\begin{aligned} &\textbf{1. Understand Zener Diode Behavior} \\ &\text{A Zener diode maintains a constant voltage } (V_z) \text{ across it when operated in the breakdown region.} \\ &\text{In this case, } V_z = 50V. \\ \\ &\textbf{2. Calculate the Current Through the 10kΩ Resistor } (I_{10k\Omega}) \\ &\text{The voltage across the 10kΩ resistor is the same as the Zener voltage, which is 50V.} \\ &\text{Using Ohm's Law } (I = V/R): \\ &I_{10k\Omega} = \frac{V_z}{R_{10k\Omega}} = \frac{50V}{10k\Omega} = 5 mA \\ \\ &\textbf{3. Calculate the Total Current Through the 5kΩ Resistor } (I_{5k\Omega}) \\ &\text{The voltage drop across the 5kΩ resistor is the difference between the source voltage and the Zener voltage:} \\ &V_{5k\Omega} = 120V - 50V = 70V \\ &\text{Using Ohm's Law: } \\ &I_{5k\Omega} = \frac{V_{5k\Omega}}{R_{5k\Omega}} = \frac{70V}{5k\Omega} = 14 mA \\ \\ &\textbf{4. Calculate the Current Through the Zener Diode } (I_{Zener}) \\ &\text{Apply Kirchhoff's Current Law (KCL) at the node where the resistors and Zener diode connect:} \\ &I_{5k\Omega} = I_{10k\Omega} + I_{Zener} \\ &\text{Rearrange to find } I_{Zener}: \\ &I_{Zener} = I_{5k\Omega} - I_{10k\Omega} = 14 mA - 5 mA = 9 mA \\ \\ &\textbf{Therefore, the current through the Zener diode is 9 mA.} \end{aligned}