For the circuit shown below, calculate the value of $$I_z$$:

$$V_{\text{open}} = V_i \left( \frac{R}{R_s + R} \right) = 100 \left( \frac{2000}{1000 + 2000} \right) = 66.67\ \text{V}$$

Since $$V_{\text{open}} > V_z$$, the Zener diode operates in the breakdown region: $$V_L = V_z = 50\ \text{V}$$

$$I_s = \frac{V_i - V_z}{R_s} = \frac{100 - 50}{1000} = 0.05\ \text{A} = 50\ \text{mA}$$

$$I_L = \frac{V_z}{R} = \frac{50}{2000} = 0.025\ \text{A} = 25\ \text{mA}$$

$$I_z = I_s - I_L = 50 - 25 = 25\ \text{mA}$$