ARCHETYPE: Transformer & Line-Loss Numericals

The Transformer module built the machine and The Grid module argued the economics. This archetype is where boards and entrance exams actually cash those two modules: numbers. A transformer question hands you three of the four quantities — turns, voltages, currents, powers — and asks for the fourth; a line-loss question hands you a power station, a resistance and a voltage and asks what survives the journey; and the full exam form chains them: generator → step-up transformer → long line → step-down transformer → consumer, with a question mark planted at every joint. The physics is finished. $V_s/V_p = N_s/N_p$ belongs to The Transformer, the $I^2R$ line-loss chain belongs to The Grid, and $P = VI$ belongs to Electrical Energy, Power & Joule Heating. What this gym adds is the bookkeeping discipline that exam papers are actually testing: currents invert the turns ratio while voltages follow it; efficiency questions must be routed through powers, never patched into the ratio formulas; and — the big one — every segment of a transmission chain carries its own current, and using the wrong segment's current in $I^2R$ is the single most reliable way to lose the marks. The ladder climbs from a two-line board pattern to a JEE-Main transmission chain and a design-to-a-loss-budget stretch. Every number is computed honestly, and every answer is audited by a power balance — the transformer's own conservation law used as a free checking machine.

Key takeaways

Voltages follow the turns ratio, currents invert it: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ but $\dfrac{I_s}{I_p} = \dfrac{N_p}{N_s}$ — the high-voltage side always carries the small current.
The word "ideal" grants $V_pI_p = V_sI_s$ ; the word "efficiency" revokes it — from that moment work only in powers: $P_{\text{out}} = \eta P_{\text{in}}$ .
In a transmission chain, each segment has its own current: line loss is $I_{\text{line}}^2 R$ with $I_{\text{line}} = P/V_{\text{line}}$ — never the generator's or the consumer's current.
Loss fraction on the line is $PR/V^2$ (The Grid's lever, applied): raise the transmission voltage $n$ times, cut the loss $n^2$ times.
Audit everything with a power ledger: watts in = watts out + watts lost, stage by stage, no exceptions.
Key idea: transformer and line-loss numericals are one bookkeeping pipeline — ratios for ideal machines, powers the moment losses appear, and each segment's own current in each segment's $I^2R$ .

The core insight

Every transformer or transmission numerical resolves into a stage-by-stage ledger: apply the turns ratio to voltages (and its inverse to currents) across each ideal transformer, route every efficiency through powers ( $P_{\text{out}} = \eta P_{\text{in}}$ ), compute line loss with the line's own current $I = P/V_{\text{line}}$ , and audit each stage with conservation of energy — the exam's favourite traps all live at the joints between stages.

Where this lives today

Every substation you pass is this archetype running in steel: a 400 kV grid line arrives, a step-down transformer drops it to 33 kV or 11 kV for local distribution, a pole- or pad-mounted transformer makes the final 230 V hop to your wall — each stage a turns ratio someone computed exactly the way the worked ladder does, each line segment budgeted in $I^2R$ before a tower was raised. Utility engineers still run the loss-budget pattern of the stretch example (with power-factor and reactance corrections layered on top — the AC Power module hints at why), and transformer nameplates still quote kVA and efficiency figures that plug straight into Chain 3. When you compute the sagged receiving-end voltage, you are doing by hand what grid operators call voltage regulation — the quantity their control rooms watch all day.

The setup — one line, two machines

Picture the canonical apparatus, because nearly every question in this archetype is a crop of it: a generator makes power at a modest voltage; a step-up transformer trades current for voltage; a long line with unavoidable resistance carries the power; a step-down transformer trades the voltage back down for the consumer. Four segments, and — this is the whole game — three different currents. The generator segment has one current, the line has a much smaller one, the consumer segment has the largest of all. A question can point its finger at any segment, and the number it wants is computed with that segment's voltage and current, nobody else's.

Figure

The canonical chain — four segments, three different currents, one power ledger

Three currents, one ledger. The generator's 300 A never touches the line; the line's slender 10 A is the entire point of the step-up transformer (The Grid module's argument); and the consumer's 295 A only exists after the step-down. Keep this picture pinned while the four chains below are drilled — every trap in this archetype is an attempt to make you carry one segment's current into another segment's formula.

The four chains you must run

Chain 1 — turns to voltage. For any transformer treated as ideal, the volts-per-turn is common property of both windings (The Transformer module's shared-flux law, cited not re-derived):

\frac{V_s}{V_p} = \frac{N_s}{N_p}

where $N_p, N_s$ are primary and secondary turn counts and $V_p, V_s$ the corresponding rms voltages. Reading: more turns, more volts, proportionally — a 1:30 turns ratio is a ×30 voltage ratio, no exceptions while the machine is ideal.

Chain 2 — the current inversion. Ideal means lossless, so $V_pI_p = V_sI_s$ , which forces

\frac{I_s}{I_p} = \frac{N_p}{N_s}

where the ratio has flipped: the winding with more turns (more volts) carries proportionally less current. Physical reading: a transformer is an exchange booth, not a power source — whatever factor it gives you in voltage, it charges you exactly that factor in current. The instant sanity check on any answer: the high-voltage side carries the small current. An answer where volts and amps rise together on the same side has invented energy and is dead on arrival.

Chain 3 — efficiency through powers. The moment a question says "efficiency", abandon the ratio formulas for anything except voltages, and run powers:

\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \qquad\Rightarrow\qquad P_{\text{out}} = \eta\, P_{\text{in}}, \qquad P_{\text{loss}} = (1-\eta)\,P_{\text{in}}

where $P_{\text{in}} = V_pI_p$ and $P_{\text{out}} = V_sI_s$ . Reading: the voltage ratio still follows the turns (the shared flux doesn't care about the losses), but the currents no longer obey the inverted ratio — the primary draws extra current to feed the loss ledger (copper heat, eddy currents, hysteresis — The Transformer module's list). Direction discipline: input is always the bigger power; going from output to input you divide by $\eta$ .

Chain 4 — line loss at the line's current. For a line of total resistance $R$ carrying power $P$ at transmission voltage $V$ (both belonging to the line segment):

I_{\text{line}} = \frac{P}{V} \qquad\Rightarrow\qquad P_{\text{loss}} = I_{\text{line}}^2\, R = \frac{P^2R}{V^2}

where the final form is The Grid module's chain, applied here as a tool. Reading: this is the $1/V^2$ lever — the whole reason the chain contains transformers at all. In this gym you will mostly use the two-step version ( $I$ first, then $I^2R$ ) because multi-part questions ask for the current anyway.

Derivation AThe Ledger of a Two Transformer Line

The goal of this derivation is the master pattern for the full chain question — generator, step-up of ratio $k$ , line of resistance $R$ , step-down, consumer — worked once in symbols so that every numerical becomes substitution. Nothing new is derived: we assemble The Transformer's ratio law, The Grid's loss chain and plain energy conservation into one ledger, and the payoff is knowing which voltage and which current live in each segment. Both transformers are ideal here; the worked ladder shows where a real efficiency slots in.

Step 1. The generator produces power $P$ at voltage $V_g$ . Its segment current is $I_g = P/V_g$ — a big current at a small voltage. Why start with the current: every trap in this archetype is a current mix-up, so the ledger tracks currents explicitly from the first line.

Step 2. The step-up transformer of turns ratio $1:k$ raises the voltage to $V_{\text{line}} = kV_g$ and, being ideal, passes the full power $P$ . Chain 2 forces the line current down to

I_{\text{line}} = \frac{P}{kV_g} = \frac{I_g}{k}

Why this is the design's whole purpose: the factor $k$ bought with turns of wire divides the current that the line will be charged for.

Step 3. The line taxes the flow at its own current: $P_{\text{loss}} = I_{\text{line}}^2 R = P^2R/(kV_g)^2$ . The power arriving at the far end is $P_{\text{arr}} = P - P_{\text{loss}}$ , and the voltage at the far end has sagged by the resistive drop to $V_{\text{arr}} = kV_g - I_{\text{line}}R$ (treating the load as resistive so current and voltage stay in phase — the standard exam idealisation). Why two separate deductions: the line steals both watts (as $I^2R$ heat) and volts (as $IR$ drop); questions ask for either, and they are computed independently from the same $I_{\text{line}}$ .

Step 4. The step-down transformer receives $P_{\text{arr}}$ at $V_{\text{arr}}$ and must deliver the consumer voltage $V_c$ , so its required ratio is $V_{\text{arr}}:V_c$ — computed from the sagged voltage, not the nominal one. The consumer current is $I_c = P_{\text{arr}}/V_c$ , the largest current in the story. Why examiners love this step: the tempting ratio $kV_g:V_c$ ignores the sag and is exactly one mark wrong.

Step 5. Close the ledger: $P = P_{\text{arr}} + P_{\text{loss}}$ , every watt accounted. If a transformer with efficiency $\eta$ replaces an ideal one anywhere in the chain, insert one multiplication ( $P_{\text{after}} = \eta P_{\text{before}}$ ) at that stage and re-run the segments downstream — nothing else changes. Verified: with $P = 66$ kW, $V_g = 220$ V, $k = 30$ , $R = 10\ \Omega$ — $I_g = 300$ A, $I_{\text{line}} = 10$ A, $P_{\text{loss}} = 1000$ W, $P_{\text{arr}} = 65$ kW, $V_{\text{arr}} = 6500$ V, $I_c = 65000/220 \approx 295$ A; ledger closes: $66{,}000 = 65{,}000 + 1000$ ✓ (all values recomputed).

The segment-current discipline

The table that kills the classic error. For the hero chain above ( $66$ kW, $220$ V generation, $1:30$ step-up, $10\ \Omega$ line):

segment	voltage	current	who sets it
generator → step-up	$220$ V	$300$ A	generator power at generator voltage
the line	$6600$ V	$10$ A	step-up ratio — this current pays the $I^2R$ tax
step-down → consumer	$\approx 220$ V	$\approx 295$ A	delivered power at consumer voltage

Three segments, three currents, a factor of thirty between neighbours. Any formula you write must take all three inputs from the same row. The moment a solution multiplies the line's resistance by the consumer's current, the answer is wrong by a factor of $k^2 \approx 900$ — spectacularly, unmissably wrong, which is why examiners can grade it at a glance.

What each knob does

The quick-verdict table for "what happens if…" one-markers, each entry one application of the four chains:

knob	change	effect
transmission voltage $V \to 2V$	line current halves	line loss $\div 4$ (the $1/V^2$ lever)
line resistance $R \to R/2$ (thicker wire)	current unchanged	loss halves — linear, once, at copper's price
demanded power $P \to 2P$ at fixed $V$	line current doubles	loss $\times 4$ ; loss fraction $\times 2$ (lines degrade as demand grows)
turns ratio $k \to 2k$ (same $P$ , $V_g$ )	line voltage doubles	same as row one — the ratio IS the voltage knob
efficiency $\eta$ appears at a stage	powers multiply by $\eta$ there	voltages keep following turns; currents stop obeying the inverted ratio
"ideal" declared	$V_pI_p = V_sI_s$ granted	full ratio laws for both $V$ and $I$ ; ledger has no leak

Figure

The inverse-square lever on the hero chain — same 66 kW, same 10 ohm line, three step-up choices

Worked examples

Board level — the exchange-rate chain, both directions (CBSE/ISC/IGCSE pattern, 3 marks)

A step-down transformer at a substation has 4000 turns on its primary and delivers 230 V to a locality from a 4600 V feeder. The locality draws 11.5 kW. Assuming an ideal transformer, find (a) the number of secondary turns, (b) the secondary current, (c) the primary current.

Step 1. Voltage ratio: $\dfrac{V_s}{V_p} = \dfrac{230}{4600} = \dfrac{1}{20}$ , so $N_s = \dfrac{4000}{20} = \boxed{200\ \text{turns}}$ .

Step 2. Secondary current from delivered power: $I_s = \dfrac{P}{V_s} = \dfrac{11{,}500}{230} = \boxed{50\ \mathrm{A}}$ .

Step 3. Primary current — ideal machine, so invert the ratio: $I_p = I_s \times \dfrac{1}{20} = \boxed{2.5\ \mathrm{A}}$ . Audit: $V_pI_p = 4600 \times 2.5 = 11{,}500$ W $= V_sI_s = 230 \times 50$ ✓. High-voltage side, small current — the sanity direction holds.

Board level — efficiency, run through powers (ISC/JEE-Main pattern, 3 marks)

A step-up transformer with turns ratio 1:5 is fed 240 V and draws 15 A from the supply. Its efficiency at this load is 88%. Find (a) the secondary voltage, (b) the power delivered by the secondary, (c) the secondary current, (d) the power lost inside the machine.

Step 1. Voltage still follows turns even in a real machine: $V_s = 240 \times 5 = \boxed{1200\ \mathrm{V}}$ .

Step 2. Input power: $P_{\text{in}} = V_pI_p = 240 \times 15 = 3600$ W. Output: $P_{\text{out}} = \eta P_{\text{in}} = 0.88 \times 3600 = \boxed{3168\ \mathrm{W}}$ .

Step 3. Secondary current from its own power and voltage: $I_s = \dfrac{3168}{1200} = \boxed{2.64\ \mathrm{A}}$ . Note what the inversion would have wrongly given: $15/5 = 3$ A is the ideal secondary current; the real machine delivers less because 432 W never make it across.

Step 4. Loss: $P_{\text{in}} - P_{\text{out}} = 3600 - 3168 = \boxed{432\ \mathrm{W}}$ — spent on the loss ledger (copper $I^2R$ , eddy currents, hysteresis; The Transformer module's table).

Competitive level — the full transmission chain (JEE-Main/NEET pattern, ~3 minutes)

A generator produces 66 kW at 220 V. The voltage is stepped up by an ideal 1:30 transformer and transmitted over a line of total resistance 10 Ω to an ideal step-down transformer that delivers power at 220 V. Find (a) the line current, (b) the power lost in the line and its percentage, (c) the voltage at the input of the step-down transformer, (d) the approximate turns ratio the step-down transformer must have.

Step 1. Line voltage $= 220 \times 30 = 6600$ V, so the line's own current is $I = \dfrac{66{,}000}{6600} = \boxed{10\ \mathrm{A}}$ (not the generator's $300$ A — segment discipline).

Step 2. Line loss: $I^2R = 10^2 \times 10 = \boxed{1000\ \mathrm{W}}$ , i.e. $\dfrac{1000}{66{,}000} \approx \boxed{1.5\%}$ . Power arriving: $65$ kW.

Step 3. Resistive drop along the line: $IR = 10 \times 10 = 100$ V, so the step-down primary sees $6600 - 100 = \boxed{6500\ \mathrm{V}}$ .

Step 4. Required ratio: $6500 : 220 \approx \boxed{29.5 : 1}$ — computed from the sagged 6500 V. The tempting "30:1 to undo the step-up" ignores the line and drops the mark. Timing note: about 3 minutes. The entire question is the ledger derivation with numbers substituted — if you drilled the ledger, each part is one line. Budget your time on part (d), the only step where a wrong voltage can sneak in.

Competitive level — design to a loss budget (JEE-Main stretch, ~4 minutes)

A plant must send 80 kW from its 250 V generator down a line of total resistance 5.0 Ω, keeping the line loss within 2.0% of the generated power. (a) Find the maximum allowed line current. (b) Find the minimum transmission voltage and the minimum integer step-up turns ratio. (c) With that integer ratio, verify the actual loss percentage. (d) If the step-up transformer is only 96% efficient (generator output unchanged), find the new line loss and the overall fraction of generated power that survives to the far end of the line.

Step 1. Loss budget: $0.020 \times 80{,}000 = 1600$ W. From $I^2 \times 5.0 \leq 1600$ : $I \leq \sqrt{320} = \boxed{17.9\ \mathrm{A}}$ .

Step 2. Minimum voltage: $V \geq \dfrac{80{,}000}{17.9} = 4472$ V, so the ratio must be at least $4472/250 = 17.9$ — the minimum integer ratio is $\boxed{1:18}$ .

Step 3. At $1:18$ : $V = 4500$ V, $I = \dfrac{80{,}000}{4500} = 17.78$ A, loss $= (17.78)^2 \times 5.0 = 1580$ W $= \boxed{1.98\%}$ ✓ — inside budget, barely; integer rounding up was the safe direction.

Step 4. Real step-up at 96%: the line now receives $0.96 \times 80{,}000 = 76.8$ kW at 4500 V. New line current $= \dfrac{76{,}800}{4500} = 17.07$ A; new line loss $= (17.07)^2 \times 5.0 \approx 1456$ W. Power surviving the line: $76{,}800 - 1456 = 75{,}344$ W — an overall $\dfrac{75{,}344}{80{,}000} \approx \boxed{94.2\%}$ of generation. Ledger: $80{,}000 = 3200\ (\text{transformer}) + 1456\ (\text{line}) + 75{,}344\ (\text{delivered})$ ✓. Timing note: about 4 minutes. Parts (a)–(c) are the loss-budget pattern: budget → current ceiling → voltage floor → integer ratio → verify. Part (d) is one $\eta$ -multiplication inserted at the right stage and the downstream segments re-run — the ledger derivation's Step 5, verbatim. Chains of efficiencies multiply: a 96% machine into a 98% line is $0.96 \times 0.98$ , never a subtraction of percentages.

Exam traps

Segment discipline above all. Line loss uses the line's current $I = P/V_{\text{line}}$ . Importing the generator's or consumer's current inflates the answer by $k^2$ — the error the whole question was built to catch.

Currents invert; efficiency breaks the inversion. Ideal: $I_s/I_p = N_p/N_s$ . Real: the voltage ratio survives, the current ratio doesn't — route currents through their own segment's power ( $I = P/V$ ).

Divide by η going upstream. $P_{\text{in}} = P_{\text{out}}/\eta$ : input is always the bigger number. Multiplying instead of dividing is the priced-in error; a one-second size check catches it.

The sagged voltage. The receiving transformer's ratio is computed from $V_{\text{line}} - IR$ , not from the nominal transmission voltage. Boards award a specific mark for this subtraction.

Percentages of what? Loss percentage is usually of generated power; delivered fraction is of generated power too. When efficiencies chain, multiply fractions ( $0.96 \times 0.98$ ), never add or subtract percentage points.

Round the ratio the safe way. A loss ceiling means the voltage is a floor: round turns ratios up. Rounding 17.9 down to 17 blows the budget and the mark.

Boards and depth. CBSE and NEET: the syllabus line "AC generator and transformer" supports the turns-ratio and efficiency patterns (the first three rungs); full grid-chain design is ISC/JEE polish. ISC 861 names transformer "efficiency and applications including transmission of power, energy losses and their minimisation" — this gym is that clause as numericals. IGCSE tests the turns-ratio and $I^2R$ -loss calculations at formula level ( $V_pI_p = V_sI_s$ given). JEE-Main sets the chain and budget patterns; JEE-Advanced's syllabus stops at LCR circuits and does not list the transformer, and A-Level 9702 leaves transformers to IGCSE as assumed background — neither is tagged. AP Physics 2 has no AC unit and IB dropped AC in 2023: for those boards this module is how-the-grid-works reading, not exam material.

Origins · 1 minOttó Titusz Bláthy and how this was discovered

Ottó Titusz Bláthy

1860–1939 · Ganz Works, Budapest

Bláthy was the calculating engineer of the trio — Zipernowsky, Bláthy, Déri — who in 1885 gave the transformer both its modern shape and its name. The induction "secondary generators" then being demonstrated across Europe had open iron cores and were strung in series, so every added lamp dimmed its neighbours; the ZBD team at Budapest's Ganz Works closed the magnetic core into an unbroken iron ring around the windings, coined the word transformer, and — Bláthy's particular insistence — connected the machines in parallel across a constant-voltage line. That last decision is why this module's arithmetic exists: in a parallel grid, every transformer sees a fixed primary voltage, the volts-per-turn law becomes dependable bookkeeping, and a substation can be designed on paper before a single coil is wound. Bláthy spent the next half-century doing exactly that kind of paper-first engineering — he calculated machines the way this gym asks you to, ratio by ratio and watt by watt, and his induction meters ended up counting the energy that ZBD-style grids delivered. The Great Barrington demonstration (The Transformer module's story) and Deptford's high-voltage gamble (The Grid's) both ran on the architecture this Hungarian trio proved first: closed core, parallel connection, arithmetic you can trust.

History

1884

— Gaulard and Gibbs demonstrate series AC distribution with open-core "secondary generators"; loads interact, regulation is poor

1885

— Zipernowsky, Bláthy and Déri patent the closed-core, parallel-connected transformer at Ganz Works and name the device

1885

— the ZBD system lights 1067 lamps at the Budapest National Exhibition from a single constant-voltage network

1886

— Stanley's Great Barrington demonstration carries the closed-core idea to America under Westinghouse

1889

— Bláthy's induction watt-hour meter begins metering AC energy — the grid learns to send bills

1891

— Lauffen to Frankfurt: 175 km at 15,000 V settles the high-voltage transmission argument for good

Going further

The two source modules are the theory behind every line here: The Transformer (shared flux, turns ratio, the losses ledger, kVA nameplates) and The Grid: Why Transmit at High Voltage (the $P^2R/V^2$ chain and the two-voltage audit). The power bookkeeping rides on Electrical Energy, Power & Joule Heating, and the rms convention on AC Signals: Peak, Average & RMS. For the craft of AC option-killing at speed — limits, phase reads, resonance spotting — the next module, Phasor & Limit Shortcuts in AC, closes the AC chapter; the track's final exam-craft pair (Circuit Solving at Speed and Option Elimination in E&M) drills this gym's patterns among its templates. A deeper field-level treatment of transformer action lives in the undergraduate article on transformers (UG deep dive). Provenance: fresh module per the HS map; worked numbers computed for this article and audited by power ledger; transformer and grid support narrative mined from the legacy HS rows on transformers and national-grid power transmission.

Why does the primary current increase when the transformer is loaded — how does the primary "know"?

Through the shared flux. With the secondary open, the primary draws only a tiny magnetising current. Close the secondary circuit and its current sets up a flux that opposes the core's flux (Lenz's Law); the primary must draw extra current to restore it, and the books balance at exactly $V_pI_p = V_sI_s$ for an ideal machine. So the "knowledge" travels through the core as flux bookkeeping — load the output and the input current rises within a cycle. The Transformer module tells this story qualitatively; in this gym you use its ledger form: powers match, currents invert.

If currents invert the ratio only for ideal transformers, what do I do with currents in a real one?

Never touch the inverted ratio once "efficiency" appears. Compute each winding's current from its own segment: $I_p = P_{\text{in}}/V_p$ and $I_s = P_{\text{out}}/V_s$ , with $P_{\text{out}} = \eta P_{\text{in}}$ linking them. The voltage ratio still follows the turns (flux doesn't care about losses), so voltages are safe; it is only the current inversion that breaks. In the worked example the ideal inversion predicted 3 A but the real machine delivered 2.64 A — that gap is the 432 W loss, made visible.

Why is line loss P²R/V² and not V²/R? Both formulas came from the power chapter.

Because the line is in series with everything else, and what is fixed about the line is the power flowing through it, not the voltage across it. The voltage across the line is just its small resistive drop ( $100$ V in the chain example), not the 6600 V transmission level. $V^2/R$ with the transmission voltage would describe a line short-circuiting the supply — a catastrophe, not a grid. The safe route is always the two-step: $I = P/V_{\text{line}}$ first, then $I^2R$ . The Grid module derives the lever; this gym applies it dozens of times.

Where do the "lost" watts actually go in each part of the chain?

Two different drains. Inside a transformer: copper heat ( $I^2R$ in the windings), eddy-current heat in the core (fought with lamination), hysteresis heat (fought with soft iron) and a little flux leakage — The Transformer module's ledger. Along the line: pure $I^2R$ in the conductors, warming a few hundred kilometres of countryside. Both drains end as heat; the ledger in this gym just tracks where the subtraction happens, because questions plant their marks at specific stages.

Why insist on rounding the turns ratio up in the design problem — 17.9 is nearly 18 anyway?

Because the constraint has a direction. The loss budget caps the current, which puts a floor under the voltage, which puts a floor under the ratio. Rounding 17.9 down to 17 gives $V = 4250$ V, $I = 18.8$ A, loss $= 1771$ W $= 2.2\%$ — over budget, question failed. Rounding up to 18 lands at 1.98%, inside. In any design-to-a-limit question, check which side of the fence each rounding lands you on; examiners choose numbers (as here) where the lazy rounding fails.

My two answers for part (d) of the chain differ: 30:1 versus 29.5:1 — which is right and why so picky?

The 29.5:1. The step-down transformer's primary is connected at the far end of the line, where the voltage has sagged to $6600 - IR = 6500$ V; designing for 6600 V would deliver $220 \times 6600/6500 \approx 223$ V instead of 220. On a board paper that's the difference between full marks and a "correct method" partial; in the real grid that 1.5% is what regulation engineering exists to manage. The general lesson of this archetype: always ask which physical point in the chain your numbers describe.