JEE Physics 2025 Question with Solution

Given:

• Supply voltage $V_s = 12 \, \text{V}$
• Series resistor $R_s = 100 \, \Omega$
• Zener breakdown voltage $V_Z = 4 \, \text{V}$
• Load resistor $R_L = 400 \, \Omega$

Find: The ammeter reading.

Since the applied voltage is greater than the Zener breakdown voltage, the Zener diode operates in breakdown and maintains a constant voltage of $4 \, \text{V}$ across itself. The Zener diode and the load resistor are in parallel, so the voltage across the load resistor is also

$$V_{R_L} = V_Z = 4 \, \text{V}$$

Using Ohm's law for the $400 \, \Omega$ resistor,

$$I = \frac{V}{R} = \frac{4}{400} = 0.01 \, \text{A} = 10 \, \text{mA}$$

The ammeter is in series with the $400 \, \Omega$ load resistor, so it measures this current.

Therefore, the reading of the ammeter is $10 \, \text{mA}$. The correct option is C.

Given:

• Source voltage $12 \, \text{V}$
• Resistors $100 \, \Omega$ and $400 \, \Omega$
• Zener voltage $4 \, \text{V}$

Find: The current through the ammeter.

First, estimate the voltage across the branch containing the Zener diode and the $400 \, \Omega$ resistor:

$$V_1 = \frac{400}{100 + 400} \times 12 = \frac{48}{5} \, \text{V}$$

Since $V_1 > V_Z$, the Zener diode goes into breakdown. Hence, the voltage across the $400 \, \Omega$ resistor becomes fixed at $4 \, \text{V}$.

Now apply Ohm's law:

$$I = \frac{4}{400} = 10 \, \text{mA}$$

Therefore, the ammeter reads $10 \, \text{mA}$. The correct option is C.