JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Semiconductors

First test whether the Zener actually reaches breakdown. Applying Kirchhoff's voltage law around the loop, $12-500(2i_1+i_2)-10=0$, so $2i_1+i_2=\frac{2}{500}=4\text{mA}$. The voltage that would appear across the diode branch stays below its breakdown value, so the Zener does not conduct and its branch current $i_2$ is essentially zero. The current through the Zener diode is therefore close to $0\text{mA}$.