QJEEM24Ph4020: JEE Main 2024 Physics - Class 12 JEEM24Ph4020 Solution: In the given circuit, the voltage across load resistance (R L ) …

Both the Germanium (Ge) and Silicon (Si) diodes are in forward bias. To find the effective voltage available for the resistors, we subtract their respective barrier potentials (0.3V and 0.7V) from the source voltage.

$$\begin{array}{rl}& \text{Effective Voltage }(V)=15\text{V}-(0.3\text{V}+0.7\text{V})=14\text{V}\\ & \text{Total Resistance }(R)=1.5\text{k}\mathrm{\Omega }+2.5\text{k}\mathrm{\Omega }=4\text{k}\mathrm{\Omega }\\ & i=\frac{V}{R}=\frac{14}{4}=3.5\text{ mA}\\ & V_L=i\times R_L=3.5\times 2.5\\ & V_L=8.75\text{ volt}\end{array}$$