ARCHETYPE: Bulbs & Ratings — Brightness Logic

Take two household bulbs — one stamped "220 V, 100 W", one stamped "220 V, 60 W" — and wire them in series across the mains. Everyone's instinct says the 100 W bulb should glow brighter; it is, after all, the more powerful lamp. Run the experiment and the 60 W bulb wins, visibly and reliably. Nothing is broken. The instinct fails because a bulb's rating is a promise, not a property: it tells you what the lamp will do at its rated voltage, wired the way the manufacturer assumed — in parallel across the mains, each bulb getting its full 220 V. Put bulbs in series and you have changed the contract; the only honest property the bulb carries into the new circuit is its resistance. That one idea — convert the rating to a resistance, then re-solve the actual circuit — is an entire industry of examination questions. CBSE and ICSE ask for the resistance and current of a rated bulb; IGCSE and A-Level dress it in fuses and filament curves; JEE-Main and NEET run the series-versus-parallel brightness duel and its lethal 440 V variant, where the dimmer-rated bulb is the one that burns out first. Every version is the same three moves: decode the rating, find the true operating point, compare powers. This archetype drills those moves: the decoding formula $R = V^2/P$ , the series and parallel brightness logic, rated versus actual power, the failure modes of a series string, and the fuse arithmetic that keeps the whole business from catching fire. The power formulas themselves belong to Electrical Energy, Power & Joule Heating; this is where they earn exam marks.

Key takeaways

A rating " $V$ volts, $P$ watts" is a promise at rated voltage; the portable property is the resistance $R = \dfrac{V_{\text{rated}}^2}{P_{\text{rated}}}$ .
Higher rated power ⇒ lower resistance: the 100 W bulb is the low-resistance bulb.
In series (same $I$ ): $P = I^2R$ — the higher-resistance, lower-rated bulb glows brighter.
In parallel (same $V$ ): $P = V^2/R$ — the lower-resistance, higher-rated bulb glows brighter, and ratings behave as stamped.
Off-rated operation scales as $P_{\text{actual}} = P_{\text{rated}}\left(\dfrac{V_{\text{actual}}}{V_{\text{rated}}}\right)^2$ — half the voltage, quarter the power (for a fixed-resistance filament).
Two rated bulbs in series across the mains: the combination delivers $P_{\text{series}} = \dfrac{P_1P_2}{P_1+P_2}$ — the "parallel formula" applied to powers.
KEY IDEA: never compare ratings across circuits — convert every rating to a resistance, solve the real circuit, then rank $I^2R$ or $V^2/R$ .

The core insight

A bulb's rating defines its resistance, $R = V_{\text{rated}}^2/P_{\text{rated}}$ ; in any other circuit the bulb keeps the resistance and abandons the rating, so brightness is ranked by recomputing the actual power — $I^2R$ when the current is shared (series), $V^2/R$ when the voltage is shared (parallel).

Where this lives today

The incandescent duel is fading from ceilings but not from engineering. Electric kettles, irons, geysers and toasters are still pure resistive loads whose nameplates read exactly like bulb ratings — and the same arithmetic sizes their fuses, cables and plugs. Voltage sag on a loaded grid dims whole neighbourhoods by the $V^2$ law: a 10% undervoltage steals 19% of every heater's output. Strings of decorative LEDs still die the fairy-light death when wired in series. And the inverse ratio at the heart of this archetype — power rating as inverse resistance — is the first sanity check an electrical engineer runs on any parallel bus design: the biggest consumers are the lowest-impedance paths, so a short circuit is simply the "infinite-watt appliance" the fuse exists to veto.

The setup

The examination bulb is a resistor wearing a label. The label — "220 V, 100 W" — means exactly one sentence: connected across 220 V, this filament will convert 100 W into heat and light. From that sentence, two numbers follow, and they are the working currency of every problem in this archetype:

R = \frac{V_{\text{rated}}^2}{P_{\text{rated}}} \qquad (1)

where: $R$ — the filament's operating resistance; $V_{\text{rated}}, P_{\text{rated}}$ — the two numbers on the glass. For the 100 W bulb: $R = 220^2/100 = 484\ \Omega$ . For the 60 W bulb: $R = 220^2/60 \approx 807\ \Omega$ . Notice the inversion that powers half the trick questions in this family: the more powerful bulb has the smaller resistance. It must — at the same mains voltage, more power means more current, means less opposition.

I_{\text{rated}} = \frac{P_{\text{rated}}}{V_{\text{rated}}} \qquad (2)

where: $I_{\text{rated}}$ — the current the filament is built to carry ( $0.455$ A for the 100 W bulb). Push the filament much beyond this and it overheats and fails: equation (2) is the bulb's safety line, and the 440 V problems below live and die by it.

Figure

The same two rated bulbs wired in series and in parallel — the brightness order inverts

The master pattern

Every bulb problem is three moves, in order, no exceptions.

Step 1 — decode. Convert every rating to a resistance via equation (1). From this point the words "60 W" and "100 W" mean nothing; only $807\ \Omega$ and $484\ \Omega$ exist. (Treat the filament resistance as fixed at its operating value — the exam convention; the real-life fine print is in the FAQ.)

Step 2 — solve the actual circuit. Series: one shared current $I = V/(R_1 + R_2 + \cdots)$ , as built in Resistors in Series & Parallel. Parallel: one shared voltage. Mixed: reduce as usual.

Step 3 — rank by recomputed power. Series (shared $I$ ): compare $P = I^2R$ — resistance order is brightness order. Parallel (shared $V$ ): compare $P = V^2/R$ — resistance order inverts. Then, if asked, check each bulb's actual numbers against its rating for survival.

Derivation AThe series brightness inversion and the two-bulb power formula

We want to prove the counter-intuitive centrepiece of this archetype: two bulbs rated $P_1$ and $P_2$ at the same voltage $V_0$ , wired in series across that same $V_0$ , glow in reverse rating order, and together deliver $P_{\text{series}} = P_1P_2/(P_1+P_2)$ . The strategy is pure pattern: decode the ratings into resistances, run the shared-current logic, and keep everything symbolic so the result works for any pair of ratings. Watch how the rated voltage cancels — the inversion is structural, not numerical.

Step 1. Decode: $R_1 = V_0^2/P_1$ and $R_2 = V_0^2/P_2$ . The higher-rated bulb carries the lower resistance.

Step 2. Series current across the same $V_0$ :

I = \frac{V_0}{R_1 + R_2} = \frac{V_0}{V_0^2\left(\frac{1}{P_1} + \frac{1}{P_2}\right)} = \frac{P_1P_2}{V_0(P_1+P_2)}

Step 3. Actual power in each bulb is $I^2R$ . Take the ratio — the current cancels:

\frac{P_{1,\text{actual}}}{P_{2,\text{actual}}} = \frac{R_1}{R_2} = \frac{P_2}{P_1}

Why this is the whole trick: with the current shared, actual power is proportional to resistance, and resistance is inversely proportional to rating — so the actual-power ratio is the rating ratio upside down. The 60 W bulb beats the 100 W bulb by exactly the factor 100/60.

Step 4. Total delivered power:

P_{\text{series}} = \frac{V_0^2}{R_1 + R_2} = \frac{V_0^2}{V_0^2\left(\frac{1}{P_1}+\frac{1}{P_2}\right)} = \frac{P_1P_2}{P_1+P_2}

The rated voltage cancels completely — the series combination of a 100 W and a 60 W bulb delivers $6000/160 = 37.5$ W across the rated mains, regardless of whether the mains is 220 V or 110 V (as long as bulbs and supply share the rating).

Step 5. Memorise the shape, not just the result: powers of series-connected same-voltage ratings combine like resistances in parallel, $\tfrac{1}{P} = \tfrac{1}{P_1} + \tfrac{1}{P_2}$ , and parallel-connected ratings simply add, $P = P_1 + P_2$ . The inversion happens because $R \propto 1/P$ . Verified: numerically for 220 V ratings — R₁ = 484 Ω, R₂ = 806.7 Ω, I = 0.1705 A, P₁₀₀ = 14.1 W, P₆₀ = 23.4 W (ratio 0.60 = 60/100 inverted), total 37.5 W = 6000/160.

Off the label — rated versus actual power

A bulb only delivers its stamped watts at its stamped volts. At any other voltage (filament resistance taken constant), the power rescales quadratically:

P_{\text{actual}} = \frac{V_{\text{actual}}^2}{R} = P_{\text{rated}}\left(\frac{V_{\text{actual}}}{V_{\text{rated}}}\right)^2 \qquad (3)

where: the square does the damage. A 100 W bulb on half its rated voltage delivers a quarter of its rating — 25 W. This is also the equation of failure: a bulb pushed to $\sqrt{2}$ times its rated voltage runs at double power, and filaments do not forgive that for long. Whenever a series string puts more than rated voltage across one bulb (see Competitive level 2), equation (3) tells you which lamp is being executed and how fast.

Figure

Actual power against applied voltage for a fixed-resistance filament — the quadratic that rewards and punishes

The same scaling underlies the fuse choice: a fuse must carry the appliance's rated current comfortably and melt not far above it. A 2.3 kW kettle on 230 V mains (the Indian and UK standard) draws $I = P/V = 2300/230 = 10$ A — so it gets a 13 A fuse, not a 3 A (instant blow) and not a 30 A (no protection). Running costs stay light here — a 100 W bulb burning for 10 hours uses $1\ \mathrm{kWh}$ , one "unit" on the meter — and the full kilowatt-hour billing machinery lives with Household Circuits, kWh & Electrical Safety.

The variants examiners actually pose

variant	governing logic	answer shape
two ratings in series, rated mains	shared $I$ , $P \propto R \propto 1/P_{\text{rated}}$	lower-rated bulb brighter; total $\dfrac{P_1P_2}{P_1+P_2}$
two ratings in parallel, rated mains	shared $V$ , $P = V^2/R$	each delivers its rating; total $P_1 + P_2$
bulb on wrong voltage	$P \propto V^2$ at fixed $R$	$P_{\text{rated}}(V/V_{\text{rated}})^2$
series string on doubled mains	find $I$ , compare with each $I_{\text{rated}}$	the lowest-rated bulb exceeds rating first
low-voltage bulb + series resistor	resistor drops the excess at rated current	$R_{\text{series}} = \dfrac{V_{\text{mains}} - V_{\text{rated}}}{I_{\text{rated}}}$
one bulb in a string blows	series: circuit opens	all go dark — the classic fairy-lights failure
fuse selection	$I = P/V$ , next standard size up	e.g. 10 A appliance → 13 A fuse

Worked examples

Board level 1 — decoding a rating (CBSE/ICSE pattern)

An electric bulb is rated $230\ \mathrm{V},\ 100\ \mathrm{W}$ . Find (a) its working resistance, (b) the current it draws at rated voltage, and (c) the power it would consume on a $115\ \mathrm{V}$ supply, assuming constant filament resistance.

Step 1. Resistance from the rating: $R = \dfrac{V^2}{P} = \dfrac{230^2}{100} = \boxed{529\ \Omega}$ .

Step 2. Rated current: $I = \dfrac{P}{V} = \dfrac{100}{230} = \boxed{0.435\ \mathrm{A}}$ . (Cross-check: $V/R = 230/529 = 0.435$ A ✓.)

Step 3. On half voltage, equation (3): $P = 100 \times \left(\dfrac{115}{230}\right)^2 = 100 \times \dfrac{1}{4} = \boxed{25\ \mathrm{W}}$ — a quarter of the rating, the square at work.

Board level 2 — running a low-voltage bulb from the mains (ISC/A-Level pattern)

A bulb rated $110\ \mathrm{V},\ 60\ \mathrm{W}$ must run at its rating from a $220\ \mathrm{V}$ supply. What resistance must be placed in series with it?

Step 1. The bulb's operating requirements, decoded: $I_{\text{rated}} = \dfrac{60}{110} = 0.545\ \mathrm{A}$ ; bulb resistance $R_{\text{bulb}} = \dfrac{110^2}{60} = 201.7\ \Omega$ .

Step 2. At rated operation the bulb takes its 110 V, so the series resistor must absorb the other $220 - 110 = 110\ \mathrm{V}$ while carrying the same $0.545$ A.

Step 3. The resistor: $R = \dfrac{110}{0.545} = \boxed{201.7\ \Omega \approx 202\ \Omega}$ — exactly equal to the bulb's own resistance, because it must drop an equal voltage at the same current. (Elegant check: doubling the supply requires doubling the total resistance.)

Step 4. Note the cost: the resistor dissipates $110 \times 0.545 = 60\ \mathrm{W}$ — as much as the bulb. This brute-force method wastes half the energy, which is why real installations use transformers instead.

Competitive level 1 — the series brightness duel (JEE-Main/NEET pattern)

Bulbs rated $220\ \mathrm{V},\ 100\ \mathrm{W}$ and $220\ \mathrm{V},\ 60\ \mathrm{W}$ are joined in series across a $220\ \mathrm{V}$ supply. Find the power delivered by each bulb and the total, and state which glows brighter.

Step 1. Decode: $R_{100} = \dfrac{220^2}{100} = 484\ \Omega$ ; $R_{60} = \dfrac{220^2}{60} = 806.7\ \Omega$ .

Step 2. Shared current: $I = \dfrac{220}{484 + 806.7} = \dfrac{220}{1290.7} = 0.1705\ \mathrm{A}$ .

Step 3. Actual powers: $P_{100} = I^2R_{100} = (0.1705)^2(484) = \boxed{14.1\ \mathrm{W}}$ ; $P_{60} = I^2R_{60} = \boxed{23.4\ \mathrm{W}}$ . The 60 W bulb glows brighter — running at less than half its rating, but still ahead.

Step 4. Total: $14.1 + 23.4 = 37.5\ \mathrm{W}$ — confirming the shortcut $\dfrac{P_1P_2}{P_1+P_2} = \dfrac{100 \times 60}{160} = 37.5\ \mathrm{W}$ . Timing note: in an MCQ, skip Steps 2–3 entirely — series actual powers stand in the inverse ratio of ratings ( $P_{60}:P_{100} = 100:60 = 5:3$ ) and the total is the "parallel of the powers", 37.5 W. Twenty seconds, no current computed.

Competitive level 2 — the over-voltage execution (JEE-Advanced stretch)

Bulbs rated $220\ \mathrm{V},\ 25\ \mathrm{W}$ and $220\ \mathrm{V},\ 100\ \mathrm{W}$ are connected in series across $440\ \mathrm{V}$ . Which bulb fuses first, and what power is it dissipating at that moment? Assume constant filament resistances.

Step 1. Decode: $R_{25} = \dfrac{220^2}{25} = 1936\ \Omega$ ; $R_{100} = 484\ \Omega$ . Rated currents: $I_{25} = \dfrac{25}{220} = 0.114\ \mathrm{A}$ ; $I_{100} = \dfrac{100}{220} = 0.455\ \mathrm{A}$ .

Step 2. Series current on 440 V: $I = \dfrac{440}{1936 + 484} = \dfrac{440}{2420} = 0.182\ \mathrm{A}$ .

Step 3. Compare with the safety lines: the 25 W bulb is carrying $0.182 > 0.114$ A — 60% over its design current — while the 100 W bulb sits comfortably below its $0.455$ A. Voltage check tells the same story: the 25 W bulb takes $IR_{25} = 352\ \mathrm{V}$ , far above its 220 V rating; the 100 W bulb gets only 88 V.

Step 4. Verdict: the 25 W bulb fuses first, dissipating $I^2R_{25} = (0.182)^2(1936) \approx \boxed{64\ \mathrm{W}}$ — over two and a half times its rating — at the moment of failure. (Epilogue for the alert: once it blows, the series circuit opens and the healthy 100 W bulb goes dark too.) Timing note: the fast read is structural — in series, voltage divides in proportion to resistance, so the highest-resistance (lowest-rated) bulb always takes the most over-voltage. Name the victim in ten seconds, then spend the minute computing its death-power. Total ~2.5 min.

Exam traps

Ranking by rating across circuits. The rating orders brightness only in parallel at rated voltage. In series it inverts. Any answer that says "100 W > 60 W, so brighter" without checking the wiring has stepped on the archetype's central landmine.

Using rated values in the wrong circuit. A "220 V, 100 W" bulb in a series string is not carrying 0.455 A and not dissipating 100 W. The only imported number is $R = 484\ \Omega$ ; everything else is recomputed from the actual circuit.

Forgetting the square in off-rated scaling. Half voltage is quarter power, not half power: $P \propto V^2$ at fixed resistance. Boards phrase this as "a 100 W bulb on 110 V mains" expecting 25 W.

Series total power formula misuse. $P_{\text{series}} = \dfrac{P_1P_2}{P_1+P_2}$ holds only when the supply voltage equals the common rated voltage of both bulbs. On any other supply, decode to resistances and recompute — the shortcut silently fails.

Fairy-lights logic. One blown bulb in a series string opens the whole circuit: all dark. In parallel each lamp lives independently — the reason houses are wired in parallel, as Household Circuits, kWh & Electrical Safety develops. Exams ask this as a one-marker; get it from the circuit topology, not memory.

Fuse ratings. The fuse protects the cable, chosen just above the appliance's operating current ( $I = P/V$ ). A fuse rated below the operating current blows at switch-on; one rated far above never protects. On 230 V mains (India/UK), a 2.3 kW appliance draws 10 A → 13 A fuse.

Constant-resistance fine print. All these problems assume the filament's resistance is fixed at its operating value. Real filaments have much lower cold resistance (an I–V curve told in Ohmic & Non-Ohmic: I–V Characteristics) — state the assumption when a question says "assume constant resistance," and don't fight it in the exam.

Origins · 1 minThomas Edison and how this was discovered

Thomas Edison

1847–1931 · Menlo Park and West Orange, New Jersey

Edison's bulb is remembered; his arithmetic built this archetype. When the Pearl Street station lit lower Manhattan in September 1882, every design choice was a bulb-rating calculation at industrial scale: lamps were standardised at a fixed voltage (~110 V) and wired in parallel, so each customer's lamp saw the same voltage and delivered its promised brightness regardless of how many neighbours switched on. That decision — voltage as the shared standard, power as the per-device promise — is exactly why bulbs carry a "V, W" rating at all, and why the rating is a parallel-circuit promise that series wiring breaks. Edison even priced electricity by the lamp before meters existed, an early ancestor of the kWh unit. The series-string exceptions he tolerated — early Christmas lights — gave the world the "one dies, all dark" failure that examiners still harvest.

History

1879

— Edison and Swan independently produce practical carbon-filament incandescent lamps

1882

— Pearl Street station: standardised-voltage parallel distribution makes the "V, W" rating meaningful

1901

— tungsten begins replacing carbon; filament resistance rises steeply from cold to hot

1962

— the compact fuse plug becomes standard in UK/Indian wiring practice, fixing the 3 A/13 A choice ritual

2010

s — LED lamps deliver the light of a "100 W" bulb from 10 W; ratings shift to lumen-equivalents, but the V–W arithmetic still governs every heater, kettle and iron

Going further

Texts. NCERT Physics Class 12, Ch. 3 (Current Electricity) · HC Verma, Concepts of Physics Vol. 2, Ch. 32–33 · CIE IGCSE 0625 §4.3 and 9702 topic 9 (rated appliances, fuses).

Provenance. Standard exam canon; every number recomputed: rating decode ( $529\ \Omega$ , $0.435$ A, quarter-power 25 W); series resistor for the 110 V bulb ( $201.7\ \Omega$ , wasting 60 W); the 100/60 duel ( $484/806.7\ \Omega$ , $I = 0.1705$ A, $14.1$ vs $23.4$ W, total $37.5$ W $= P_1P_2/(P_1+P_2)$ ); the 440 V execution ( $I = 0.182$ A vs rated $0.114$ A, 352 V across the 25 W bulb, 64 W at failure); kettle fuse ( $10$ A → 13 A).

Prerequisites. Electrical Energy, Power & Joule Heating · Resistors in Series & Parallel. Peers. Battery Problems — Terminal V, Max Power & Efficiency (the source-side power ledger) · Household Circuits, kWh & Electrical Safety (where the kWh billing and wiring detail live) · Ohmic & Non-Ohmic: I–V Characteristics (the real filament's curve). Coming up. The technique module "P = V²/R Ratio Craft" turns this archetype's comparisons into a 30-second MCQ system.

Why does the higher-wattage bulb have the lower resistance? That feels backwards.

Read the rating as a contract at fixed voltage: to convert more power at the same 220 V, the bulb must draw more current, and more current at the same voltage means less resistance — $R = V^2/P$ puts $P$ in the denominator. The intuition "more powerful = more substantial = more resistance" is the exact inversion the examiners are counting on. A useful anchor: a heater (2000 W) has a resistance of only ~26 Ω on 230 V, far below a 100 W bulb's 529 Ω.

In the series duel, is the 60 W bulb actually delivering 60 watts?

No — it delivers 23.4 W, well under half its rating, and the 100 W bulb manages only 14.1 W. Both bulbs are dim compared with their parallel selves; the 60 W bulb is merely the less-starved of the two. The rating tells you what each bulb would do with the full 220 V to itself; in series they share the voltage unequally (in proportion to resistance), and neither gets its promise honoured.

Does a real filament actually keep its resistance constant when the voltage changes?

No — and exams know it, which is why careful questions say "assume constant resistance." A tungsten filament's resistance climbs steeply with temperature: cold, it can be a tenth of its operating value (the reason bulbs usually blow at switch-on, when the inrush current is huge). The constant- $R$ model is the standard approximation for rating arithmetic, and it is what every board expects unless the question supplies an I–V curve — the curved reality is the subject of Ohmic & Non-Ohmic: I–V Characteristics.

Why do houses wire everything in parallel if series wiring uses less power?

Series wiring uses less power precisely by starving every device — the "saving" is the failure. Parallel wiring gives every appliance the full mains voltage so each delivers its rated performance, lets you switch devices independently, and survives any single failure. Series wiring shares out the voltage in proportion to resistance (so nothing works as designed), dies entirely when one element opens, and changes every device's behaviour whenever another is plugged in. The full story — ring mains, earthing, why the fuse sits on the live wire — belongs to Household Circuits, kWh & Electrical Safety.

Is there a quick way to remember which formula ranks brightness?

Ask what the bulbs share. Series shares current → use $P = I^2R$ → bigger $R$ (smaller rating) wins. Parallel shares voltage → use $P = V^2/R$ → smaller $R$ (bigger rating) wins. One sentence covers every variant: the shared quantity picks the formula, and the formula picks the winner. The upcoming technique module on power-ratio craft builds a whole speed system on this one sentence.

How do I choose a fuse rating in an exam question?

Compute the appliance's normal current, $I = P/V$ , then pick the next standard fuse size above it. A 2.3 kW kettle on 230 V draws 10 A → 13 A fuse (standard sizes being 3 A, 5 A, 13 A). Choosing 3 A blows at switch-on; choosing 30 A lets a fault cook the cable before the fuse notices. State the reasoning — "just above normal operating current" — for the explanation mark on IGCSE and ICSE papers.

What happens to the other bulbs when one bulb in a parallel bank blows?

Nothing — each parallel branch has its own private path across the supply, so the survivors keep their full voltage and full brightness. The total current drawn from the supply simply drops by the dead branch's share. This independence is the entire case for parallel domestic wiring, and its absence is the fairy-light tragedy: in series, one open filament breaks the only path, and the whole string goes dark.