JEE Main 2025 Apr 7 Shift 1, Physics Q11: Semiconductors, p-n Junction Diode and Applications

Step 1: Check whether the Zener conducts. Without the Zener, the 100 $\Omega$ and 400 $\Omega$ resistors would form a voltage divider across 12 V, placing across the 400 $\Omega$:

$V_{400}=\frac{400}{100+400}\times 12=\frac{4}{5}\times 12=9.6\text{ V}$

This exceeds the Zener breakdown voltage $V_z=4$ V, so the Zener conducts in the breakdown region and clamps the parallel branch at exactly 4 V.

Step 2: Ammeter current. The ammeter reads the current through the 400 $\Omega$ branch, which now sees exactly 4 V:

$I=\frac{V_z}{400\Omega }=\frac{4}{400}=0.01\text{ A}=10\text{ mA}$

$\therefore$ Option (3).