In the circuit shown below, is working as a 8 V dc regulated voltage source. When 12 V is used as an input, the power dissipated (in mW) in each diode is (Considering both zener diodes are identical) ___________.

Total series resistance connected to the input source: $$R_s = 200 + 200 = 400\,\Omega$$

Total current $$I$$ from the $$12\text{ V}$$ input source:

$$I = \frac{V_{\text{in}} - V_0}{R_s} = \frac{12 - 8}{400} = \frac{4}{400} = 0.01\text{ A} = 10\text{ mA}$$

Since no load resistor is connected at $$V_0$$, the entire current passes through the Zener network:

$$I_Z = I = 10\text{ mA}$$

For two identical Zener diodes in series providing a combined regulated output of $$8\text{ V}$$:

$$V_{Z1} + V_{Z2} = 8\text{ V} \implies 2V_Z = 8 \implies V_Z = 4\text{ V}$$

Power dissipated in each Zener diode: $$P_Z = V_Z \times I_Z = 4\text{ V} \times 10\text{ mA} = 40\text{ mW}$$