Q 12-14-077JEE MainJEE Main 2024 (30 Jan, Shift 2)Medium
In the given circuit, a $15\ \text{V}$ battery is connected in series with a germanium diode $D_1$ and a silicon diode $D_2$ (both forward biased), a $1.5\ \text{k}\Omega$ resistor and a load resistance $R_L = 2.5\ \text{k}\Omega$. The voltage across the load resistance $R_L$ is:
Answer: (A) $8.75\ \text{V}$
Forward drops: Ge $\approx 0.3\ \text{V}$, Si $\approx 0.7\ \text{V}$, total $1.0\ \text{V}$.
$$I = \frac{15 - 1}{1.5 + 2.5}\ \text{mA} = 3.5\ \text{mA}$$
$$V_L = 3.5\ \text{mA}\times2.5\ \text{k}\Omega = 8.75\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics