JEE Main 2020 · PhysicsNumerical answerNumerical valueMediumMulti-step

JEE Main 6 September 2020, Shift 2, Physics Q23

Question 23 of 75 in this shift, Physics question 23 of 25.

In a series LR circuit, power of 400 W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as

\left(\frac{n}{3\pi}\right)\mu\mathrm{F}

, then value of n is ________.

Official answer

400.00

NTA final key (Sep 2020).

Chapter: Alternating Current
Topic: Phasor Diagrams
Idea tested: Power Factor from Phasors

Same idea in other shifts

Asked 4× in all

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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.