In connection with the circuit drawn below, the value of current flowing through 2 k$$\Omega$$ resistor is ______ $$\times 10^{-4}$$ A.

$$\text{The } 10\text{ V DC source reverse-biases the Zener diode.}$$

$$\text{Assuming the Zener diode enters breakdown, it fixes the load voltage.}$$

$$\text{Voltage across the parallel branch: } V_L = V_Z = 5\text{ V}$$

$$\text{Verify breakdown: } V_{\text{open}} = 10\text{ V} \times \frac{2\text{ k}\Omega}{1\text{ k}\Omega + 2\text{ k}\Omega} = 6.67\text{ V} > 5\text{ V } \implies \text{Active breakdown.}$$

$$\text{Current through the } 2\text{ k}\Omega \text{ load resistor: } I_L = \frac{V_L}{R_L}$$

$$I_L = \frac{5\text{ V}}{2\text{ k}\Omega} = \frac{5}{2000}\text{ A} = 2.5 \times 10^{-3}\text{ A}$$

$$I_L = 25 \times 10^{-4}\text{ A}$$