JEE Main 2025 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 7 April 2025, Shift 2, Physics Q48

Question 48 of 75 in this shift, Physics question 23 of 25, Section B.

An inductor of reactance

100\ \Omega

, a capacitor of reactance

50\ \Omega

, and a resistor of resistance

50\ \Omega

are connected in series with an AC source of 10 V, 50 Hz. Average power dissipated by the circuit is __________ W.

Official answer

NTA final key.

Chapter: Alternating Current
Topic: RMS and Peak Values
Idea tested: Power Calculation using RMS

Same topic in other shifts

All RMS and Peak Values questions

3 Apr 2025, Shift 2 · Q37An electric bulb rated as 100W-220V is connected to an ac source of rms voltage 220V. The peak value of current through the bulb is :EasySingle correct
7 Apr 2025, Shift 1 · Q35An ac current is represented as $i = 5\sqrt{2} + 10\cos\left(650\pi t + \frac{\pi}{6}\right)$ Amp The r.m.s value of the current isEasySingle correct
30 Jan 2024, Shift 2 · Q43An alternating voltage $V(t) = 220\sin 100\pi t$ volt is applied to a purely resistive load of $50\ \Omega$ . The time taken for the…MediumSingle correct
24 Jun 2022, Shift 1 · Q41A resistance of $40\ \Omega$ is connected to a source of alternating current rated 220 V, 50 Hz. Find the time taken by the current to…EasySingle correct
18 Mar 2021, Shift 1 · Q6An AC source rated 220 V, 50 Hz is connected to a resistor. The time taken by the current to change from its maximum to the rms value is :EasySingle correct

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.