ARCHETYPE: Battery Problems — Terminal V, Max Power & Efficiency

In 1839 a crowd on the banks of the Neva in St Petersburg watched a boat carry fourteen passengers upriver with no sails and no steam — just a stack of battery cells and Moritz von Jacobi's electric motor. Jacobi then did what every honest engineer does after a triumph: the accounting. However he arranged his cells, the motor refused to deliver more than a fixed ceiling of power, and at that ceiling exactly half the batteries' energy heated the batteries themselves. He had discovered, forty years before anyone could explain a battery's insides, that a real source has a built-in speed limit — and that hitting it costs you half of everything. That accounting is now one of the most reliable point-scorers in circuit examinations. Boards ask you to run the ledger: given an emf and an internal resistance, find the terminal voltage, the current, the useful power, the wasted power. JEE and NEET sharpen it: which load extracts maximum power, what efficiency do you pay there, and what happens when a charger forces current backwards through the battery so the familiar minus sign flips. Every one of these is the same five-line pattern — one loop equation, one voltage walk, one power audit. The concepts here belong to their own modules: EMF, Terminal Voltage & Internal Resistance builds the source model, and Electrical Energy, Power & Joule Heating builds the $P = VI$ family. This archetype is the gym where those tools get drilled into exam speed: the standard battery-problem setups, the maximum-power theorem proved with pure algebra, the efficiency fine print, and the charging variant that catches thousands of students every year.

Key takeaways

Every battery problem is one loop: $I = \dfrac{\varepsilon}{R + r}$ , then the terminal voltage $V = \varepsilon - Ir$ while discharging.
The power ledger always balances: $\varepsilon I = I^2R + I^2r$ — chemistry's output splits into load power and internal heat.
Maximum power reaches the load when $R = r$ , giving $P_{\max} = \dfrac{\varepsilon^2}{4r}$ — proved by completing the square, no calculus needed.
At maximum power the efficiency $\eta = \dfrac{R}{R+r}$ is exactly $50\%$ : half the energy cooks the battery. That is why power grids deliberately run $R \gg r$ .
Charging flips the sign: current is forced backwards, so the terminals must sit above the emf, $V = \varepsilon + Ir$ .
KEY IDEA: solve the loop for $I$ , walk the loop for $V$ , audit the powers — and remember maximum power and maximum efficiency are different questions with different answers.

The core insight

A real battery is an emf $\varepsilon$ in series with an internal resistance $r$ ; one loop equation gives the current, the walk across the terminals gives $V = \varepsilon \mp Ir$ (minus discharging, plus charging), and the ledger $\varepsilon I = I^2R + I^2r$ settles every power and efficiency question — with the load-matching peak $R = r$ costing exactly half the energy as internal heat.

Where this lives today

Your phone's "battery health" percentage is largely a report on $r$ : as a lithium cell ages its internal resistance creeps up, and the terminal voltage sags harder under the same load. Electric-vehicle packs run this archetype thousands of times a second — the management system pulses a known current, reads the sag, extracts $r$ per cell group, and caps the fast-charge rate, because the $I^2r$ term in the charging ledger is pure heat, and heat kills cells. And the theorem's twin regimes still split the electrical world: headphone amplifiers impedance-match to squeeze signal power across, while the grid feeding them sits as far from the match as copper allows.

The setup

Draw it before anything else, because in this archetype the circuit is the physics. A battery of emf $\varepsilon$ and internal resistance $r$ drives a current $I$ around a single loop through an external resistance $R$ . The dashed box in the figure is the battery case: everything inside it — the ideal pump $\varepsilon$ and the resistance $r$ — travels together, and the only voltage you can ever measure is the one between the terminals $A$ and $B$ where the case ends.

Figure

The battery problem — emf and internal resistance inside the case, load outside, one current around the loop

The module EMF, Terminal Voltage & Internal Resistance explains why the model looks like this — the chemistry fixes $\varepsilon$ , the electrolyte and plates supply $r$ . Here we take the model and race it. Two readings orient every problem:

The loop equation. Each coulomb is lifted $\varepsilon$ joules by the chemistry and must spend all of it crossing the two resistances in its path:

I = \frac{\varepsilon}{R + r} \qquad (1)

where: $I$ — the one current in the loop; $R + r$ — the total resistance, external plus internal. Physically: the battery's own body is in series with everything you connect. Forgetting the $+\,r$ in the denominator is the single most common error in this archetype.

The terminal walk. Walk from terminal $B$ to terminal $A$ through the inside of the battery: up $\varepsilon$ through the pump, down $Ir$ through the internal resistance. What survives at the terminals is

V = \varepsilon - Ir \qquad (2)

where: $V$ — the terminal voltage, the only thing a voltmeter can see. The battery keeps $Ir$ for itself — the "lost volts" — and hands the rest to the outside. Draw no current and the loss vanishes: an idle battery shows its full emf.

The master pattern

Every battery problem — board or JEE — yields to the same four moves.

Step 1. Draw the loop with $r$ drawn explicitly as a series resistor inside the battery case. Half the errors in this archetype die at this step.

Step 2. Find the current: from equation (1) if you know the resistances, or from given meter readings.

Step 3. Walk the loop for whatever voltage is asked: $V = \varepsilon - Ir$ across the terminals while discharging.

Step 4. Audit the powers. Multiply equation (2) by $I$ :

\varepsilon I = I^2 R + I^2 r \qquad (3)

where: $\varepsilon I$ — the chemistry's total spend; $I^2R = VI$ — the payload delivered to the load; $I^2 r$ — the internal heating. This ledger must balance to the last watt, which makes it a free self-check: if your three powers don't add up, a number upstream is wrong. The formulas themselves come from Electrical Energy, Power & Joule Heating — this archetype is where you learn to close the books with them. The efficiency of the delivery follows immediately by dividing payload by spend:

\eta = \frac{I^2R}{\varepsilon I} = \frac{R}{R + r} \qquad (4)

where: $\eta$ — the fraction of chemical energy that escapes the case. Read the structure: $\eta \to 1$ when $R \gg r$ (light load, near-ideal delivery) and $\eta \to 0$ as $R \to 0$ (short circuit: everything burns inside). Efficiency always rises with $R$ . Hold that thought — it is about to collide with the power question.

Maximum power — the peak and its price

Ask a different question: not "how efficiently," but "how much?" Which load $R$ pulls the largest wattage out of a given battery? Substituting equation (1) into $P = I^2R$ :

P(R) = \frac{\varepsilon^2 R}{(R + r)^2} \qquad (5)

where: $P(R)$ — the power landing in the load, as a function of the load itself. Both extremes fail: $R \to 0$ sends the current soaring but the payload share to zero, and $R \to \infty$ chokes the current entirely. Somewhere between sits a peak — and at high-school level we can find it exactly, with algebra alone.

Derivation AMaximum power transfer by completing the square

We want the load resistance that maximises $P(R) = \varepsilon^2 R/(R+r)^2$ , and the value of the maximum, using nothing beyond algebra — the trick is to flip the fraction and complete a square in $\sqrt{R}$ . This is the proof CBSE and ISC accept and the one that works when calculus is off the table. The plan: minimising the denominator-over-numerator is the same as maximising $P$ , and the flipped expression splits into a perfect square plus a constant.

Step 1. Flip the problem. Since $\varepsilon^2$ is fixed, $P$ is maximum exactly when $\dfrac{(R+r)^2}{R}$ is minimum. Expand it:

\frac{(R+r)^2}{R} = \frac{R^2 + 2Rr + r^2}{R} = R + \frac{r^2}{R} + 2r

Why flip: the awkward part of $P(R)$ is the squared denominator; flipping turns the question into minimising a simple sum.

Step 2. Complete the square in $\sqrt{R}$ . Notice that

R + \frac{r^2}{R} = \left(\sqrt{R} - \frac{r}{\sqrt{R}}\right)^2 + 2r

because expanding the square gives $R - 2r + r^2/R$ , and adding back $2r$ restores the left side exactly.

Step 3. Read off the minimum. A squared quantity is never negative, so

\frac{(R+r)^2}{R} = \left(\sqrt{R} - \frac{r}{\sqrt{R}}\right)^2 + 4r \;\ge\; 4r

with equality exactly when the bracket vanishes: $\sqrt{R} = r/\sqrt{R}$ , i.e. $R = r$ . Why this is the physics: below $r$ the internal toll dominates; above $r$ the current starves. The peak balances the two losses — at the match, load and battery dissipate equally.

Step 4. The maximum power follows at once:

P_{\max} = \frac{\varepsilon^2}{4r}

Step 5. Audit the price. At $R = r$ , equation (4) gives $\eta = r/(r+r) = \tfrac{1}{2}$ . Exactly half the chemistry's output heats the battery at the moment of maximum delivery — the famous 50% fine print. Verified: for ε = 6 V, r = 0.5 Ω the formula gives P_max = 36/2 = 18 W at R = 0.5 Ω; direct substitution into equation (5) at R = 0.4 and 0.6 Ω gives 17.8 W both sides of the peak — recomputed numerically.

Figure

Load power peaks at R equals r but efficiency keeps climbing — the two curves answer different questions

Two exam-grade consequences hide in that figure. First, the peak is blunt and symmetric in a special way: any load $R$ and its partner $r^2/R$ deliver exactly equal power — check with equation (5): swapping $R \to r^2/R$ leaves $\varepsilon^2 R/(R+r)^2$ unchanged after clearing fractions. Multiple-choice papers exploit this by offering two resistances that give the same power and asking for $r$ ; the answer is $r = \sqrt{R_1 R_2}$ . Second, the peak is where power engineering refuses to operate. A grid running matched would waste half of every power station inside the station; real grids run $R \gg r$ , far up the green curve, at efficiencies above 90%. Matching is for the opposite regime — weak signal sources where the watts are precious and the waste is trivial. (One scope note: the maximum-power theorem itself is JEE/A-Level territory — the AP Physics 2 exam uses the non-ideal battery $V = \varepsilon - Ir$ but does not test the $R = r$ optimisation.)

The charging variant — the sign that flips

Connect a charger — or an alternator — across the battery and force the current backwards, into the positive terminal. The pump now works against the traffic, and so does the toll booth. Walk the loop again and the internal drop adds instead of subtracting:

V_{\text{charging}} = \varepsilon + Ir \qquad (6)

where: $V_{\text{charging}}$ — the terminal voltage during charging, now above the emf. It must be: the charger has to overpower the battery's own pump and push the current through $r$ . This is why a car's electrical system sits near 14 V while the engine runs, even though the battery's emf is only about 12.6 V — the extra volts are the $Ir$ headroom for the charging current.

Go deeper · Derivation B · 1 minThe charging battery and its energy ledger

Derivation BThe charging battery and its energy ledger

We want the terminal voltage and the complete power accounting for a battery of emf $\varepsilon$ and internal resistance $r$ being charged at current $I$ , to see exactly where the charger's energy goes and why $V$ exceeds $\varepsilon$ . The method is the same loop walk as discharge with one physical change — the current direction — and the ledger will split the charger's input into stored chemical energy plus internal heat. Keep the sign convention honest: we walk across the battery in the direction the charging current actually flows.

Step 1. Walk from the − terminal to the + terminal along the charging current's path inside the battery. The current now enters at + and is forced through the pump the wrong way: crossing the emf costs the charger $\varepsilon$ per coulomb (energy banked into chemistry), and crossing $r$ costs a further $Ir$ per coulomb (energy lost as heat).

Step 2. Add the two costs. The charger must supply, per coulomb, the sum:

V = \varepsilon + Ir

Why the sign flipped: in discharge the coulomb collects ε from the pump and pays Ir; in charge it pays both. Whoever drives the current pays the toll — the toll itself never changes direction preference, it just charges whoever passes.

Step 3. Multiply by $I$ to get the power ledger of charging:

VI = \varepsilon I + I^2 r

The charger's full input $VI$ splits into $\varepsilon I$ — the rate at which energy is banked as chemical energy — and $I^2r$ , the unavoidable heating of the battery's own body.

Step 4. Run the standard numbers: $\varepsilon = 12\ \mathrm{V}$ , $r = 0.25\ \Omega$ , charging at $I = 8\ \mathrm{A}$ . Terminal voltage $V = 12 + 8(0.25) = 14\ \mathrm{V}$ . Input $VI = 112\ \mathrm{W}$ ; banked $\varepsilon I = 96\ \mathrm{W}$ ; heat $I^2r = 16\ \mathrm{W}$ . The books close: $96 + 16 = 112$ . Verified: ledger recomputed numerically — 14 V terminals, 112 = 96 + 16 W; heating fraction 16/112 ≈ 14%, which is why fast-charging batteries need thermal management.

The variants examiners actually pose

situation	governing relations	what to watch
discharging into $R$	$I = \dfrac{\varepsilon}{R+r}$ , $V = \varepsilon - Ir$	$V < \varepsilon$ always, gap grows with $I$
open circuit ( $I = 0$ )	$V = \varepsilon$	the only time a voltmeter reads the emf
short circuit ( $R = 0$ )	$I_{sc} = \dfrac{\varepsilon}{r}$ , $V = 0$	the battery's absolute current ceiling
maximum power	$R = r$ , $P_{\max} = \dfrac{\varepsilon^2}{4r}$	efficiency exactly 50% there
twin loads, equal power	$R_1 R_2 = r^2$	MCQ favourite: $r = \sqrt{R_1R_2}$
charging at $I$	$V = \varepsilon + Ir$	terminals sit above the emf
efficiency	$\eta = \dfrac{R}{R+r}$	rises with $R$ forever — no peak

Worked examples

Board level 1 — the full ledger of a loaded battery (CBSE/ISC pattern)

A battery of emf $12\ \mathrm{V}$ and internal resistance $0.5\ \Omega$ is connected to a $5.5\ \Omega$ resistor. Find the current, the terminal voltage, the power delivered to the resistor, the power wasted inside the battery, and the efficiency.

Step 1. The loop: $I = \dfrac{\varepsilon}{R + r} = \dfrac{12}{5.5 + 0.5} = \boxed{2.0\ \mathrm{A}}$ .

Step 2. The terminal walk: $V = \varepsilon - Ir = 12 - 2.0(0.5) = \boxed{11\ \mathrm{V}}$ . (Cross-check: $V = IR = 2.0 \times 5.5 = 11$ V ✓ — the terminal voltage is the load voltage in a single loop.)

Step 3. Powers: load $P_R = I^2R = 4 \times 5.5 = \boxed{22\ \mathrm{W}}$ ; internal $P_r = I^2r = 4 \times 0.5 = 2.0\ \mathrm{W}$ ; total $\varepsilon I = 24\ \mathrm{W}$ . Ledger: $22 + 2 = 24$ ✓.

Step 4. Efficiency: $\eta = \dfrac{R}{R+r} = \dfrac{5.5}{6.0} = \boxed{91.7\%}$ — a healthy, deliberately unmatched load.

Board level 2 — measuring a battery you cannot open (ISC/A-Level pattern)

A cell drives $0.10\ \mathrm{A}$ through one load and its terminal voltage reads $1.45\ \mathrm{V}$ ; with a heavier load it drives $0.40\ \mathrm{A}$ at $1.30\ \mathrm{V}$ . Find the internal resistance, the emf, and the short-circuit current.

Step 1. Write $V = \varepsilon - Ir$ at both operating points: $1.45 = \varepsilon - 0.10\,r$ and $1.30 = \varepsilon - 0.40\,r$ .

Step 2. Subtract — the emf cancels: $0.15 = 0.30\,r \Rightarrow r = \boxed{0.50\ \Omega}$ . The internal resistance is the slope of terminal voltage against current.

Step 3. Back-substitute: $\varepsilon = 1.45 + 0.10(0.50) = \boxed{1.5\ \mathrm{V}}$ — the AA cell on its label, recovered from loaded readings alone.

Step 4. Short-circuit ceiling: $I_{sc} = \varepsilon/r = 1.5/0.5 = \boxed{3.0\ \mathrm{A}}$ . Sanity: both given currents are far below the ceiling ✓.

Competitive level 1 — the maximum-power question, with the twin-load trap (JEE-Main pattern)

A source of emf $6.0\ \mathrm{V}$ and internal resistance $0.5\ \Omega$ can be connected to a variable load. (a) Find the maximum power it can deliver to any load and the efficiency at that point. (b) A student finds that loads of $0.25\ \Omega$ and $1.0\ \Omega$ deliver the same power. Show this is consistent, and find that power.

Step 1. Maximum power at the match $R = r = 0.5\ \Omega$ : $P_{\max} = \dfrac{\varepsilon^2}{4r} = \dfrac{36}{2} = \boxed{18\ \mathrm{W}}$ , at efficiency exactly $\boxed{50\%}$ .

Step 2. Twin-load check: equal power requires $R_1R_2 = r^2$ . Here $0.25 \times 1.0 = 0.25 = (0.5)^2$ ✓ — the two loads sit symmetrically astride the peak.

Step 3. The shared power, from either load: $P = \dfrac{\varepsilon^2 R}{(R+r)^2} = \dfrac{36 \times 1.0}{(1.5)^2} = \boxed{16\ \mathrm{W}}$ . (Check with $R = 0.25$ : $36 \times 0.25/(0.75)^2 = 16$ W ✓.)

Step 4. Note the MCQ inversion: given two equal-power loads, the internal resistance is $r = \sqrt{R_1R_2}$ — here $\sqrt{0.25} = 0.5\ \Omega$ , recovering the given value. Timing note: parts like (a) are 30-second lookups if $P_{\max} = \varepsilon^2/4r$ is at your fingertips; budget the saved time for (b), and remember $r = \sqrt{R_1R_2}$ rather than solving the quadratic live. Two minutes total.

Competitive level 2 — charging a battery, the reversed-sign audit (NEET/JEE-Advanced pattern)

A storage battery of emf $12\ \mathrm{V}$ and internal resistance $0.25\ \Omega$ is charged by a supply that drives $8.0\ \mathrm{A}$ into it. Find (a) the battery's terminal voltage during charging, (b) the power drawn from the supply at the terminals, and (c) how that power splits between stored chemical energy and heat.

Step 1. Charging walk — the drop across $r$ now adds: $V = \varepsilon + Ir = 12 + 8.0(0.25) = \boxed{14\ \mathrm{V}}$ . The terminals sit above the emf, as they must for current to be forced backwards.

Step 2. Power entering the battery at its terminals: $P_{\text{in}} = VI = 14 \times 8.0 = \boxed{112\ \mathrm{W}}$ .

Step 3. The split: banked chemically at the pump, $\varepsilon I = 12 \times 8 = \boxed{96\ \mathrm{W}}$ ; lost as internal heat, $I^2r = 64 \times 0.25 = \boxed{16\ \mathrm{W}}$ . Ledger: $96 + 16 = 112$ ✓.

Step 4. Physical read: about 14% of the charger's energy never becomes stored charge — it warms the battery, which is why fast charging is a thermal problem and why the charging current, not the voltage, is what battery-management systems limit. Timing note: the entire problem is one sign flip — write $V = \varepsilon + Ir$ the moment you read "charged," and the three powers fall out in under two minutes. If your ledger refuses to balance, you almost certainly kept the discharge minus sign.

Exam traps

The minus sign that should be a plus. "A battery is being charged…" and the answer $V = \varepsilon - Ir$ is already wrong. Discharge: terminals below emf. Charge: terminals above emf, $V = \varepsilon + Ir$ . Decide which way the current flows through the battery before writing anything.

Dropping $r$ from the denominator. $I = \varepsilon/R$ is an ideal-battery formula. The moment a question grants the battery an internal resistance, the loop current is $\varepsilon/(R+r)$ — and every downstream number inherits the error if you forget.

$P = \varepsilon I$ versus $P = VI$ . $\varepsilon I$ is the chemistry's total spend; $VI$ is what reaches the outside world. Using the emf where the terminal voltage belongs overstates the delivered power by exactly the internal loss. When in doubt, run the full ledger $\varepsilon I = I^2R + I^2r$ and let it balance.

Maximum power ≠ maximum efficiency. The peak of $P(R)$ is at $R = r$ with $\eta = 50\%$ ; the efficiency $\eta = R/(R+r)$ has no peak — it climbs forever. Questions that ask "for efficient transfer" want $R \gg r$ , not $R = r$ .

The twin-load shortcut. Two loads with equal power satisfy $R_1R_2 = r^2$ : offered $1\ \Omega$ and $4\ \Omega$ , then $r = 2\ \Omega$ — no quadratic required.

Board conventions. CBSE and ISC expect the completing-the-square (or verification-by-substitution) proof of $R = r$ — quoting the result bare loses method marks on a "show that" question. AP Physics 2 tests $V = \varepsilon - Ir$ and energy reasoning but not the maximum-power theorem; JEE-Main, JEE-Advanced, NEET and A-Level all consider $P_{\max} = \varepsilon^2/4r$ fair game. Where mains voltage enters battery-charger problems, Indian and UK papers use 230 V.

Calculator habit. Keep the current unrounded before squaring — $I^2r$ doubles any rounding error. Square first, round last.

Origins · 1 minMoritz von Jacobi and how this was discovered

Moritz von Jacobi

1801–1874 · St Petersburg Academy of Sciences

Moritz Hermann von Jacobi — engineer brother of the mathematician Carl Jacobi, and easy to confuse with him on a fast exam read — built the first practical electric motor in 1834 and, in 1839, mounted it in a 28-foot boat that carried fourteen passengers along the Neva against the current, powered by hundreds of battery cells. Auditing why the boat was so costly to run, he formulated what is still called Jacobi's law: a source of internal resistance $r$ delivers its maximum power to a matched load $R = r$ , and does so at only 50% efficiency. The result was initially read as a death sentence for electric power — engineers concluded half of all battery energy must always be wasted — until it was understood that the theorem is a statement about maximum extraction, not a mandate: run unmatched, $R \gg r$ , and efficiency is yours. Jacobi also invented electrotyping, the first industrial process of electrochemistry, making him one of the few people to have both a fundamental theorem and a factory technique to his name.

History

1800

— Volta's pile gives physics its first steady source, complete with an unsuspected internal resistance

1834

— Jacobi builds the first rotating electric motor powerful enough to do useful work

1839

— Jacobi's electric boat carries fourteen passengers on the Neva; the battery bill prompts the power audit

1840

— Jacobi's law: maximum power to a matched load, at exactly 50% efficiency

1859

— Planté's lead–acid cell makes the battery rechargeable — and makes the charging equation V = ε + Ir an engineering daily

1882

— Edison's Pearl Street station designs generators for minimum internal resistance, running deliberately far from the match

2010

s — electric-vehicle battery management systems track internal resistance live, cell by cell, as the primary health signal

Going further

Texts. NCERT Physics Class 12, Ch. 3 (Current Electricity) · HC Verma, Concepts of Physics Vol. 2, Ch. 32 · CIE 9702 topic 10 (D.C. circuits).

Provenance. All results are standard exam canon; every number recomputed: Board 1 ledger ( $2.0$ A, $11$ V, $22 + 2 = 24$ W, $\eta = 91.7\%$ ); Board 2 slope method ( $r = 0.15/0.30 = 0.50\ \Omega$ , $\varepsilon = 1.5$ V, $I_{sc} = 3.0$ A); Competitive 1 ( $P_{\max} = 18$ W, twins $0.25/1.0\ \Omega$ both $16$ W); Competitive 2 charging ledger ( $14$ V, $112 = 96 + 16$ W).

Prerequisites. EMF, Terminal Voltage & Internal Resistance · Electrical Energy, Power & Joule Heating. Peers. Combining Cells: Series & Parallel (stack the sources, then run this archetype's ledger on the stack) · Kirchhoff's Rules (when one loop becomes several). Leads to. Bulbs & Ratings — Brightness Logic (the power ledger goes shopping). Deep dive. EMF and Internal Resistance (UG deep dive) — the load-line view and the calculus proof of the matching theorem.

Why does the battery get to keep some of its own voltage? Where do the lost volts physically go?

They become heat inside the battery's body. The current has no way around the electrolyte and plates — every coulomb physically crosses them, dropping $Ir$ on the way, exactly as it would crossing any resistor. That energy warms the cell (touch a hard-worked battery and feel it). "Lost volts" is bookkeeping language for a real dissipation: $I^2r$ watts of it, continuously, whenever current flows in either direction.

Is the maximum power transfer theorem saying I should always use R = r?

No — it answers exactly one question: "what is the biggest wattage this source can possibly deliver?" If watts are scarce and waste is irrelevant (a microphone signal, an antenna), match and take the 50% hit. If energy efficiency is the point (chargers, grids, anything with a power bill), the theorem marks the operating point to avoid: run $R \gg r$ and keep $\eta$ near 1. Same theorem, opposite engineering verdicts.

How can two different resistors receive the same power from one battery?

Because $P(R) = \varepsilon^2R/(R+r)^2$ rises to a peak at $R = r$ and falls again — so almost every power level below the peak is achieved twice, once on the way up and once on the way down. The two loads pair off neatly: $R_1R_2 = r^2$ . The small load wastes energy in heavy current through $r$ ; the large load starves the current; both end up at the same payload.

Why is the terminal voltage above the emf during charging? That feels like getting something for nothing.

Nothing is free — the charger is paying. To force current backwards, the charger must beat the battery's own pump ( $\varepsilon$ per coulomb, banked as chemical energy) and push through the internal resistance ( $Ir$ per coulomb, lost as heat). The terminal voltage is the sum of both bills: $V = \varepsilon + Ir$ . That is also why a car's system voltage (~14 V) sits above the battery's emf (~12.6 V) whenever the alternator is charging.

Does the efficiency formula η = R/(R+r) work during charging too?

The formula as written is for discharge (payload over chemistry's spend). Charging has its own ledger: input $VI$ splits into banked energy $\varepsilon I$ and heat $I^2r$ , so the charging efficiency is $\varepsilon I / VI = \varepsilon/(\varepsilon + Ir)$ . Same physics, different denominator — the useful output during charging is the stored energy, not the load power. In the worked example: $96/112 \approx 86\%$ .

If a short circuit gives the maximum current, why is it not the maximum power to the load?

Because at $R = 0$ the load receives $I^2 \times 0 = 0$ — all the enormous power ( $\varepsilon^2/r$ of it) burns inside the battery itself. Maximum current and maximum load power are different questions: the current ceiling is at $R = 0$ , the load-power peak is at $R = r$ , and the efficiency optimum is at $R \to \infty$ . Three questions, three answers — examiners rotate through all three.

My two-readings calculation gave a negative internal resistance. What went wrong?

Almost certainly the readings were subtracted in mismatched order: $r = (V_1 - V_2)/(I_2 - I_1)$ needs the same pairing top and bottom — the voltage that dropped goes with the current that rose. A genuine negative $r$ would mean terminal voltage rising as you draw more current, which no chemical cell does. If the arithmetic is clean and $r$ still comes out negative on real data, suspect the readings themselves (meters swapped, or the "heavier load" reading actually taken at lighter load).