Q 12-14-178JEE MainJEE Main 2019 (10 Jan, Shift 2)Medium
A $120\ \text{V}$ source is connected through a $5\ \text{k}\Omega$ series resistor to a Zener diode of breakdown voltage $50\ \text{V}$ (reverse biased). A $10\ \text{k}\Omega$ load resistor is connected across the Zener diode. The current through the Zener diode is
Answer: (C) $9\ \text{mA}$
The Zener holds $50\ \text{V}$ across the load.
Series resistor current: $\dfrac{120-50}{5\ \text{k}\Omega} = 14\ \text{mA}$. Load current: $\dfrac{50}{10\ \text{k}\Omega} = 5\ \text{mA}$.
Zener current $= 14 - 5 = 9\ \text{mA}$.
Solution by Sreeraj P, M.Sc Physics