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Saturday, 31 October

JEE Main 2021 · PhysicsMultiple choiceDiagram basedMediumMulti-step

JEE Main 26 August 2021, Shift 2, Physics Q15

Question 15 of 90 in this shift, Physics question 15 of 30, Section A. NTA dropped this question from the final key; it is left out of every count on this site.

In the given circuit the AC source has ω=100 rad s−1\omega=100\ \mathrm{rad\ s^{-1}}. Considering the inductor and capacitor to be ideal, what will be the current I flowing through the circuit ?

The figure, in words

Circuit: an AC source (~) labelled 200 V in the bottom wire, with current I marked by an upward arrow on the left wire. Two parallel branches: upper branch has a capacitor 100 μF in series with a resistor 100 Ω; lower branch has an inductor 0.50 H in series with a resistor 50 Ω.
  1. (1)6 A
  2. (2)4.24 A
  3. (3)0.94 A
  4. (4)5.9 A

Official answer

Dropped by NTA (marks given to all)

NTA final key: NTA final key: DROP (question dropped).

Chapter
Alternating Current
Topic
Phasor Diagrams
Idea tested
No close match in our taxonomy

Same topic in other shifts

All Phasor Diagrams questions
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  5. 4 Apr 2025, Shift 2 · Q48An inductor of self inductance 1 H is connected in series with a resistor of 100 π100\,\pi ohm and an ac supply of 100 π100\,\pi volt, 50 Hz.…MediumNumerical value
  6. 7 Apr 2025, Shift 1 · Q49For ac circuit shown in figure, R = 100 kΩ\Omega and C = 100 pF and the phase difference between Vin\mathrm{V_{in}} and…HardNumerical 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. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.