1. Given data

• Maximum zener voltage: $V_Z = 10\,\text{V}$
• Maximum power dissipation: $P_{\max} = 0.4\,\text{W}$
• Variable supply: $0$ to $15\,\text{V}$

We need the series protection resistor $R_P$ so that the zener diode is not damaged.

2. Maximum current allowed through zener diode

Using power relation:

$P = V I$

So the maximum safe current is

$I_{Z,\max} = \frac{P_{\max}}{V_Z} = \frac{0.4}{10} = 0.04\,\text{A}$

Thus,

$I_{Z,\max} = 40\,\text{mA}$

3. Maximum voltage across resistor

The supply can go up to $15\,\text{V}$. When the zener is in breakdown, it maintains about $10\,\text{V}$ across itself.

So voltage across the series resistor at maximum supply is

$V_R = 15 - 10 = 5\,\text{V}$

4. Calculate required series resistance

By Ohm’s law,

$R_P = \frac{V_R}{I_{Z,\max}} = \frac{5}{0.04} = 125\,\Omega$

5. Final answer

$\boxed{125\,\Omega}$

6. Comparison with stored answer

Stored correct answer = $125$

Our derived answer matches the stored answer.