JEE Physics 2025 Question with Solution

Given: $V_{in} = 25 \, \text{V}$, $V_Z = 5 \, \text{V}$, series resistance $R_S = 400 \, \Omega$, and zener current $I_Z = 4I_L$.

Find: The load current $I_L$ and load resistance $R_L$.

Since the zener diode regulates the voltage, the load voltage is equal to the zener voltage:

$$V_L = V_Z = 5 \, \text{V}$$

The voltage across the series resistor is:

$$V_{R_S} = V_{in} - V_Z = 25 - 5 = 20 \, \text{V}$$

So the current through the series resistor is:

$$I_S = \frac{V_{R_S}}{R_S} = \frac{20}{400} = 0.05 \, \text{A} = 50 \, \text{mA}$$

Using Kirchhoff's current law:

$$I_S = I_Z + I_L$$

Given $I_Z = 4I_L$, therefore:

$$I_S = 4I_L + I_L = 5I_L$$

So,

$$50 \, \text{mA} = 5I_L$$ $$I_L = 10 \, \text{mA}$$

Now the load resistance is:

$$R_L = \frac{V_L}{I_L} = \frac{5 \, \text{V}}{0.01 \, \text{A}} = 500 \, \Omega$$

Therefore, the load current is $10 \, \text{mA}$ and the load resistance is $500 \, \Omega$. The correct option is D.

Given: $V_{in} = 25 \, \text{V}$, $V_Z = 5 \, \text{V}$, $R_S = 400 \, \Omega$, and $I_Z = 4I_L$.

Find: $I_L$ and $R_L$.

First find the current supplied through the series resistor:

$$I_S = \frac{25 - 5}{400} = \frac{20}{400} = 0.05 \, \text{A} = 50 \, \text{mA}$$

Because $I_Z : I_L = 4 : 1$, the total current splits into 5 equal parts. Hence the load current is one part:

$$I_L = \frac{50}{5} = 10 \, \text{mA}$$

Then,

$$R_L = \frac{5}{10 \times 10^{-3}} = 500 \, \Omega$$

Therefore, the correct option is D.