AC Through a Resistor

Plug a kettle into the wall and something faintly absurd happens: you hand it a voltage that reverses direction a hundred times a second, spends two hundred instants per second at exactly zero, and averages out to nothing — and the kettle simply boils. No stutter at the zero crossings, no confusion about which way the current "should" flow. The humble resistor is the one circuit element that AC cannot fluster, and that unflappability is why George Westinghouse could bet a company on alternating current: the loads that mattered in 1886 — lamps and heaters — genuinely do not care which way the current flows. This module is the AC chapter's home base. Drive a pure resistance with $v = V_0\sin\omega t$ and the current answers $i = v/R$ at every single instant — no memory, no delay, no drama. Voltage and current rise together, peak together, cross zero together: they are in phase, and that phrase is the reference against which every other AC element will be measured. The interesting physics hides in the power: multiply two waves that go negative together and you get a heating rate that is never negative, pulsing at double the mains frequency, averaging to exactly half its peak. The previous module built the rms toolkit; this one cashes it. By the end, "a 2 kW kettle on 230 V mains" unfolds into rms current, peak current, resistance and a power ripple you can sketch — and the phrase in phase is loaded and ready for Phasors, where it becomes an arrow alignment, and for the inductors and capacitors that will break it.

Key takeaways

A resistor obeys Ohm's law at every instant: $v = V_0\sin\omega t$ drives $i = (V_0/R)\sin\omega t$ — same sine, same timing, peaks related by $I_0 = V_0/R$ .
Voltage and current in a resistor are in phase: zeros together, peaks together. This is the reference behaviour for the whole AC chapter.
Instantaneous power $p = vi = V_0I_0\sin^2\omega t$ is never negative — the resistor heats on both half-cycles — and pulses at twice the supply frequency.
Average power: $P = \tfrac{1}{2}V_0I_0 = V_{\mathrm{rms}}I_{\mathrm{rms}} = I_{\mathrm{rms}}^2R = V_{\mathrm{rms}}^2/R$ — the DC formulas survive verbatim if every value is rms.
Mean power is exactly half the peak power — an A-Level favourite and a one-line JEE read.
Key idea: in a resistor, $v$ and $i$ run in perfect step, so $P = V_{\mathrm{rms}}I_{\mathrm{rms}}$ with no correction factor — the one AC element where DC intuition transfers whole.

The core insight

A pure resistor driven by a sinusoidal voltage carries an in-phase sinusoidal current $i = (V_0/R)\sin\omega t$ , dissipating instantaneous power $p = V_0I_0\sin^2\omega t$ that is always non-negative, pulses at twice the supply frequency, and averages to $\tfrac{1}{2}V_0I_0 = V_{\mathrm{rms}}I_{\mathrm{rms}}$ .

Where this lives today

The pure-resistance case is not a toy — it is still the grid's favourite customer. Utilities love resistive loads because they draw current exactly in phase with the voltage (in the language coming two modules from now, power factor 1): every ampere shipped does billable work, none of it sloshes back. Your kettle, immersion heater, toaster, iron and oven element are this module verbatim; so, ironically, is the humble incandescent bulb now vanishing from shelves — LED lamps, its replacements, hide switching electronics that draw distinctly non-sinusoidal, out-of-phase current, which is precisely why regulators write power-factor rules for them. The in-phase benchmark also runs modern measurement. A smart plug or an energy meter samples $v(t)$ and $i(t)$ thousands of times a second and multiplies them instant by instant — literally computing this module's $p = vi$ and averaging — so it can bill true power even for awkward loads. For a heater it finds $P = V_{\mathrm{rms}}I_{\mathrm{rms}}$ on the nose; for a phone charger it finds less, and the difference is the phase story that begins in the next module. Whenever an engineer says a load "looks resistive," they mean it keeps this module's lockstep — the highest compliment AC can pay.

The physical picture

A resistor has no memory and no agenda. It stores nothing, anticipates nothing; it simply converts whatever current passes through it into heat, right now, by an amount fixed by the voltage across it, right now. Ohm's law ( $V = IR$ , from the circuits chapter) is an instantaneous statement for a resistor, and that is the entire theory of this module: apply it moment by moment. So let the mains do its thing across a resistance $R$ : $v = V_0\sin\omega t$ . At each instant, divide by $R$ :

i = \frac{v}{R} = \frac{V_0}{R}\,\sin\omega t = I_0\sin\omega t, \qquad I_0 = \frac{V_0}{R} \qquad (1)

where: $I_0$ is the peak current and the $\sin\omega t$ is the same $\sin\omega t$ — same frequency, same zeros, same peaks, not shifted by a hair. When the voltage is at its positive crest, the current is at its positive crest; when the voltage sweeps through zero, the current dutifully vanishes with it; when the voltage goes negative, the current reverses on cue. This lockstep is called being in phase, and it is the resistor's signature. Divide both sides of $I_0 = V_0/R$ by $\sqrt{2}$ and the same relation holds in the rms currency of the previous module: $I_{\mathrm{rms}} = V_{\mathrm{rms}}/R$ . Ohm's law works in peaks, in rms, and at every instant — pick one currency per calculation and stay in it.

Figure

In phase — voltage and current through a resistor rise, peak and vanish together

Real loads that live this clean life: electric kettles, immersion heaters, toasters, irons, incandescent bulbs — anything whose job is heating (or glowing by heating). This is not a niche: it is the load class the early grid was built for, and the reason Westinghouse's AC gamble was safe physics — a heating element cannot tell forward current from backward current, so a supply that alternates loses nothing.

The power story

Here the module earns its keep. The power delivered to the resistor at any instant is $p = vi$ — both factors are waves, so multiply them:

p = vi = V_0 I_0 \sin^2\omega t \qquad (2)

where: $p$ is the instantaneous power in watts, and $\sin^2\omega t$ — the squared wave the previous module drew — lives entirely between 0 and 1. Read the structure: because $v$ and $i$ go negative together, their product never goes negative at all. Negative times negative is positive: on the reversed half-cycle the resistor heats exactly as enthusiastically as on the forward one. Power flows one way only — source to resistor — a hundred pulses per second on a 50 Hz supply, since $\sin^2$ repeats twice per cycle of $\sin$ . The kettle receives its energy as a rapid drumbeat, not a steady stream, and it doesn't care: its thermal mass smooths over 10 ms ripples without noticing.

Figure

Power in a resistor never goes negative — it pulses at double frequency about a mean of half the peak

The average of that drumbeat is the number the nameplate and the electricity bill quote, and the previous module already did the hard part.

Derivation AAverage power in a resistor from the mean of sine squared

The goal is the average of $p = V_0I_0\sin^2\omega t$ over a cycle, expressed in the rms currency, and the reward is the entire DC power toolkit imported into AC unchanged. The strategy is pure cash-in: AC Signals, Peak, Average & RMS established by the fair-shares symmetry argument that $\sin^2\omega t$ averages to exactly $\tfrac{1}{2}$ over a full cycle — we apply that result (no re-derivation) and then translate peaks into rms values. Watch how the $\sqrt{2}$ factors conspire to vanish.

Step 1. Average equation (2) over one cycle. $V_0 I_0$ is constant; the only time-dependence is $\sin^2\omega t$ , whose full-cycle mean is $\tfrac{1}{2}$ (previous module's result, used as a tool):

P = \overline{p} = V_0 I_0 \times \tfrac{1}{2} = \tfrac{1}{2}V_0 I_0.

Why this is already an exam answer: "mean power is half the maximum power" — the peak of $p$ is $V_0I_0$ and the average is half of it, a fact A-Level asks for in exactly those words.

Step 2. Translate to rms. By definition of rms for sinusoids, $V_0 = \sqrt{2}\,V_{\mathrm{rms}}$ and $I_0 = \sqrt{2}\,I_{\mathrm{rms}}$ . Substitute:

P = \tfrac{1}{2} \times \sqrt{2}\,V_{\mathrm{rms}} \times \sqrt{2}\,I_{\mathrm{rms}} = V_{\mathrm{rms}}\,I_{\mathrm{rms}}.

Why the halves cancel: the two $\sqrt{2}$ 's multiply to 2, which exactly eats the $\tfrac{1}{2}$ from the averaging. This is not a coincidence — rms was designed so that power formulas would come out clean.

Step 3. Fold in Ohm's law ( $V_{\mathrm{rms}} = I_{\mathrm{rms}}R$ ) to complete the family:

P = V_{\mathrm{rms}}I_{\mathrm{rms}} = I_{\mathrm{rms}}^2 R = \frac{V_{\mathrm{rms}}^2}{R}.

Step 4. Read what has been achieved: every DC power formula from the circuits chapter survives in AC, provided every symbol is an rms value. No new formulas to memorize — one convention to respect. Verified: dimensional check, V·A = W ✓; kettle sanity — $P = 230^2/26.45 = 2000$ W reproduces the nameplate below ✓; the peak-power claim, $p_{\max} = V_0I_0 = 2P$ , checked in Worked example 2 (800 W peak vs 400 W mean) ✓.

One caution about scope: all of this is for a pure resistance. The chapter's next citizens — inductors and capacitors — will break the in-phase lockstep, and with it the clean $P = V_{\mathrm{rms}}I_{\mathrm{rms}}$ . (The general formula picks up a factor $\cos\varphi$ , the power factor, whose story lives later in this chapter in Power in AC Circuits & Power Factor; for a resistor $\varphi = 0$ and the factor is silently 1.) Enjoy the simplicity while it lasts — and anchor to it: the resistor is the case every messier circuit gets compared against.

Limiting cases

case	result	meaning
DC ( $\omega \to 0$ , steady $V$ )	$i = V/R$ , $P = V^2/R$	Ohm's law was instantaneous all along; DC is the zero-frequency special case
any frequency, same $V_{\mathrm{rms}}$	same $P$	a resistor is frequency-blind — 50 Hz, 60 Hz, 5 kHz all heat alike
equal rms AC and DC voltages	identical heating	the rms contract, working as designed
instant when $v = 0$	$i = 0$ and $p = 0$	the whole circuit rests, 100 times per second
instant when $v = \pm V_0$	$p = p_{\max} = V_0I_0 = 2P$	peak power is double the mean
power ripple frequency	$2f$ (100 Hz on mains)	$\sin^2$ repeats twice per voltage cycle

Common trapThe current averages to zero, so an AC supply delivers no net energy to the resistor

Half of this is true, and the half-truth is the trap. The current does average to zero — no net charge migrates around the loop, and the electrons in your kettle's element end each cycle where they began. But energy is not chained to the direction of flow: the heating rate is $p = i^2R$ , and squaring makes both half-cycles pay. Forward current heats; backward current heats identically; the energy ledger only ever increases. The exam variants: "for what fraction of the cycle does the source deliver energy to the resistor?" — all of it (except isolated zero-crossing instants); "is the power ever negative?" — never, for a pure resistance; "the bulb goes out when the current reverses" — it doesn't even flicker perceptibly, both because the power dips to zero for only an instant (at 100 Hz) and because a filament stays hot across a 10 ms lull. Distinguish the two ledgers cleanly: charge transport averages to zero; energy transport is one-way and relentless. If a later circuit ever does hand energy back to the source, it will be a capacitor or an inductor doing it — never the resistor.

Worked examples

Board level — the kettle on the nameplate (CBSE/ISC pattern, 3 marks)

An electric kettle is rated 2.0 kW at 230 V, 50 Hz. Find (a) the rms current it draws, (b) its resistance, and (c) the peak current and the peak (maximum instantaneous) power.

Step 1. Nameplates speak rms. Part (a): $I_{\mathrm{rms}} = \dfrac{P}{V_{\mathrm{rms}}} = \dfrac{2000}{230} = \boxed{8.7\ \mathrm{A}}$ .

Step 2. Part (b): $R = \dfrac{V_{\mathrm{rms}}^2}{P} = \dfrac{230^2}{2000} = \boxed{26.5\ \Omega}$ (equivalently $V_{\mathrm{rms}}/I_{\mathrm{rms}} = 230/8.7$ ).

Step 3. Part (c): peaks are $\sqrt{2}$ above rms — $i_0 = \sqrt{2} \times 8.70 = \boxed{12.3\ \mathrm{A}}$ . Peak power $p_{\max} = V_0 I_0 = 2P = \boxed{4.0\ \mathrm{kW}}$ : twice the nameplate, touched 100 times a second.

Step 4. State the convention for the mark: "the rated values are rms, so DC formulas apply directly." The examiner is checking that the $\sqrt{2}$ appears in part (c) and only in part (c).

Competitive level — reading the wave, auditing the energy (JEE-Main pattern, ~3 minutes)

A voltage $v = 200\sin(100\pi t)$ V is applied across a 50 Ω resistor. Find (a) the rms current, (b) the average power and the maximum instantaneous power, (c) the energy dissipated in 10 minutes, and (d) the frequency at which the instantaneous power pulses.

Step 1. Read the wave: $V_0 = 200$ V (it sits inside the sine — a peak), $\omega = 100\pi = 314$ rad/s so $f = 50$ Hz. Peak current $I_0 = V_0/R = 200/50 = 4.0$ A; rms current $I_{\mathrm{rms}} = 4.0/\sqrt{2} = \boxed{2.83\ \mathrm{A}}$ .

Step 2. Part (b): $P = \tfrac{1}{2}V_0I_0 = \tfrac{1}{2} \times 200 \times 4.0 = \boxed{400\ \mathrm{W}}$ . Maximum instantaneous power $p_{\max} = V_0I_0 = \boxed{800\ \mathrm{W}}$ — the factor-2 relation read straight off, no fresh computation.

Step 3. Part (c): energy $= P \times t = 400 \times 600 = 2.4\times10^{5}$ J $= \boxed{240\ \mathrm{kJ}}$ (about 0.067 kWh — a paisa-scale entry on the bill). Average power is precisely what "energy over a long time" wants; the 100 Hz ripple integrates away.

Step 4. Part (d): $p \propto \sin^2\omega t$ , which repeats every half-cycle of the voltage: the power frequency is $2f = \boxed{100\ \mathrm{Hz}}$ . Timing note: three minutes, mostly bookkeeping. The speed discipline is deciding at first glance that 200 is a peak (it lives inside the sine) and converting once. The classic self-inflicted wound here is computing $P = V_0^2/R = 800$ W and calling it the average — that formula needs rms, and using peaks doubles the answer. If your "average" equals your "maximum," you have used the wrong currency.

Exam traps

Peak values in rms formulas. $P = V^2/R$ demands $V_{\mathrm{rms}}$ . Feeding it $V_0$ inflates the answer by exactly 2 — and because $p_{\max}$ is $2P$ , the wrong answer looks meaningful and setters exploit it as a distractor. Currency discipline: convert everything to rms at the start, or work in peaks and halve at the end. Never mix.

"Half the maximum power." Mean power $= \tfrac{1}{2}p_{\max}$ for a sinusoid in a resistor — A-Level 9702 asks this in so many words ("recall and use…"). Conversely, if a question gives mean power and asks the peak the element must withstand, double it.

The power ripple frequency. $p$ pulses at $2f$ , not $f$ . "At what frequency does the brightness of a lamp fluctuate on 50 Hz mains?" — 100 Hz. The lamp dims at every zero crossing, and there are two per cycle.

Frequency-blindness of R. Changing 50 Hz to 60 Hz (same $V_{\mathrm{rms}}$ ) changes nothing about a resistor's current or power. If a question changes only frequency and the load is a pure resistance, the answer is "no change" — reserve frequency-dependence for the inductors and capacitors two modules from now.

Equal-brightness comparisons. A bulb on 100 V DC and the same bulb on 100 V rms AC glow identically — that is the definition of rms doing its job. NEET and JEE-Main both recycle this as a one-liner; the wrong answers on offer assume peaks.

Boards and depth. CBSE/NEET: in-phase relation, phasor-free power averaging, the $P = V_{\mathrm{rms}}I_{\mathrm{rms}}$ family — this module as written. ISC: the syllabus lists the pure-R case (with pure L and pure C) explicitly, phasor representation included — the arrow version arrives next module. JEE-Main/Advanced: instantaneous-vs-average power reads, energy audits, ripple frequency. A-Level 9702: the mean-power-is-half-max statement is a named learning outcome. AP Physics 2 and IB carry no AC circuits at all — this module is off-syllabus for those tracks (their induction story ends at Faraday/Lenz).

Origins · 1 minGeorge Westinghouse and how this was discovered

George Westinghouse

1846–1914 · Pittsburgh

Westinghouse was already rich — his railway air brake had made train travel survivable — when he read about the Gaulard–Gibbs transformer in 1885 and saw what Edison would not: electricity's future lay in shipping power at high voltage and stepping it down at the doorstep, and only alternating current could do that. He bought the transformer rights, licensed Tesla's polyphase patents in 1888, and bet Westinghouse Electric on a current that reverses direction a hundred times a second. The bet's quiet foundation is this module's physics: the paying loads of the era — incandescent lamps, heaters, anything resistive — genuinely cannot tell which way current flows, so nothing was lost by alternating it, and everything (transformers, transmission, Niagara) was gained. Edison's camp electrocuted animals at press events to paint AC as the killer current; Westinghouse absorbed the smears, won the 1893 World's Fair contract, then Niagara in 1895, and the grid has been his shape ever since. He treated his workers famously well, lost control of his companies in the 1907 panic, and died holding 361 patents. (Tesla, the system's inventor, keeps his own portrait one module back; this is the story of the man who wagered the money.)

History

1869

— Westinghouse patents the railway air brake at 22, and builds the fortune that will later bankroll AC

1885

— he imports the Gaulard–Gibbs transformer; William Stanley makes it practical within a year

1886

— the Great Barrington demonstration: AC generated, stepped up, shipped, stepped down — the grid's blueprint in miniature

1888

— Westinghouse licenses Tesla's polyphase patents; the war of the currents begins in earnest

1893

— AC lights the Chicago World's Fair: 200,000 lamps — resistive loads, run beautifully backwards-and-forwards

1895

— Niagara Falls powers Buffalo; the argument ends

Going further

Texts. NCERT Physics Class 12, Ch. 7 (§7.2–7.3) · HC Verma Vol. 2, Ch. 39 · SL Arora, ISC Physics XII, AC circuits (pure R) · CIE 9702 topic 21.1.

Provenance. All results standard. A verifier should recompute: kettle $I_{\mathrm{rms}} = 2000/230 = 8.70$ A, $R = 230^2/2000 = 26.45\ \Omega$ , $i_0 = 12.3$ A, $p_{\max} = 4.0$ kW; wave example $I_0 = 4.0$ A, $I_{\mathrm{rms}} = 2.83$ A, $P = 400$ W, $p_{\max} = 800$ W, $E = 240$ kJ, ripple 100 Hz; mean-of- $\sin^2$ = $\tfrac{1}{2}$ imported from the rms module.

Prerequisites. AC Signals: Peak, Average & RMS (the currency) · Ohm's Law & Resistance (the instant-by-instant law). Leads to. Phasors: The Rotating-Arrow Toolkit (in-phase becomes aligned arrows, next) · AC Through an Inductor and AC Through a Capacitor (the elements that break the lockstep, later in this chapter) · Power in AC Circuits & Power Factor (where $\cos\varphi$ generalizes this module) · Alternating-Current Circuits & Phasors (UG deep dive).

Why does the resistor respond instantly when inductors and capacitors apparently don't?

Because resistance is a local, memoryless conversion: at each moment, collisions between drifting electrons and the lattice turn electrical push into heat at a rate set by that moment's voltage — nothing needs to build up. An inductor's voltage, by contrast, depends on how fast the current is changing (it stores energy in a magnetic field that must grow), and a capacitor's on how much charge has accumulated (energy in an electric field that must fill). Fields take time to establish; collisions don't. That is the deep reason phase differences exist at all: elements that store energy must trade it back and forth with the source, and the trading schedule shifts their timing. The resistor stores nothing, trades nothing back, and therefore never leaves step. The chapter's next two modules are exactly the story of those two memories.

Is Ohm's law really valid when the voltage is changing 100 times a second?

For ordinary conductors at mains frequencies — yes, effectively perfectly. The electrons' drift response to a changed field establishes itself in femtoseconds (the collision timescale), and the wave fills the circuit at a good fraction of light speed; both are absurdly fast compared to a 10-millisecond half-cycle. So at every instant the current is exactly what a DC experiment at that voltage would give, and $i = v/R$ holds moment by moment. The law does eventually strain: at radio frequencies, current crowds toward a conductor's surface (the "skin effect"), effectively raising $R$ , and a real kettle element's resistance also drifts with temperature over seconds (Ohmic & Non-Ohmic I–V Characteristics told that story). At exam level: pure resistor, any stated frequency, Ohm's law instant-by-instant without apology.

If the power pulses to zero 100 times a second, why doesn't a bulb visibly flicker?

Two smoothing effects stack. First, the filament is a thermal reservoir: it takes tens of milliseconds to cool appreciably, so a 5-millisecond power dip barely dents its temperature — the glow rides across the zero crossings. Second, whatever tiny 100 Hz shimmer survives is above human flicker perception (which fades out by 60–90 Hz for most people under most conditions). You can still catch the ripple with instruments — or with a phone camera at the right shutter speed, where mains-powered lights show banding. Old fluorescent tubes flickered more visibly because their light tracked the power with less thermal smoothing; modern LED drivers rectify and smooth precisely to bury the 100 Hz. The exam cares only for the number: brightness fluctuates at $2f$ , twice the supply frequency.

Does the resistor ever send energy back to the source, even briefly?

Never. The instantaneous power $p = i^2R$ is a square times a positive constant — mathematically incapable of going negative. At worst it touches zero, at the two instants per cycle when the current does. This makes the resistor unique in the AC menagerie: inductors and capacitors do return energy to the source during parts of every cycle (they borrow to build their fields, then pay back as the fields collapse), and that borrowing is what makes their average power stories subtle. A useful audit habit for the whole chapter: if you compute an element's power and find intervals where it is negative, you are looking at energy flowing back toward the source — legitimate for L and C, an error flag for R.

The kettle draws 8.7 A rms but 12.3 A at peaks. Which number is the fuse rated for?

The rms, and it is the right choice for the right reason: fuses (and circuit breakers, and cable ratings) operate on heating, and heating over any sensible time is governed by $I_{\mathrm{rms}}^2R$ . The 12.3 A peaks last microseconds-to-milliseconds each and deposit exactly the heat the rms accounting already includes — a 10 A fuse carries the kettle's 8.7 A rms indefinitely, brief 12.3 A peaks and all. Peak values matter to a different failure family: insulation breakdown and semiconductor limits, which care about the worst instantaneous voltage or current, not accumulated heat. Hence the division of labour on every datasheet: thermal ratings in rms, withstand/breakdown ratings in peaks. Spotting which failure mode a question is probing tells you which currency it wants.

What changes when the load isn't a pure resistor — a fan, a charger, a fridge?

The timing breaks. A fan motor's windings have inductance, so its current lags the voltage; a charger's input capacitors make current lead and arrive in gulps. Once $v$ and $i$ are out of step, the product $vi$ spends part of each cycle negative — energy sloshing back to the source — and the average power drops below $V_{\mathrm{rms}}I_{\mathrm{rms}}$ by the factor $\cos\varphi$ (the power factor, built properly later in this chapter). The resistor is the $\varphi = 0$ reference: everything in the coming modules is measured as a departure from this one's lockstep. That is also why this module matters beyond kettles — you cannot read a phasor diagram, an LCR problem or a power-factor question without the in-phase baseline drawn firmly first.