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Friday, 9 October

JEE Main 2020 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 9 January 2020, Shift 2, Physics Q25

Question 25 of 75 in this shift, Physics question 25 of 25, Section B.

The 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 diode is; (considering both zener diodes are identical) ______.

The figure, in words

Series 200 ohm resistor from V_in = 12 V, then a 200 ohm resistor in a vertical branch to the output node V_o = 8 V, with two identical zener diodes in series across the output (to the bottom rail).

Official answer

12.00

NTA final key (Jan 2020).

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. 2 Apr 2025, Shift 1 · Q45A 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…MediumSingle correct
  2. 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
  3. 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

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. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.