Q 12-14-214JEE MainJEE Main 2025 (7 Apr, Shift 1)Medium
In the circuit shown in the figure, a $12\ \text{V}$ battery is connected through a $100\ \Omega$ resistor to a zener diode (breakdown voltage $4\ \text{V}$), a voltmeter, and a $400\ \Omega$ resistor in series with an ammeter, all three in parallel. The reading of the ammeter will be:
Answer: (C) $10\ \text{mA}$
First check whether the zener breaks down. Without it, the $400\ \Omega$ load would get
$$V = \frac{400}{100 + 400}\times12 = 9.6\ \text{V} > 4\ \text{V}$$
So the zener conducts and holds the parallel part at $4\ \text{V}$. The ammeter is in series with $400\ \Omega$:
$$I = \frac{4}{400} = 0.01\ \text{A} = 10\ \text{mA}$$
Solution by Sreeraj P, M.Sc Physics