JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Semiconductors

Read the regulator this way. The Zener clamps the output at its breakdown value of $6$ V, so whatever the unregulated input does, the load always sits at $6$ V. The Zener carries only the current left over after the load has taken its share, so the Zener current is largest when the input is at its maximum of $16$ V. The series resistor then drops $16-6=10$ V, and with the $2\text{k}\Omega$ series resistor the current pulled from the supply is $\frac{10}{2000}=5$ mA. The $4\text{k}\Omega$ load, held at $6$ V, draws $\frac{6}{4000}=1.5$ mA. The Zener swallows the difference, $5-1.5=3.5$ mA.