JEE Mains Physics · Alternating Current

RMS Values and Timing in AC

The rms value of an alternating current is the steady current that heats a resistor at the same rate: I₀/√2 for a sine wave, and for a constant plus a sinusoid the squares add.

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Why this matters

Fifteen PYQs, thirteen of them multiple choice, and five from 2025 or 2026. Eight read a single sine wave: its rms value, its frequency, the heat it makes, or the time it takes to move between two values. Seven ask for the rms of a sum, a constant plus a sinusoid or a sine plus a cosine. None needs more than two lines once you square before you average.

Concept 1 of 2: RMS value, meters and timing on a sine wave

Heat in a resistor goes as the square of the current, so the useful average of an alternating current is the root of the mean of its square. For a sine wave that is the peak divided by √2. Meters and ratings quote this rms value. For timing, remember that the phase ωt grows at a steady rate ω, so the time between two points on the wave is their phase gap divided by ω.

Definition

$i = I_0\sin\omega t$ : $I_{rms} = \dfrac{I_0}{\sqrt{2}} \approx 0.707\,I_0$ ; in the same way $V_{rms} = \dfrac{V_0}{\sqrt{2}}$ .
Read $\omega$ from the argument: $\sin(200\pi t)$ has $\omega = 200\pi\ \text{rad/s}$ , so $f = \dfrac{\omega}{2\pi} = 100\ \text{Hz}$ and $T = 10\ \text{ms}$ .
A.c. ammeters, voltmeters and hot-wire meters read rms values. A supply 'rated 230 V' means $V_{rms} = 230\ \text{V}$ , with peak $230\sqrt{2} \approx 325\ \text{V}$ .
Heat in time t: $H = I_{rms}^{2}Rt$ . A d.c. current I and an a.c. current of PEAK value I heat equal resistors in the ratio $2 : 1$ .
A lamp rated P at V: $I_{rms} = \dfrac{P}{V}$ , $R = \dfrac{V^{2}}{P}$ , and the peak current is $\sqrt{2}$ times $I_{rms}$ .
Time between two values $= \dfrac{\text{phase gap}}{\omega}$ . From zero: to half peak $\dfrac{T}{12}$ , to the rms value $\dfrac{T}{8}$ , to the peak $\dfrac{T}{4}$ . From half peak to peak $\dfrac{T}{6}$ ; from peak down to rms $\dfrac{T}{8}$ .

RMS value and time on the wave

I_{rms} = \frac{I_0}{\sqrt{2}} \qquad t = \frac{\Delta(\omega t)}{\omega}

Worked example

A current

i = 10\sin(200\pi t)\ \text{A}

flows through a resistor. Find the rms current, the frequency, and the time the current takes to fall from its peak to its rms value.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q98Moderate

Example 1 · Alternating Current · RMS Values and Timing in AC

An alternating voltage

V(t) = 220\sin100\pi t

volt is applied to a purely resistive load of

50\Omega

. The time taken for the current to rise from half of the peak value to the peak value is:

Peak to rms is T/8, not T/4

From the peak, the wave must lose a phase of π/4 to fall to 1/√2 of the peak. That is one eighth of a cycle. A quarter cycle takes it all the way to zero.

Taking the coefficient of t as the frequency

In sin(200πt) the coefficient 200π is ω in rad/s. The frequency is ω/2π = 100 Hz. Do not multiply by 2π again.

Using the peak where the meter reads rms

A meter reading or a rating like '220 V' is an rms value. Multiply by √2 only when the question asks for a peak.

Drill 7 more on rms value, meters and timing on a sine wave

Concept 2 of 2: RMS of a sum: d.c. plus a.c., sine plus cosine

The rms value comes from the average of i², never of i. Square the sum and average it over a cycle. When the pieces are a constant and a sinusoid, or a sine and a cosine of the same ω, the cross term averages to zero. So the mean squares simply add, and you take the root at the end.

Definition

$i = I_{dc} + I_0\sin(\omega t + \phi)$ : $I_{rms} = \sqrt{I_{dc}^{2} + \dfrac{I_0^{2}}{2}}$ . The phase $\phi$ does not matter.
$i = I_1\sin\omega t + I_2\cos\omega t$ is one sinusoid of amplitude $\sqrt{I_1^{2} + I_2^{2}}$ , so $I_{rms} = \sqrt{\dfrac{I_1^{2} + I_2^{2}}{2}}$ . A hot-wire ammeter reads this.
Any shape: $I_{rms}^{2} = \dfrac{1}{T}\displaystyle\int_0^{T} i^{2}\,dt$ . Write i as a function of t over one period, square, integrate, divide by T.
A wave that is $I_0$ for half of each period and zero for the other half has $I_{rms} = \dfrac{I_0}{\sqrt{2}}$ , but its average is $\dfrac{I_0}{2}$ .
The rms of a sum is NOT the sum of the rms values.

Mean squares add

I_{rms} = \sqrt{I_{dc}^{2} + \frac{I_0^{2}}{2}} \qquad I_{rms}^{2} = \frac{1}{T}\int_0^{T} i^{2}\,dt

Worked example

Find the rms value of

i = \left[3 + 4\sqrt{2}\sin(200\pi t)\right]\ \text{A}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q24Moderate

Example 2 · Alternating Current · RMS Values and Timing in AC

A alternating current at any instant is given by

i =\left\lbrack 6 +\sqrt{56}\sin\left( 100\pi t +\frac{\pi}{3} \right) \right\rbrack A

. The rms value of the current is _____ A.

Adding rms values

For 3 + 4√2 sin ωt the rms is √(9 + 16) = 5, not 3 + 4 = 7. Mean squares add; rms values do not.

Averaging i instead of i²

The plain average of a sinusoid over a cycle is zero. The rms comes from the average of the SQUARE, and the root is taken last.

Adding the amplitudes of a sine and a cosine

sin ωt and cos ωt are 90° apart, so their amplitudes combine like the sides of a right triangle: √(I₁² + I₂²), not I₁ + I₂.

Drill 6 more on rms of a sum: d.c. plus a.c., sine plus cosine

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

RMS value, meters and timing on a sine wave
RMS value and time on the wave
$I_{rms} = \frac{I_0}{\sqrt{2}} \qquad t = \frac{\Delta(\omega t)}{\omega}$
RMS of a sum: d.c. plus a.c., sine plus cosine
Mean squares add
$I_{rms} = \sqrt{I_{dc}^{2} + \frac{I_0^{2}}{2}} \qquad I_{rms}^{2} = \frac{1}{T}\int_0^{T} i^{2}\,dt$

Watch out for (6)

Peak to rms is T/8, not T/4
→ RMS value, meters and timing on a sine wave
Taking the coefficient of t as the frequency
→ RMS value, meters and timing on a sine wave
Using the peak where the meter reads rms
→ RMS value, meters and timing on a sine wave
Adding rms values
→ RMS of a sum: d.c. plus a.c., sine plus cosine
Averaging i instead of i²
→ RMS of a sum: d.c. plus a.c., sine plus cosine
Adding the amplitudes of a sine and a cosine
→ RMS of a sum: d.c. plus a.c., sine plus cosine

Test yourself on Alternating Current

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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