A Zener diode of breakdown voltage $$10 \text{ V}$$ is used as a voltage regulator as shown in the figure. The current through the Zener diode is

First, we verify if the Zener diode is operating in the breakdown region by checking the open-circuit voltage ($$V_Z'$$) across it:

$$V_Z' = V_{\text{in}} \times \frac{R_L}{R_S + R_L} = 20 \times \frac{500}{200 + 500} = \frac{100}{7} \approx 14.3\text{ V}$$

Since $$V_Z' > V_Z$$ ($$14.3\text{ V} > 10\text{ V}$$), the Zener diode is in the breakdown region and maintains a constant voltage drop of $$10\text{ V}$$ across it.

Voltage across the load resistor ($$R_L = 500\ \Omega$$): $$V_L = V_Z = 10\text{ V}$$

Load current ($$I_L$$): $$I_L = \frac{V_L}{R_L} = \frac{10}{500} = 0.02\text{ A} = 20\text{ mA}$$

Voltage across the series resistor ($$R_S = 200\ \Omega$$): $$V_S = V_{\text{in}} - V_Z = 20 - 10 = 10\text{ V}$$

Total source current ($$I_S$$): $$I_S = \frac{V_S}{R_S} = \frac{10}{200} = 0.05\text{ A} = 50\text{ mA}$$

Using Kirchhoff's Current Law at the node above the Zener diode ($$I_S = I_Z + I_L$$): $$I_Z = I_S - I_L = 50 - 20 = 30\text{ mA}$$