In the given circuit, the voltage across load resistance $$(R_L)$$ is:

Both $$D_1$$ (Germanium) and $$D_2$$ (Silicon) are forward-biased.

$$\text{Knee voltages: } V_{\text{Ge}} = 0.3\text{ V}, \quad V_{\text{Si}} = 0.7\text{ V}$$

Net loop voltage remaining for the resistive network:

$$V_{\text{net}} = V_{\text{in}} - V_{\text{Ge}} - V_{\text{Si}} \implies V_{\text{net}} = 15 - 0.3 - 0.7 = 14\text{ V}$$

Applying the voltage divider rule across the remaining series resistors to find $$V_{R_L}$$:

$$V_{R_L} = V_{\text{net}} \times \frac{R_L}{R + R_L}$$

$$\implies V_{R_L} = 14 \times \frac{2.5}{1.5 + 2.5} = 14 \times \frac{2.5}{4} = 8.75\text{ V}$$