JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Current Electricity

A diode carries current only when it is forward biased. With the battery polarity shown, the diodes in the $I_4$ and $I_2$ branches are forward biased and conduct, while the oppositely connected diode in the $I_3$ branch is reverse biased and blocks, so $I_3=0$. Each conducting branch has the diode's forward resistance in series with its resistor: $25+125=150\Omega$. The two conducting branches are in parallel, $\frac{150\times 150}{300}=75\Omega$, and the source branch adds $25\Omega$ in series, giving $100\Omega$. The total current is $I_1=\frac{5}{100}=0.05\text{A}$, which splits equally between the two identical branches so that $I_2=I_4=0.025\text{A}$. Hence $\frac{I_1}{I_2}=\frac{0.05}{0.025}=2$.