Interpret the circuit

Since this is a Zener diode question, the key idea is:

• In reverse bias, the Zener diode maintains approximately its breakdown voltage once breakdown occurs.
• Given Zener breakdown voltage: $V_Z = 4\text{ V}$.

Voltage across the resistor in series with the ammeter

In the given circuit, the Zener diode fixes the voltage across the parallel branch at $4\text{ V}$. Therefore, the remaining voltage drops across the resistor connected with the ammeter.

From the circuit values, the voltage available across the resistor branch is such that the current through that branch becomes $I = \frac{V}{R} = \frac{4}{400} = 0.01\text{ A}$

Hence, $I = 10\text{ mA}$

Ammeter reading

Therefore, the ammeter reads: $\boxed{10\text{ mA}}$

Check with options

• A: $60\text{ mA}$
• B: $80\text{ mA}$
• C: $10\text{ mA}$
• D: $24\text{ mA}$

Correct option is C.