JEE Physics 2026 Question with Solution

Given: $V_z = 5 \, \text{V}$, load current $I_L = 5 \, \text{mA}$, maximum unregulated input $V_{in} = 25 \, \text{V}$. The Zener diode can withstand four times the load current, so $I_z = 20 \, \text{mA}$.

Find: The value of series resistor $R_s$.

A Zener regulator keeps the load voltage nearly equal to $V_z$. Therefore the current through $R_s$ is the sum of load current and Zener current:

$$I = I_z + I_L = 20 \, \text{mA} + 5 \, \text{mA} = 25 \, \text{mA}$$

The voltage across the series resistor is:

$$V_s = V_{in} - V_z = 25 - 5 = 20 \, \text{V}$$

Using Ohm's law:

$$R_s = \frac{V_s}{I} = \frac{20}{25 \times 10^{-3}} = 800 \, \Omega$$

The working gives $800 \, \Omega$, but this value is not present in the options. The solution states that $1000 \, \Omega$ is the intended competitive-exam selection based on the given options.

Therefore, the correct option is D.

Given: Same circuit data as above.

Find: Which option should be selected from the listed choices.

From the regulator condition, the maximum current through the Zener is four times the load current:

$$I_z = 4I_L = 4 \times 5 \, \text{mA} = 20 \, \text{mA}$$

So the total current through $R_s$ is:

$$I = I_z + I_L = 20 \, \text{mA} + 5 \, \text{mA} = 25 \, \text{mA}$$

Required resistor value:

$$R_s = \frac{25 - 5}{25 \times 10^{-3}} = 800 \, \Omega$$

Since $800 \, \Omega$ is absent from the options $100$, $10$, $4000$, and $1000$, the closest defensible listed option is $1000 \, \Omega$. This matches the marked answer on the solution's.

Therefore, the intended answer is D.