Q 12-14-069JEE MainJEE Main 2024 (29 Jan, Shift 1)Easy
In the given circuit, the breakdown voltage of the Zener diode is $3.0\ \text{V}$. What is the value of $I_z$?
Answer: (B) $5.5\ \text{mA}$
The Zener holds $3\ \text{V}$ across the load.
Current through the $1\ \text{k}\Omega$ resistor: $I = \dfrac{10 - 3}{1000} = 7\ \text{mA}$
Load current: $I_L = \dfrac{3}{2000} = 1.5\ \text{mA}$
$$I_z = I - I_L = 7 - 1.5 = 5.5\ \text{mA}$$
Solution by Sreeraj P, M.Sc Physics