JEE Main 2021 · 20 July · morning · Single correct · +4 / −1
For the circuit shown below, calculate the value of I z :
- (a) 25 mA
- (b) 0.1 A
- (c) 0.15 A
- (d) 0.05 A
Answer and solution
Answer (a) 25 mA
Solution
Physics sir · 4 steps
Once the Zener conducts, the load voltage is fixed at 50 V. Find both currents from that and subtract.
- Is it in breakdown. Remove the Zener and the load would get 100 × 2000/3000 ≈ 66.7 V. That is above 50 V, so the Zener breaks down and holds the load at V_z = 50 V.
- Series current. The series resistor drops 100 - 50 = 50 V, so Iₛ = 50/1000 = 0.05 A = 50 mA.
- Load current. The load has 50 V across it, so I_L = 50/2000 = 0.025 A = 25 mA.
I_z = Iₛ - I_L = 50 mA - 25 mA = 25 mA
- So. The current through the Zener is I_z = 25 mA, which is option (a).
The idea to keep. For a Zener regulator, first check that it is in breakdown. Then the load voltage is V_z, the series current is (Vᵢ - V_z)/Rₛ, and the Zener takes whatever the load does not.
Answer from the NTA final answer key. Solution by PrepWiser.