PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 12 · Chapter 7 · JEE Main

Alternating Current

Chapter-wise practice and revision for JEE Main: concise notes on rms values, reactance and impedance, resonance, power and power factor, LC oscillations and the transformer, concept figures, and 52 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

52 selected questions6 topics3 concept figuresPreview · first 3 of 52 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 124 previous-year JEE questions in the PiTheory bank

AC circuits40
Resonance29
Power23
RMS values15
Transformer11
LC oscillations5

Contents

  1. 1Peak, rms and instantaneous valuesQ1–Q8
  2. 2Reactance, impedance and phaseQ9–Q22
  3. 3Series resonance and quality factorQ23–Q34
  4. 4Power and power factorQ35–Q43
  5. 5LC oscillationsQ44–Q47
  6. 6TransformerQ48–Q52
  7. ✓Last-day checklist
01

Topic 1 · Q1–Q8

Peak, rms and instantaneous values

15 of the 124 JEE PYQs in the bank test this as their main idea

Quick notes

$$I_{\text{rms}}=\sqrt{\langle i^2\rangle},\qquad \text{sinusoid: }I_{\text{rms}}=\frac{I_0}{\sqrt2},\qquad \langle i\rangle_{\text{half cycle}}=\frac{2I_0}{\pi}$$

Meters and ratings (220 V mains, bulb ratings) are rms values. Heat uses $I_{\text{rms}}^2R$.

  • $a\sin\omega t+b\cos\omega t$: amplitude $\sqrt{a^2+b^2}$.
  • DC plus AC: $I_{\text{rms}}^2=I_{\text{dc}}^2+\dfrac{I_0^2}{2}$.
  • Sawtooth $i_0t/T$: $\dfrac{i_0}{\sqrt3}$; square wave: $i_0$.
  • Times from phases: zero to peak $T/4$; peak to rms $T/8$; half-peak to peak $T/6$.
Trap. $\omega=2\pi f$: $100\pi t$ means $f=50$ Hz.

Practice: first 3 of 8 questions

Q1JEE Main 2026 (28 Jan, Shift 1)Easy

The electric current in the circuit is given as $i = i_o(t/T)$. The r.m.s current for the period $t = 0$ to $t = T$ is ______ .

  1. A$\dfrac{i_o}{\sqrt2}$
  2. B$\dfrac{i_o}{\sqrt3}$
  3. C$i_o$
  4. D$\dfrac{i_o}{\sqrt6}$
Answer and solution

Answer: Option B: $\dfrac{i_o}{\sqrt3}$

Concept. rms value $=\sqrt{\langle i^2\rangle}$ over the period.

Step 1. Mean square. $\dfrac1T\displaystyle\int_0^Ti_0^2\frac{t^2}{T^2}dt=\frac{i_0^2}{3}$.

Answer. $\dfrac{i_0}{\sqrt3}$ (Option B).

Point to note. $\langle i^2\rangle=i_0^2/3$.

Q2JEE Main 2025 (24 Jan, Shift 1)Easy

An alternating current is given by $I = I_A\sin\omega t + I_B\cos\omega t$. The r.m.s. current will be

  1. A$\dfrac{|I_A + I_B|}{\sqrt2}$
  2. B$\sqrt{\dfrac{I_A^2 + I_B^2}{2}}$
  3. C$\sqrt{I_A^2 + I_B^2}$
  4. D$\dfrac{\sqrt{I_A^2 + I_B^2}}{2}$
Answer and solution

Answer: Option B: $\sqrt{\dfrac{I_A^2 + I_B^2}{2}}$

Concept. $I_A\sin\omega t+I_B\cos\omega t$ is one sinusoid of amplitude $\sqrt{I_A^2+I_B^2}$; rms $=\dfrac{\text{amplitude}}{\sqrt2}$.

Answer. $\sqrt{\dfrac{I_A^2+I_B^2}{2}}$ (Option B).

Point to note. Amplitude $\sqrt{I_A^2+I_B^2}$.

Q3JEE Main 2024 (4 Apr, Shift 1)Medium

An alternating current at any instant is given by $i = \left[6 + \sqrt{56}\sin(100\pi t + \pi/3)\right]\ \text{A}$. The rms value of the current is ______ A.

Numerical answer type

Answer and solution

Answer: 8

Concept. DC plus sinusoid: the cross term averages to zero, so $I_{\text{rms}}^2=I_{\text{dc}}^2+\dfrac{I_0^2}{2}$.

Step 1. Apply. $36+\dfrac{56}{2}=64$.

Answer. $8$ A.

Point to note. $I_{dc}^2+I_0^2/2$.

49 more questions and 5 more topics in the full chapter

  1. 2Reactance, impedance and phase14 questions with solutions
  2. 3Series resonance and quality factor12 questions with solutions
  3. 4Power and power factor9 questions with solutions
  4. 5LC oscillations4 questions with solutions
  5. 6Transformer5 questions with solutions
  6. ✓Last-day checklist

The complete Alternating Current chapter is in the Class 12 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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