In the following circuit, the reading of the ammeter will be (Take Zener breakdown voltage = 4 V)

Using voltage division to test if Zener breakdown occurs: $$V_{\text{open}} = V_s \left(\frac{R_L}{R_s + R_L}\right)$$

Given: $$V_s = 12\text{ V}$$, $$R_s = 100\ \Omega$$, $$R_L = 400\ \Omega$$

$$V_{\text{open}} = 12 \times \left(\frac{400}{100 + 400}\right) = 12 \times \frac{4}{5} = 9.6\text{ V}$$

Since $$V_{\text{open}} > V_Z$$ ($$9.6\text{ V} > 4\text{ V}$$), the Zener diode operates in the breakdown region:

$$V_L = V_Z = 4\text{ V}$$

Using Ohm's law for the load branch containing the ammeter:

$$I_A = \frac{V_L}{R_L} = \frac{4}{400} = 0.01\text{ A} = 10\text{ mA}$$