JEE Physics 2024 Question with Solution

Given: Zener breakdown voltage $V_Z = 10 \, \text{V}$, source voltage $V_{in} = 20 \, \text{V}$, series resistor $R_S = 200 \, \Omega$, load resistor $R_L = 500 \, \Omega$.

Find: Current through the Zener diode $I_Z$.

First verify that the Zener diode is in breakdown region. If the diode were not conducting, the open-circuit voltage across the load branch would be

$$V_{OC} = V_{in} \times \frac{R_{load}}{R_{series} + R_{load}} = 20 \, \text{V} \times \frac{500 \, \Omega}{200 \, \Omega + 500 \, \Omega} \approx 14.3 \, \text{V}$$

Since $14.3 \, \text{V} > 10 \, \text{V}$, the Zener diode is in breakdown and maintains $10 \, \text{V}$ across itself and the parallel load resistor.

Voltage across the series resistor is

$$V_{R_S} = V_{in} - V_Z = 20 \, \text{V} - 10 \, \text{V} = 10 \, \text{V}$$

Hence total current through the series resistor is

$$I_{total} = \frac{V_{R_S}}{R_S} = \frac{10 \, \text{V}}{200 \, \Omega} = 0.05 \, \text{A} = 50 \, \text{mA}$$

Current through the load resistor is

$$I_L = \frac{V_Z}{R_L} = \frac{10 \, \text{V}}{500 \, \Omega} = 0.02 \, \text{A} = 20 \, \text{mA}$$

Applying Kirchhoff's Current Law at the junction,

$$I_{total} = I_Z + I_L$$

so

$$I_Z = I_{total} - I_L = 50 \, \text{mA} - 20 \, \text{mA} = 30 \, \text{mA}$$

Therefore, the current through the Zener diode is $30 \, \text{mA}$. The correct option is C.

Given: A Zener regulator circuit with $V_Z = 10 \, \text{V}$, $V_{in} = 20 \, \text{V}$, $R_S = 200 \, \Omega$, and $R_L = 500 \, \Omega$.

Find: Zener current $I_Z$.

Use the regulator idea: once the Zener is in breakdown, the branch voltage becomes fixed at $10 \, \text{V}$.

1. Check whether breakdown is possible by comparing the no-load branch voltage with $V_Z$.
2. Find the current through the series resistor from the remaining voltage drop.
3. Find the load current through $500 \, \Omega$.
4. Subtract load current from source current to get Zener current.

The open-circuit voltage across the branch is greater than the Zener voltage, so the diode conducts in breakdown. Then branch voltage is $10 \, \text{V}$.

Series current:

$$I_{total} = \frac{20 - 10}{200} = \frac{10}{200} = 50 \, \text{mA}$$

Load current:

$$I_L = \frac{10}{500} = 20 \, \text{mA}$$

Hence,

$$I_Z = 50 - 20 = 30 \, \text{mA}$$

So the Zener diode carries $30 \, \text{mA}$.