JEE Physics 2024 Question with Solution

Given: Supply voltage is $10 \, \text{V}$, Zener breakdown voltage is $V_Z = 3.0 \, \text{V}$, series resistor is $1 \, \text{k}\Omega$, and the parallel resistor is $2 \, \text{k}\Omega$.

Find: The Zener current $I_Z$.

When the Zener diode is in breakdown, the voltage across it remains $3.0 \, \text{V}$. Therefore, the voltage across the $1 \, \text{k}\Omega$ resistor is

$$V_R = 10 - 3 = 7 \, \text{V}$$

So the current through the series resistor is

$$I = \frac{V_R}{R} = \frac{7}{1000} = 7 \, \text{mA}$$

The current through the $2 \, \text{k}\Omega$ resistor is

$$I_{2k} = \frac{3}{2000} = 1.5 \, \text{mA}$$

Using Kirchhoff's current law,

$$I = I_Z + I_{2k}$$

Hence,

$$I_Z = 7 - 1.5 = 5.5 \, \text{mA}$$

Therefore, the current through the Zener diode is $5.5 \, \text{mA}$. The correct option is B.

The solution mentions option C, but the worked calculation and final value clearly give $5.5 \, \text{mA}$, which matches option B.

Given: The Zener diode maintains a constant potential difference of $3 \, \text{V}$ across the parallel branch.

Find: The current through the Zener diode.

Take the lower node as $0 \, \text{V}$. Then the node across the Zener branch is at $3 \, \text{V}$, while the supply node is at $10 \, \text{V}$.

So the current entering the junction through the $1 \, \text{k}\Omega$ resistor is

$$I = \frac{10 - 3}{1000} = 7 \, \text{mA}$$

The current leaving through the $2 \, \text{k}\Omega$ resistor is

$$I_1 = \frac{3}{2000} = 1.5 \, \text{mA}$$

The remaining current must flow through the Zener diode:

$$I_Z = I - I_1 = 7 - 1.5 = 5.5 \, \text{mA}$$

Therefore, the correct answer is $5.5 \, \text{mA}$.