MHT-CET Physics · AC Circuits

Alternating Current: Instantaneous, Peak and RMS Values

An alternating current follows I = I₀ sin(ωt + φ); its peak is I₀, and its r.m.s. value I₀/√2 is the steady current that would heat a resistor equally.

All AC Circuits notes MHT-CET Physics strategy

Why this matters

7 PYQs, one HARD. Two things are asked: when a sinusoid first reaches its peak, half its peak or zero, and converting between peak and r.m.s. values — what an a.c. ammeter or voltmeter reads.

Concept 1 of 2: Reading a Sinusoid: Peak Times, Zeros and Phase

The phase ωt + φ tells you where in its cycle the signal is. It reaches its peak when the phase is π/2, half its peak (from zero) when the phase is π/6, and it crosses zero twice every cycle — so f cycles a second means 2f zeros.

Definition

$i = I_0\sin(\omega t + \phi)$ , $\omega = 2\pi f = \dfrac{2\pi}{T}$ .
First peak: $\omega t + \phi = \dfrac{\pi}{2}$ . With $\phi = \dfrac{\pi}{3}$ : $t = \dfrac{T}{12}$ .
From zero to half the peak: $\omega t = \dfrac{\pi}{6}$ , $t = \dfrac{T}{12}$ ; to the peak: $\dfrac{T}{4}$ .
Zero crossings per second $= 2f$ : $\sin(50\pi t)$ has $f = 25$ Hz and 50 zeros a second.

Instantaneous value

i = I_0\sin(\omega t + \phi), \qquad \omega = 2\pi f

Worked example

v = 10\sin\left(100\pi t + \dfrac{\pi}{6}\right)

. When does it first reach its peak?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q46Easy

Example 1 · AC Circuits · RMS, Peak, and AC Source Characteristics

The alternating voltage is given by

v = v_0\sin(\omega t + \pi/3)

, when will the voltage be maximum for first time?

Forgetting the starting phase

With a phase

\phi

already in the equation, the peak comes EARLIER: solve

\omega t + \phi = \frac{\pi}{2}

, not

\omega t = \frac{\pi}{2}

\frac{T}{4}

is the planted answer.

Drill 4 more on reading a sinusoid: peak times, zeros and phase

Concept 2 of 2: Peak and R.M.S. Values

Meters read r.m.s. values, and the mains '230 V' is an r.m.s. value too. For a sine wave the r.m.s. is the peak divided by √2 — the equation gives the peak, the meter gives the r.m.s.

Definition

$I_{\text{rms}} = \dfrac{I_0}{\sqrt{2}}$ , $V_{\text{rms}} = \dfrac{V_0}{\sqrt{2}}$ ; $e = 200\sin 50t$ has $V_{\text{rms}} = 100\sqrt{2}$ V.
A resistor: $I_{\text{rms}} = \dfrac{V_{\text{rms}}}{R}$ , in phase, so $P = V_{\text{rms}}I_{\text{rms}}$ .
A phase written into the voltage alone ( $\sin(\omega t + 60^\circ)$ ) is a starting point, not a phase DIFFERENCE — a lamp still has power factor 1.

R.m.s. value

I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \approx 0.707\,I_0

Worked example

e = 311\sin 314t

volt drives a

110\,\Omega

resistor. R.m.s. voltage and current?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q24Easy

Example 2 · AC Circuits · RMS, Peak, and AC Source Characteristics

The a.c. source of e.m.f. with instantaneous value 'e' is given by

e = 200\sin(50t)

volt. The r.m.s value of current in a circuit of resistance

50\,\Omega

Using the peak where the r.m.s. is meant

The number in front of

\sin

is the PEAK. A meter reading, a power, or a '220 V supply' all mean r.m.s.; divide by

\sqrt{2}

first.

Drill 1 more on peak and r.m.s. values

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)

Reading a Sinusoid: Peak Times, Zeros and Phase
Instantaneous value
$i = I_0\sin(\omega t + \phi), \qquad \omega = 2\pi f$
Peak and R.M.S. Values
R.m.s. value
$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \approx 0.707\,I_0$

Watch out for (2)

Forgetting the starting phase
→ Reading a Sinusoid: Peak Times, Zeros and Phase
Using the peak where the r.m.s. is meant
→ Peak and R.M.S. Values

Test yourself on AC Circuits

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Related notes

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