TECHNIQUE: P = V²/R Ratio Craft

The same-V/same-I gate is a live engineering decision, not just exam craft. Electric-vehicle makers moved their packs from 400 V to 800 V architectures for one quadratic reason: delivering the same charging power at double the voltage halves the current and quarters the $I^2R$ heat in every cable and connector — the series-jurisdiction costume applied to a highway charger. Data centres budget racks the same way, distributing at higher voltage and converting late to keep conductor losses in the $I^2R$ lane small. And your phone's fast charger renegotiates voltage with the phone in real time (that is what the USB-PD handshake is) precisely so the cable — the fixed-current, series-like element — never becomes the biggest spender in the chain. Grid operators still quote the classic form of the argument: transmission at hundreds of kilovolts exists because line loss is $I^2R$, and the only way to move gigawatts with small $I$ is enormous $V$.

Every power problem opens with the same fork in the road, and most marks are lost at the fork rather than on the arithmetic. Ask one question before touching any formula: what do these two devices share — the voltage across them, or the current through them?

The physics under the gate is ownership. In a parallel connection every branch owns the full supply voltage but chooses its own current; the narrow lane (high $R$) lets little through and works gently. In a series chain every element owns the same current but charges its own toll; the biggest obstacle (high $R$) extracts the most energy from each passing coulomb. Same devices, opposite verdicts — which is why the gate must come before the formula, every single time. The second tool is the decode. A label like "230 V, 1000 W" is a promise about performance at rated voltage, and Bulbs & Ratings — Brightness Logic established what survives when the promise is broken: the resistance,

where: $V_{\text{rated}}, P_{\text{rated}}$ — the two numbers on the label; $R$ — the operating resistance, the only property the device carries into any circuit. A 1000 W kettle at 230 V decodes to $52.9\ \Omega$; a 100 W bulb to $529\ \Omega$. Notice the inversion at birth: more rated watts means fewer ohms, and that single fact powers half the trick questions in the archive.

Boards love the time version: a heater must deliver a fixed job — boil this water, melt that wax — and the question trades in minutes rather than watts. Time is just energy divided by power, and since the job $H$ is fixed, $t = H/P$: time inverts power. At fixed mains this stacks with equation (2) into $t \propto R$ — time tracks resistance directly, which makes series/parallel timetables absurdly clean.

The whole toolkit in one table — the costume, the proportionality, and the reflex:

Board level — the cut coil (CBSE/NEET pattern, 2-3 marks)

A heater coil boils a fixed quantity of water in 10 minutes on the mains. The coil is cut into two equal halves, and both halves are connected in parallel across the same mains. How long does the same job take?

The slow way. Assign $V = 230$ V, invent a resistance, compute the original power, halve the resistance twice, recompute, divide into the job… two minutes and three invented numbers.

The ratio way. Step 1. Half the coil = half the resistance: each piece is $R/2$. Step 2. Two $R/2$ pieces in parallel: $R/4$. Step 3. Same mains ⇒ $P \propto 1/R$ ⇒ power quadruples. Step 4. Fixed job ⇒ $t \propto 1/P$: $t = \dfrac{10}{4} = \boxed{2.5\ \text{minutes}}$. Timing note: 20–30 seconds. Every number the slow way invents cancels — the mark scheme wants exactly the chain R → R/4 ⇒ P → 4P ⇒ t → t/4. State each arrow.

Board level — the three-bulb string

Three identical bulbs, each rated 60 W at the mains voltage, are connected in series across that mains. Find the power delivered by each bulb and by the whole string.

The slow way. Decode $R = V^2/60$, triple it, recompute the current and each $I^2R$ — safe but slow, and every step drags $V$ along only to cancel it at the end.

The ratio way. Step 1. Each bulb sees a third of the mains voltage. Step 2. Quadratic scaling: $x = \tfrac13 \Rightarrow$ each bulb delivers $60/9 \approx \boxed{6.7\ \mathrm{W}}$. Step 3. String total: three of those, $60/3 = \boxed{20\ \mathrm{W}}$ — the $P_0/n$ rule read straight off the string theorem. Timing note: 30 seconds. Sanity check the pair: each-bulb $P_0/n^2$ times $n$ bulbs equals total $P_0/n$ ✓. The planted distractor is 20 W per bulb — the total misread as the share.

Competitive level — two coils, one deadline (NEET/JEE-Main pattern, ~75 seconds)

Coil A boils a pot of water in 15 minutes; coil B boils the same pot in 30 minutes, both on the same mains. Find the boiling time with the coils in series, and in parallel. In the parallel case, what fraction of the heat does each coil supply?

Step 1. Series: times add — $t_s = 15 + 30 = \boxed{45\ \text{minutes}}$. Step 2. Parallel: product over sum — $t_p = \dfrac{15 \times 30}{45} = \boxed{10\ \text{minutes}}$. Step 3. The split: powers invert times, $P_A : P_B = 30 : 15 = 2 : 1$, so in the shared 10 minutes coil A delivers $\boxed{\tfrac23}$ of the heat and coil B $\tfrac13$. Step 4. Sanity: coil A alone delivers the whole job in 15 min, so in 10 min it delivers $\tfrac{10}{15} = \tfrac23$ — the two routes agree exactly. Timing note: 60–75 seconds including the check. Nothing was decoded, no voltage was named. The trap answers are 22.5 min (averaging the times) and 20 min (adding rates wrongly); the sanity check in Step 4 kills both for free.

Competitive level — the series mismatch (JEE-Main pattern, ~90 seconds)

A bulb rated 100 W at 230 V and a heater rated 1000 W at 230 V are connected in series across a 230 V supply. Find the power dissipated in each, and state which device runs hotter relative to its rating.

Step 1 — decode. Equation (1): $R_{\text{bulb}} = \dfrac{230^2}{100} = 529\ \Omega$; $R_{\text{heater}} = \dfrac{230^2}{1000} = 52.9\ \Omega$.

Step 2 — the gate. Series ⇒ same current ⇒ $P \propto R$: the bulb takes ten times the heater's power. The verdict arrives before any arithmetic — on a ratio-only MCQ you would bubble now.

Step 3 — price it. $I = \dfrac{230}{529 + 52.9} = \dfrac{230}{581.9} = 0.395\ \mathrm{A}$. Then $P_{\text{bulb}} = I^2 R_{\text{bulb}} = (0.395)^2 \times 529 \approx \boxed{82.6\ \mathrm{W}}$ and $P_{\text{heater}} = (0.395)^2 \times 52.9 \approx \boxed{8.3\ \mathrm{W}}$ — ratio $10 : 1$, as the gate promised.

Step 4 — the reading. The bulb runs at $82.6/100 \approx 83\%$ of rating: near-normal glow. The kilowatt heater runs at $8.3/1000 = 0.83\%$: stone cold. The powerful device lost by the exact factor its own label predicts, because the label priced its resistance low. Timing note: 90 seconds fully priced, 20 seconds for the ratio verdict alone. The favourite wrong answer is 90.9 W apiece from $\frac{P_1P_2}{P_1+P_2}$ misapplied as an each-device figure — that formula (from Bulbs & Ratings — Brightness Logic) gives the series TOTAL: here $\frac{100 \times 1000}{1100} = 90.9$ W, which indeed equals $82.6 + 8.3$ ✓ — a free consistency check, not a share.

Costume without jurisdiction. Writing $P = V^2/R$ with the rated voltage for a device that is no longer across the mains — the number-one killer. The rated voltage belongs to the label; the actual voltage belongs to the circuit. Decode the label to ohms (equation 1) and re-derive everything else from the wiring.

Averaging the two verdicts. In mixed or two-source networks where neither $V$ nor $I$ is shared, equations (2) and (3) are both unlicensed. There is no "splitting the difference" — Redraw the Circuit, solve for the actual current or node voltages, then price the power.

Series total mistaken for a share. $\dfrac{P_1P_2}{P_1+P_2}$ is the series combination's total at rated mains, never what one device gets. The shares follow equation (3): proportional to resistance, inverse to rating.

Linear scaling of sag. A 10% voltage drop is not a 10% power drop — the quadratic gives $0.9^2 = 0.81$, a 19% drop. Boards plant the linear answer every time.

Forgetting what cutting a coil does. Shorter coil, lower resistance, more power on the same mains — counterintuitive enough that "the shorter coil heats less" appears as a distractor annually. $R \propto L$ (Resistivity & Geometry), then $P \propto 1/R$.

Timetable formulas without the fixed-job check. $t_s = t_1 + t_2$ and equation (4) require the same job and the same mains for all configurations. A question that changes the water quantity or the supply between parts voids the shortcut — recompute from $t = H/P$.

Texts. NCERT Physics XII, Ch. 3 (heating effect exercises) · HC Verma, Concepts of Physics Vol. 2, Ch. 33 · NCERT Exemplar XII Ch. 3 for the heater-timetable family in board dress.

Provenance. Fresh module (exam-technique class); the drill families — cut coils, bulb strings, heater timetables, series mismatches — are the standard board and JEE/NEET archive lore, recomputed here from scratch. A verifier should recheck: the decodes $230^2/100 = 529\ \Omega$ and $230^2/1000 = 52.9\ \Omega$; the series mismatch $I = 230/581.9 = 0.395$ A, $P = 82.6$ W and $8.3$ W (ratio 10, total matching $\frac{100\times1000}{1100} = 90.9$ W); the timetable $15, 30 \to 45$ series and $10$ parallel with the $2{:}1$ split; the string $60/9 = 6.7$ W each, $20$ W total; the sag $(0.9)^2 = 0.81$.

Prerequisites. Electrical Energy, Power & Joule Heating · ARCHETYPE: Bulbs & Ratings — Brightness Logic. Companions. TECHNIQUE: Redraw the Circuit — when the gate says "neither shared," the redraw is the lawful next move; ARCHETYPE: Battery Problems — Terminal V, Max Power & Efficiency owns the internal-resistance regime. The full-speed synthesis returns in Circuit Solving at Speed (coming up in the E8 exam-craft band).