Semiconductor Electronics question for JEE Main (JEE Main 2023 (10 Apr, Shift 2)), with solution

Answer: (C) $\dfrac{I_1}{I_2}=2$

$D_1$ and $D_3$ are forward biased; $D_2$ is reverse biased, so $I_3=0$.

Each conducting branch has $25+125=150\ \Omega$; two in parallel give $75\ \Omega$. With the $25\ \Omega$ in series:
$$I_1=\frac{5}{100}=0.05\ \text{A},\qquad I_2=I_4=0.025\ \text{A}$$
So $\dfrac{I_1}{I_2}=2$.

Solution by Sreeraj P, M.Sc Physics