Q 12-14-208JEE MainJEE Main 2025 (2 Apr, Shift 1)Medium
A zener diode with $5\ \text{V}$ zener voltage is used to regulate an unregulated dc voltage input of $25\ \text{V}$. For a $400\ \Omega$ resistor connected in series, the zener current is found to be 4 times the load current. The load current ($I_L$) and load resistance ($R_L$) are:
Answer: (D) $I_L = 10\ \text{mA};\ R_L = 500\ \Omega$
The zener holds the load at $5\ \text{V}$, so the series resistor drops $25 - 5 = 20\ \text{V}$:
$$I = \frac{20}{400} = 0.05\ \text{A} = 50\ \text{mA}$$
This current splits as $I = I_Z + I_L = 4I_L + I_L = 5I_L$, so $I_L = 10\ \text{mA}$.
$$R_L = \frac{5\ \text{V}}{10\ \text{mA}} = 500\ \Omega$$
Solution by Sreeraj P, M.Sc Physics