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

Solution

Physics sir · 4 steps

Check first that the Zener breaks down. Once it does, the load has a fixed 4 V across it.

1. Is it reverse biased. The top wire connects to the positive terminal of the battery and to the cathode (bar end) of the Zener. So the Zener is reverse biased and can act as a voltage regulator.
2. Test for breakdown. Remove the Zener for a moment. The 400 Ω load would then get 12 × 400/500 = 9.6 V. This is more than 4 V, so the Zener breaks down and holds the load at 4 V.

   I_A = V_Z/R_L = (4 V)/(400 Ω) = 0.01 A

3. Where the rest goes. The other 8 V drops across 100 Ω, so 80 mA leaves the battery. Of this, 10 mA goes through the load and the remaining 70 mA goes through the Zener.
4. So. The ammeter is in series with the 400 Ω load, so it reads 10 mA, option (c).

The idea to keep. For a Zener regulator, first find the load voltage with the Zener removed. If it exceeds V_Z, the load sits at exactly V_Z, and its current is V_Z/R_L.

Answer from the NTA final answer key. Solution by PrepWiser.