The Circuit and the Equations
Consider a single resistor of resistance $R$ connected across an AC source. The source produces a sinusoidally varying potential difference across its terminals, written as
$$ v = v_m \sin \omega t $$
where $v_m$ is the amplitude (peak value) of the oscillating voltage and $\omega$ is its angular frequency. To find the current, NCERT applies Kirchhoff's loop rule to the circuit, $\varepsilon(t) = \sum V = 0$, which for a single resistor gives $v_m \sin \omega t = iR$. Rearranging,
$$ i = \frac{v_m}{R}\sin \omega t = i_m \sin \omega t \qquad \text{with} \qquad i_m = \frac{v_m}{R} $$
The relation $i_m = v_m / R$ is just Ohm's law, which holds equally well for AC and DC voltages across a resistor. The current is therefore a sinusoid of the same frequency as the voltage, scaled down by the factor $1/R$ and carrying no phase shift.
A pure resistor across an AC source. The same instantaneous current flows everywhere in the single loop.
Voltage and Current in Phase
Because $v$ and $i$ are both sinusoids of the form $\sin \omega t$, they reach their zero, minimum and maximum values at the same instants. There is no time lag between them: the phase angle between the voltage and the current in a purely resistive circuit is zero. This is the defining behaviour of a resistor and the benchmark against which inductors and capacitors are later compared, where the current leads or lags the voltage by a quarter cycle.
The graph below plots $v$ and $i$ against time. The two curves rise and fall together; only their amplitudes differ, set by the factor $R$.
Both curves cross zero and reach their extrema at the same instants — the phase difference is zero.
Instantaneous and Average Power
The current is sinusoidal, so over one complete cycle it spends equal time positive and negative, and the sum of its instantaneous values over a cycle is zero. The average current is therefore zero. This does not mean no energy is dissipated. Joule heating is given by $i^2 R$, and $i^2$ is always positive whether $i$ is positive or negative. There is genuine dissipation of electrical energy every instant the current is non-zero.
The instantaneous power dissipated in the resistor is
$$ p = i^2 R = i_m^2 R \, \sin^2 \omega t $$
To get the average power over a cycle, take the average of $\sin^2 \omega t$. Using the identity $\sin^2 \omega t = \tfrac12 (1 - \cos 2\omega t)$ and the fact that the average of $\cos 2\omega t$ over a cycle is zero, the average of $\sin^2 \omega t$ is $\tfrac12$. Hence the average power is
$$ \bar{p} = \tfrac12\, i_m^2 R $$
Power is a $\sin^2$ curve — never negative. Its average over a cycle sits at exactly half the peak.
The structure $\bar{p} = \tfrac12 i_m^2 R$ is awkward compared with the DC expression $P = I^2 R$. To bring AC power into the same algebraic form as DC, a special "steady-equivalent" value of current is defined.
The RMS Value
The root-mean-square (RMS) or effective current $I$ is the square root of the average of $i^2$ over a cycle. Since the average of $i^2 = i_m^2 \sin^2 \omega t$ is $\tfrac12 i_m^2$, taking the square root gives
$$ I = I_{rms} = \frac{i_m}{\sqrt{2}} = 0.707\, i_m $$
The name encodes the recipe in reverse: take the root of the mean of the square. With this definition the average power becomes $\bar{p} = I^2 R$, identical in form to the DC heating law. Physically, $I$ is the equivalent steady DC current that would produce the same average power loss in the resistor as the alternating current does.
Peak, average and RMS are three different numbers
The simple average of $\sin \omega t$ over a full cycle is zero, but the average of $\sin^2 \omega t$ over a full cycle is $\tfrac12$. Students who confuse the two write the average power as $i_m^2 R$ (forgetting the factor $\tfrac12$) or claim the average current is $i_m/\sqrt2$ when in fact the mean current is zero and it is the RMS current that equals $i_m/\sqrt2$.
Mean of $\sin\omega t$ = 0. Mean of $\sin^2\omega t$ = $\tfrac12$. RMS current $= i_m/\sqrt2$, never the plain average.
| Quantity | Symbol | For a sinusoid | Use |
|---|---|---|---|
| Peak (amplitude) | iₘ, vₘ | maximum reached each cycle | insulation, breakdown limits |
| Average over a cycle | — | 0 (equal positive & negative) | cannot describe heating |
| RMS (effective) | I, V | peak / √2 = 0.707 × peak | power, Ohm's law, meters |
The zero phase angle here is the baseline for everything that follows. See how a quarter-cycle lag appears once reactance enters in AC Voltage Applied to an Inductor.
RMS Voltage and Ohm's Law
The RMS voltage, or effective voltage, is defined the same way as the current:
$$ V = V_{rms} = \frac{v_m}{\sqrt{2}} = 0.707\, v_m $$
Starting from $v_m = i_m R$ and dividing both sides by $\sqrt2$ gives $V = IR$. So Ohm's law for AC, written in RMS terms, has exactly the same form as the DC case. This is the practical advantage of RMS values: the AC power and current–voltage relations reduce to the familiar DC equations.
| Relation | DC form | AC form (RMS) |
|---|---|---|
| Ohm's law | V = IR | V = IR |
| Power | P = I²R | P = I²R = V²/R = IV |
| Peak ↔ effective | — | vₘ = √2 V, iₘ = √2 I |
It is customary to measure and quote RMS values for AC quantities. The household line voltage of 220 V is an RMS value; its peak is $v_m = \sqrt2 \times V = 1.414 \times 220\ \text{V} \approx 311\ \text{V}$. Meters calibrated for AC read RMS, and appliance ratings are RMS, which is why the equation $P = V^2/R = IV$ can be applied directly to nameplate figures.
Worked Examples
A light bulb is rated at 100 W for a 220 V supply. Find (a) the resistance of the bulb, (b) the peak voltage of the source, and (c) the RMS current through the bulb.
(a) With $P = 100\ \text{W}$ and $V = 220\ \text{V}$, the resistance is $$ R = \frac{V^2}{P} = \frac{(220)^2}{100} = 484\ \Omega $$
(b) The peak voltage of the source is $$ v_m = \sqrt2\, V = 1.414 \times 220 \approx 311\ \text{V} $$
(c) Since $P = IV$, $$ I = \frac{P}{V} = \frac{100}{220} \approx 0.454\ \text{A} $$
An AC source $v = 311 \sin(314\,t)$ V is applied across a $100\ \Omega$ resistor. Find the RMS current and the average power dissipated.
The peak voltage is $v_m = 311\ \text{V}$, so the RMS voltage is $V = v_m/\sqrt2 = 311/1.414 \approx 220\ \text{V}$. The RMS current is $$ I = \frac{V}{R} = \frac{220}{100} = 2.2\ \text{A} $$ The average power, using the resistive form $P = I^2 R$, is $$ P = (2.2)^2 \times 100 \approx 484\ \text{W} $$ Equivalently $P = V^2/R = (220)^2/100 = 484\ \text{W}$. Note the current and voltage are in phase, so the power factor is 1 and no $\cos\varphi$ correction is needed.
AC Voltage Applied to a Resistor
- For $v = v_m \sin \omega t$ across $R$, the current is $i = i_m \sin \omega t$ with $i_m = v_m/R$ (Ohm's law).
- Voltage and current are in phase; the phase angle is zero, and they peak and cross zero together.
- Average current over a cycle is zero, but heating $\propto i^2$ is always positive, so power is dissipated.
- Instantaneous power $p = i_m^2 R \sin^2\omega t$; average power $\bar{p} = \tfrac12 i_m^2 R = I^2 R$.
- RMS current $I = i_m/\sqrt2 = 0.707\, i_m$; RMS voltage $V = v_m/\sqrt2$; in RMS terms $V = IR$.
- Mains 220 V is RMS; peak $\approx 311$ V. Mean of $\sin\omega t = 0$, mean of $\sin^2\omega t = \tfrac12$.