AC applied to a resistor

Plug a heater into an ordinary wall socket rated "230 V" and something odd is true: the voltage across it is never actually 230 volts. It swings — up to about 325 volts, back down through zero, down to about minus 325 volts, and back up again, fifty times every second. Yet the heater glows steadily, the wire heats up exactly as a textbook DC calculation would predict, and the number stamped on the appliance ("230 V, 1000 W") behaves as if the voltage really were a constant 230. That is not a contradiction — it is the whole point of this article. A resistor is the simplest possible load an alternating source can drive, and it is the best place to build the vocabulary that the rest of the chapter Alternating Current leans on: peak values, RMS values, phase, and average power. Because a resistor obeys Ohm's law at every single instant — with no memory of the past and no delay — the current it draws simply mirrors the applied voltage, scaled by $1/R$ . There is no phase lag to track, no reactance to compute, no energy quietly stored and released. That simplicity is precisely why the resistor is the right first stop: every complication that inductors and capacitors will later introduce (phase angles, reactance, impedance, power factor) is measured against the plain, in-phase behaviour worked out here. This section walks through why an alternating voltage is written as a sine function, why a resistor's current tracks it exactly in step, why engineers quote RMS values rather than peak values on every appliance label, and how the average power dissipated in a resistor connects the AC world back to the familiar DC formula $P = I^2R$ you already trust from Current Electricity.

Key takeaways

An alternating voltage is written $v = V_0\sin(\omega t)$ , where $V_0$ is the peak value and $\omega = 2\pi f$ is the angular frequency.
A pure resistor obeys Ohm's law instant by instant, so the current $i = v/R = I_0\sin(\omega t)$ is a perfect sine wave in phase with the voltage — no lag, no lead.
RMS (root-mean-square) values, $I_{rms} = I_0/\sqrt2$ and $V_{rms} = V_0/\sqrt2$ , are defined so that a resistor dissipates the same average power under AC as it would under a steady DC current equal to $I_{rms}$ .
Instantaneous power $p = i^2R$ oscillates between $0$ and $I_0^2R$ ; its time-average over a full cycle is $P_{avg} = \tfrac12 I_0^2R = I_{rms}^2R = V_{rms}I_{rms}$ .
Appliance and mains ratings ("230 V", "1000 W") are always RMS/average quantities by convention — the instantaneous voltage and current are larger, and instantaneous power spikes to twice the average.
A resistor's phase angle between current and voltage is $\phi = 0$ , so its power factor $\cos\phi = 1$ — the maximum possible; this is the reference case every other AC element in the chapter is compared to.
Key idea: in a pure resistor, current tracks the alternating voltage in exact phase via Ohm's law at every instant, and RMS values are defined precisely so that the everyday power formula $P = VI$ survives unchanged from DC into AC.

The core insight

A resistor's current is an in-phase sine wave scaled by $1/R$ from the applied AC voltage, and RMS values are built so that $P_{avg} = V_{rms}I_{rms} = I_{rms}^2R$ reproduces the familiar DC power law.

Where it fits in the syllabus

This is the opening topic of the chapter Alternating Current, which follows Electromagnetic Induction in CBSE Class 12 Physics (Unit VI, Electromagnetic Induction and Alternating Currents). Everything else in the chapter — reactance, impedance, phasor diagrams for the LCR series circuit, resonance, power factor, the AC generator, and the transformer — is built on the two ideas fixed here: how peak and RMS values relate, and how a resistor's current stays exactly in step with the driving voltage. This article stays firmly inside the 2025-26 CBSE syllabus: it uses only the phasor-free, purely algebraic Ohm's-law argument for a resistor (no differential equations), and it does not touch inductors, capacitors, or the LCR circuit — those are separate chapter-siblings that build on what is established here.

The physical picture

An alternating voltage supplied by a household or laboratory AC source varies smoothly and periodically with time. It is written as

v = V_0\sin(\omega t)

where: $v$ is the instantaneous voltage at time $t$ , $V_0$ is the peak (maximum) voltage, and $\omega = 2\pi f$ is the angular frequency in radians per second, with $f$ the frequency in hertz ( $f = 50\ \text{Hz}$ for Indian household mains). Physically, $v$ rises from zero to $+V_0$ , falls back through zero to $-V_0$ , and returns to zero, completing one full cycle every $T = 1/f = 2\pi/\omega$ seconds — for mains supply, one cycle every $20$ milliseconds. Connect this source directly across a pure resistor of resistance $R$ — a heater coil or an incandescent filament are close real-world approximations, since they store negligible magnetic or electric energy. Because Ohm's law $v = iR$ holds at every single instant for a resistor (it has no capacity to store energy the way an inductor or capacitor does), the current follows immediately:

i = \frac{v}{R} = \frac{V_0}{R}\sin(\omega t) = I_0\sin(\omega t)

where: $i$ is the instantaneous current and $I_0 = V_0/R$ is the peak current. Reading these two equations side by side is the entire physical content of this topic: $v$ and $i$ are both plain sine functions of $\omega t$ , with no phase offset between them. Whenever $v$ is at its positive peak, $i$ is too; whenever $v$ crosses zero heading downward, so does $i$ . Current and voltage in a resistor are said to be in phase.

Figure

A graph with two in phase sine curves for voltage and current across a resistor, both starting at zero and rising together, peaking together, and crossing zero together, with a small resistor symbol and an AC source symbol shown in a simple circuit loop above the graph

Because $i$ oscillates between $-I_0$ and $+I_0$ , a plain time-average over one full cycle is exactly zero — a DC ammeter would read nothing useful, and simply reporting "the current" as a single sine amplitude is not how appliances or wiring are rated. What matters for heating, lighting, and every energy-delivery purpose is the RMS (root-mean-square) value, defined so that a steady DC current equal to it would deliver the same average power to the same resistor:

I_{rms} = \frac{I_0}{\sqrt2}, \qquad V_{rms} = \frac{V_0}{\sqrt2}

where: the $\sqrt2$ arises from averaging $\sin^2(\omega t)$ over a cycle (worked out step by step below). These are the values printed on every appliance and quoted for mains supply: "230 V AC" always means $V_{rms} = 230\ \text{V}$ , so the actual peak voltage reaching the appliance is $V_0 = 230\sqrt2 \approx 325\ \text{V}$ — noticeably higher than the number on the label, a fact that matters for insulation and fuse ratings.

Derivation AFrom peak values to average power: why RMS carries the half

The goal is to show, using nothing beyond the identity $\sin^2\theta = \tfrac12(1-\cos2\theta)$ and the fact that a cosine averages to zero over a whole number of cycles, exactly why the factor $1/\sqrt2$ appears in the RMS definition and why the average power in a resistor comes out to $P_{avg} = \tfrac12 I_0^2R$ — and hence why $P_{avg} = I_{rms}^2R = V_{rms}I_{rms}$ reproduces the ordinary DC power law once RMS values are used in place of constant DC values.

Step 1. Write the instantaneous power. For current $i = I_0\sin(\omega t)$ flowing through resistance $R$ , the instantaneous power dissipated is $p = i^2R = I_0^2R\sin^2(\omega t)$ . This quantity is never negative (a resistor always absorbs power, whichever way the current flows), and it oscillates between $0$ (when $i=0$ ) and $I_0^2R$ (when $i = \pm I_0$ ).

Step 2. Rewrite $\sin^2(\omega t)$ using the double-angle identity. Since $\sin^2\theta = \tfrac12(1-\cos2\theta)$ , the power becomes $p = I_0^2R\left[\tfrac12 - \tfrac12\cos(2\omega t)\right]$ . The power has been split into a constant part, $\tfrac12 I_0^2R$ , plus a part that oscillates twice as fast as the original current, $-\tfrac12 I_0^2R\cos(2\omega t)$ .

Step 3. Average over one full cycle. Averaging any pure cosine or sine term over a whole number of complete cycles gives exactly zero, because it spends equal time above and below the axis. So the oscillating term $-\tfrac12 I_0^2R\cos(2\omega t)$ averages to zero, and only the constant term survives: $P_{avg} = \tfrac12 I_0^2R$ .

Step 4. Express this as an RMS current squared times $R$ . Define $I_{rms}$ by demanding $P_{avg} = I_{rms}^2R$ , matching the ordinary DC power formula. Comparing with Step 3, $I_{rms}^2R = \tfrac12 I_0^2R$ , so $I_{rms}^2 = I_0^2/2$ , giving $I_{rms} = I_0/\sqrt2$ . The identical argument applied to $v = V_0\sin(\omega t)$ gives $V_{rms} = V_0/\sqrt2$ . Since $I_{rms} = V_{rms}/R$ for a resistor, $P_{avg} = I_{rms}^2R$ can equally be written as $P_{avg} = V_{rms}I_{rms}$ — the AC power formula that mirrors DC's $P = VI$ exactly, once RMS values replace constant values.

Worked examples

Board level - a resistor on a laboratory AC source

A resistor of $R = 50\ \Omega$ is connected to an AC source with peak voltage $V_0 = 200\ \text{V}$ and frequency $f = 50\ \text{Hz}$ . Find (a) the peak current, (b) the RMS voltage and RMS current, and (c) the average power dissipated.

Step 1. Find the angular frequency (useful for writing the full time-dependence, even though it is not needed for the magnitudes asked here): $\omega = 2\pi f = 2\pi(50) = 314\ \text{rad/s}$ .

Step 2. Find the peak current using Ohm's law on the peak values: $I_0 = V_0/R = 200/50 = 4\ \text{A}$ .

Step 3. Convert both peak values to RMS values using $X_{rms} = X_0/\sqrt2$ : $V_{rms} = 200/1.414 \approx 141.4\ \text{V}$ and $I_{rms} = 4/1.414 \approx 2.83\ \text{A}$ .

Step 4. Compute the average power two independent ways and check they agree. Using $P_{avg} = V_{rms}I_{rms} = 141.4 \times 2.83 \approx 400\ \text{W}$ . Using $P_{avg} = I_{rms}^2R = (2.83)^2 \times 50 \approx 400\ \text{W}$ . Both routes land on $P_{avg} \approx 400\ \text{W}$ , confirming the two forms of the power formula are consistent.

Competitive level - sizing a rated heater from its label

An electric heater is rated "230 V, 1000 W" for use on Indian AC mains, where $V_{rms} = 230\ \text{V}$ and $f = 50\ \text{Hz}$ by convention. Find the resistance of the heater element, the peak voltage actually appearing across it, and the peak current it draws at switch-on.

Step 1. Recover the RMS current from the power rating, using $P_{avg} = V_{rms}I_{rms}$ : $I_{rms} = P_{avg}/V_{rms} = 1000/230 \approx 4.35\ \text{A}$ .

Step 2. Find the resistance from $R = V_{rms}/I_{rms} = 230/4.35 \approx 52.9\ \Omega$ ; equivalently, directly from the rating, $R = V_{rms}^2/P_{avg} = 230^2/1000 \approx 52.9\ \Omega$ — a useful shortcut worth memorising for this exact question type.

Step 3. Convert both RMS quantities to their peak values using $X_0 = X_{rms}\sqrt2$ : $V_0 = 230\sqrt2 \approx 325.3\ \text{V}$ and $I_0 = 4.35\sqrt2 \approx 6.15\ \text{A}$ .

Step 4. Interpret the result physically. Although the "rated" current used for average-power bookkeeping is only $4.35\ \text{A}$ , the actual instantaneous current in the coil swings up to $I_0 \approx 6.15\ \text{A}$ twice in every $20\ \text{ms}$ cycle — a fact that fuse and wire-gauge selection must respect, since a fuse rated only marginally above $4.35\ \text{A}$ could still be safe (it is thermal, and responds to the RMS heating effect) while a switch or wire insulation rated by peak voltage must clear $325\ \text{V}$ , not $230\ \text{V}$ . Timing note: in a board exam this pair of conversions (rating to $I_{rms}$ , $I_{rms}$ to $R$ , $R$ -or- $I_{rms}$ to peak values) should take under three minutes — recognise it as a fixed four-line template rather than re-deriving each step from scratch.

Common traps

Writing $P_{avg} = V_0I_0$ using peak values directly — this overstates the true average power by a factor of exactly $2$ , since $P_{avg} = \tfrac12 V_0I_0 = V_{rms}I_{rms}$ .
Inserting a phase angle or a $\cos\phi$ factor for a resistor out of habit carried over from inductor/capacitor problems later in the chapter — for a pure resistor $\phi = 0$ always, so $\cos\phi = 1$ and no correction factor is needed.
Confusing "RMS" with the plain time-average of $\sin(\omega t)$ — the ordinary average of a full sine cycle is exactly zero; RMS is a different, always-positive quantity built from the square root of the mean of the square.
Assuming appliance-label numbers ("230 V", "1000 W") are peak or instantaneous values — by universal convention they are RMS/average values, so any peak-value question about a rated appliance requires an extra $\times\sqrt2$ step that is easy to forget under exam pressure.
Forgetting that instantaneous power in a resistor is never negative, unlike instantaneous power in an inductor or capacitor (previewed later in the chapter) — a resistor always absorbs energy, it never returns any to the source.

Common trapPeak values are not what the label says, and RMS is not a simple average

The single costliest error in this topic is treating $V_0$ and $I_0$ as if they were the numbers quoted for household or rated equipment. They are not: "230 V" mains and a "1000 W" heater rating are always RMS and average values respectively, by international convention, precisely because those are the quantities that correctly predict heating effect through the ordinary-looking formula $P_{avg} = V_{rms}I_{rms}$ . The peak voltage genuinely reaching the appliance is about $41\%$ higher, $V_0 = V_{rms}\sqrt2 \approx 325\ \text{V}$ for 230 V mains — a fact that matters for anything rated by breakdown voltage. The companion error is calling RMS an "average": the true time-average of a sinusoid over a full cycle is zero (it spends equal time positive and negative), while RMS is the square root of the mean of the squared value, which is always positive and captures the heating effect correctly. Keep the three quantities distinct in your mind: instantaneous value (varies every instant), plain average (zero for a full sine cycle), and RMS (the physically meaningful, always-positive "effective" value).

You should be able to…

Write $v = V_0\sin(\omega t)$ for an AC source and derive $i = I_0\sin(\omega t) = (V_0/R)\sin(\omega t)$ for a resistor, explicitly showing zero phase difference.
Convert confidently between peak and RMS values in both directions using $X_{rms} = X_0/\sqrt2$ and $X_0 = X_{rms}\sqrt2$ , for both voltage and current.
Explain, in words a classmate would accept, why RMS values rather than peak or plain-average values are used to specify AC ratings.
Derive or reproduce the average power result $P_{avg} = \tfrac12 I_0^2R = I_{rms}^2R = V_{rms}I_{rms}$ starting from the double-angle identity for $\sin^2$ .
Work backward from an appliance's rated voltage and power to its resistance and its peak current.
State the resistor's phase angle ( $\phi = 0$ ) and power factor ( $\cos\phi = 1$ ) as the reference case for the rest of the AC chapter.

Ask the tutor: if the double-angle averaging step in the derivation feels rushed, ask the AI tutor to plot $\sin^2(\omega t)$ alongside its constant average and walk through why the oscillating part cancels out.

Why do we even bother with RMS instead of just always using peak values?

Because peak values only tell you the single highest instant of the cycle, and that instant lasts for zero time — it does not tell you how much heat a resistor actually produces over a stretch of time. RMS is built specifically to answer the useful question "what steady DC current would deliver the same average power?" so that all the familiar DC formulas ( $P = I^2R$ , $P = VI$ ) keep working unchanged once you plug in RMS values. That is exactly why every appliance label, wiring code, and safety rating in the real world is written in RMS, not peak.

If current and voltage are exactly in phase for a resistor, does that mean resistors are somehow special compared to inductors and capacitors?

Yes, in exactly this one respect. A resistor converts electrical energy to heat instantly and irreversibly, with no mechanism to store energy in a magnetic or electric field, so its response has no time delay at all — Ohm's law applies at every single instant, giving zero phase difference. Inductors and capacitors, which you will meet later in this chapter, do store energy temporarily (in a magnetic field and an electric field respectively), which is exactly what produces the 90-degree phase differences you will calculate there. The resistor's zero-phase, "instantaneous response" behaviour is the reference case every later phase calculation in the LCR circuit is compared against.

Why does the instantaneous power in a resistor never go negative, and does that matter?

Instantaneous power is $p = i^2R$ , and since $i^2$ is a square, it can never be negative regardless of whether the current happens to be flowing in the positive or negative direction at that instant — a resistor is always converting electrical energy into heat, never giving energy back to the source. This does matter later in the chapter: for an inductor or capacitor, instantaneous power genuinely does swing negative for part of each cycle (energy flows back to the source), which is exactly why their long-run average power is zero even though current is flowing — a fact this article's steady, always-positive resistor power sets up as the contrast.

Is this the same 230 V, 50 Hz number I should just memorise for every mains question?

For CBSE numericals, yes — unless a question explicitly states different values, assume Indian domestic AC supply is $V_{rms} = 220$ - $230\ \text{V}$ (different past papers round differently, so always use whatever the question states) and $f = 50\ \text{Hz}$ , giving $\omega = 2\pi(50) \approx 314\ \text{rad/s}$ and a peak voltage near $311$ - $325\ \text{V}$ . Always double-check whether a given problem's "voltage" is peak or RMS before starting the calculation — that single reading-comprehension step is where most marks are lost on this topic, not the arithmetic itself.