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Friday, 9 October

JEE Main 2025 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 2 April 2025, Shift 1, Physics Q45

Question 45 of 75 in this shift, Physics question 20 of 25, Section A.

A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 Ω400\ \Omega resistor connected in series, the zener current is found to be 4 times load current. The load current (ILI_L) and load resistance (RLR_L) are :
  1. (1)IL=0.02I_L=0.02 mA; RL=250 ΩR_L=250\ \Omega
  2. (2)IL=20I_L=20 mA; RL=250 ΩR_L=250\ \Omega
  3. (3)IL=10I_L=10 mA; RL=500 ΩR_L=500\ \OmegaOfficial answer
  4. (4)IL=10I_L=10 A; RL=0.5 ΩR_L=0.5\ \Omega

Official answer

Option 3

NTA final key.

Chapter
Semiconductor Electronics: Materials, Devices and Simple Circuits
Topic
Semiconductor Diodes
Idea tested
Zener Diode Regulation

Same idea in other shifts

Asked 4× in all
  1. 7 Apr 2025, Shift 1 · Q45In the following circuit, the reading of the ammeter will be (Take Zener breakdown voltage = 4 V)MediumDiagram basedHas a figure
  2. 29 Jan 2024, Shift 1 · Q48In the given circuit, the breakdown voltage of the Zener diode is 3.0 V. What is the value of IzI_z?MediumDiagram basedHas a figure
  3. 9 Jan 2020, Shift 2 · Q25The circuit shown below is working as a 8 V dc regulated voltage source. When 12 V is used as input, the power dissipated (in mW) in each…MediumNumerical valueHas a figure

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.