z}} = I_{\text{Z(max)}}$$. Hence

$$R_S = \frac{V_S - V_Z}{I_{\text{Z(max)}}} = \frac{25\text{ V} - 10\text{ V}}{0.05\text{ A}} = \frac{15\text{ V}}{0.05\text{ A}} = 300\ \Omega.$$

Therefore, the minimum series resistance required is 300 Ω.