Given Zener voltage VZ=10 V and maximum power dissipation Pmax=0.5 W.
The maximum safe current through the Zener diode is given by Imax=VZPmax=100.5=0.05 A.
The voltage drop across the series resistor is VR=VS−VZ=25−10=15 V.
For safety, the current through the circuit should not exceed Imax. Thus, the minimum resistance R is R=ImaxVR=0.0515=300 Ω.
Answer: 300