EMF, Terminal Voltage & Internal Resistance

Internal resistance has become one of the most closely watched numbers in consumer technology. The "battery health" percentage on a phone is, behind the marketing, largely a statement about $r$: as lithium-ion cells age, their internal resistance creeps upward, terminal voltage sags harder under load, and the phone shuts down early even though plenty of charge remains — the modern version of the torch that dims with "fresh-looking" cells. Electric-vehicle battery-management systems run the two-reading measurement of this article continuously: they pulse a known current, watch the voltage sag, compute the slope, and log the internal resistance of every cell group as the earliest honest indicator of ageing. The cold-morning car start is still the classic demonstration. Near freezing, a lead–acid battery's electrolyte chemistry slows and its internal resistance multiplies; the starter motor's enormous current then drags the terminal voltage down so far that the dashboard dims — and, if $r$ has grown enough with age, below what the ignition electronics tolerate. Same emf, same voltmeter reading on the bench, failed start: a pure internal-resistance effect, diagnosed by exactly the physics of this module.

Picture the cell as a water pump inside a fenced compound, with one narrow pipe between the pump and the gate. The pump pressure is fixed by its machinery — that is the emf, set by chemistry alone. But every litre that leaves the compound must first squeeze through the narrow internal pipe, losing some pressure on the way. The pressure you measure at the gate is the pump's pressure minus the internal loss — and the faster the flow, the bigger that loss. Close the gate entirely (no flow) and the gate pressure equals the full pump pressure. Electrically: the chemical reactions inside a cell lift each coulomb through $\varepsilon$ joules of energy — for the AA cell, 1.5 joules per coulomb. But the coulomb then has to travel out through the cell's own innards: electrolyte, electrode surfaces, contacts. These materials conduct, imperfectly, and their combined resistance is the internal resistance $r$. It is not a component someone installed; it is the physical interior of the source, and no coulomb has a way around it.

Hold two consequences of the picture before any algebra. First, the toll is traffic-dependent: the internal drop is $Ir$, so it vanishes when no current flows and grows with demand. A voltmeter, which draws almost nothing, therefore reads nearly the full emf — which is precisely why a voltmeter alone cannot diagnose a worn-out battery. Second, the toll is paid as heat inside the source: a battery working hard warms up, and that warmth is exactly the missing $Ir$ volts' worth of energy per coulomb. Nothing is lost from the books; it is just spent in the wrong shop. The emf itself comes from chemistry in a cell, but the concept is broader: a solar panel's emf comes from light, a generator's from motion (that story arrives with Electromagnetic Induction in E6). In every case the source hands each coulomb a fixed energy budget, and in every case its own interior takes a cut proportional to the current.

Start from the definition you met in Electric Potential & Potential Difference: potential difference is energy per charge. The emf is defined the same way, but for the source's contribution:

where: $W$ — the work done (energy transferred) by the source, in joules; $Q$ — the charge driven through it, in coulombs; $\varepsilon$ — the emf, in volts. A 12 V battery gives each coulomb 12 J. Despite the historical name "electromotive force", equation (1) is an energy-per-charge statement, in volts — writing "emf is the force that pushes charges" is a guaranteed lost mark on every board. Now do the accounting for a coulomb making one full lap of a simple circuit: out through the cell's interior, around the external load, and back. The chemistry pays it $\varepsilon$. Crossing the internal resistance costs $Ir$ (that is just Ohm's Law & Resistance applied to $r$). What remains when it emerges at the terminals — the energy per coulomb available to the outside — is the terminal voltage:

where: $V$ — terminal voltage, what a voltmeter across the terminals reads, in volts; $I$ — the current being drawn, in amperes; $r$ — internal resistance, in ohms. The product $Ir$ is the lost volts — "lost" only to the external circuit, since it reappears as heat inside the cell. Read the two limits directly from the equation: at $I = 0$, $V = \varepsilon$ (open circuit reads the emf); as $I$ grows, $V$ sags linearly. Connect a load resistance $R$ across the terminals and the same coulomb must also spend $IR$ crossing it. The budget must balance — $\varepsilon = IR + Ir$ — which fixes the current the circuit settles at:

where: $R + r$ — the total resistance of the loop, external plus internal, in series because every coulomb crosses both. This is the master equation of single-loop battery problems. Two special cases orient it: with $R \gg r$ the internal toll is negligible and $I \approx \varepsilon/R$ — the ideal-battery regime your earlier circuit work quietly assumed. With $R = 0$ (a short circuit) the current hits its ceiling $I_{sc} = \varepsilon/r$: even a perfect wire across the terminals cannot draw more, because the source's own interior throttles it.

One more reading of equation (2) earns marks surprisingly often: solve it for the efficiency of delivery. The fraction of the chemistry's energy that reaches the outside is $V/\varepsilon = R/(R+r)$. A torch running a $5\ \Omega$ bulb from a cell with $r = 0.5\ \Omega$ delivers $5/5.5 \approx 91\%$ of its energy usefully; short the same cell and the efficiency is zero — every joule cooks the cell. (Which load extracts the maximum power is a different, sneakier question — the classic answer $R = r$ — and it gets its full treatment in the battery-problems archetype coming later in this chapter.)

Equation (2) is a straight line, and examiners love straight lines. Plot terminal voltage $V$ (vertical) against current drawn $I$ (horizontal) and a real battery traces a falling line: it starts at $\varepsilon$ on the voltage axis (the open-circuit reading), drops with slope $-r$, and would hit the current axis at $\varepsilon/r$ (the short-circuit current). Every battery question with a graph in it is asking you to read one or more of those three features.

The graph is also a measuring instrument. Take two readings under different loads — $(I_1, V_1)$ and $(I_2, V_2)$ — and equation (2) written twice gives two simultaneous equations in the two unknowns $\varepsilon$ and $r$. Subtracting them cancels $\varepsilon$:

where: the ordering matters — the reading with the larger current has the smaller voltage, so both differences are positive when you pair them correctly. In words: internal resistance is the sag in voltage per extra ampere of demand. Back-substitute either reading to get $\varepsilon = V_1 + I_1 r$. This two-point method is the standard school practical, and the competitive worked example below runs it end to end. One direction of travel remains. Push current backwards through the cell — which is exactly what a charger does — and the internal drop changes sides: the charger must overcome both the emf and the toll, so the terminals read

where: $I$ — the charging current forced through the cell. This is the only situation in this chapter where the terminal voltage exceeds the emf, and it is why a phone battery whose cell emf is 3.7 V shows about 4.0 V across it while fast-charging, and why a running car holds its 12.6 V battery at roughly 14 V. Spotting "charging" in a question and flipping the sign is a one-mark decision that many candidates miss under time pressure.

Board level — the torch circuit budget (CBSE/ICSE pattern, 3 marks)

A battery of emf $6.0\ \mathrm{V}$ and internal resistance $0.50\ \Omega$ is connected to a lamp of resistance $11.5\ \Omega$. Find (a) the current in the circuit, (b) the terminal voltage of the battery, and (c) the largest current this battery could ever supply.

Step 1. Part (a), from equation (3): the loop's total resistance is $R + r = 11.5 + 0.50 = 12.0\ \Omega$, so $I = \dfrac{\varepsilon}{R+r} = \dfrac{6.0}{12.0} = \boxed{0.50\ \mathrm{A}}$.

Step 2. Part (b), two equivalent routes — use one, check with the other. Budget route: $V = \varepsilon - Ir = 6.0 - 0.50 \times 0.50 = \boxed{5.75\ \mathrm{V}}$. Check via the load: $V = IR = 0.50 \times 11.5 = 5.75\ \mathrm{V}$ ✓. The lost volts are $0.25\ \mathrm{V}$ — energy heating the battery, not the lamp.

Step 3. Part (c): the ceiling is the short-circuit current, $I_{sc} = \dfrac{\varepsilon}{r} = \dfrac{6.0}{0.50} = \boxed{12\ \mathrm{A}}$ — no external circuit, however greedy, can pull more.

Step 4. The sentence that earns the method mark: "The battery delivers $V = \varepsilon - Ir$; the difference between emf and terminal voltage is the potential drop across the internal resistance." Stating equation (2) in words, with each symbol placed in its hardware, is what the scheme is looking for.

Competitive level — two readings and a backwards battery (JEE-Main pattern, ~3 minutes)

A cell drives $0.20\ \mathrm{A}$ through a $9.5\ \Omega$ resistor. When the resistor is replaced by a $4.5\ \Omega$ one, the current rises to $0.40\ \mathrm{A}$. Find (a) the emf and internal resistance of the cell, (b) the terminal voltage in the second case, and (c) the terminal voltage if this cell is instead charged at $2.0\ \mathrm{A}$.

Step 1. Write equation (3) twice, as $\varepsilon = I(R + r)$: $\varepsilon = 0.20(9.5 + r)$ and $\varepsilon = 0.40(4.5 + r)$.

Step 2. Set them equal and solve: $1.9 + 0.20r = 1.8 + 0.40r$, so $0.10 = 0.20r$, giving $\boxed{r = 0.50\ \Omega}$. Back-substitute: $\varepsilon = 0.20(9.5 + 0.50) = \boxed{2.0\ \mathrm{V}}$.

Step 3. Part (b): $V = IR = 0.40 \times 4.5 = \boxed{1.8\ \mathrm{V}}$ — or via the budget, $V = 2.0 - 0.40 \times 0.50 = 1.8\ \mathrm{V}$ ✓. Notice the sanity pattern: the heavier load (smaller $R$, larger $I$) shows the lower terminal voltage. If your simultaneous equations ever produce a negative $r$, you have paired the readings wrongly — larger current must go with smaller resistance.

Step 4. Part (c) is the sign flip: charging reverses the current through the cell, so equation (5) applies: $V = \varepsilon + Ir = 2.0 + 2.0 \times 0.50 = \boxed{3.0\ \mathrm{V}}$. The charger must present 3.0 V to force 2.0 A backwards through this cell. Timing note: about 3 minutes. The subtraction trick — divide the two ε-equations or subtract them so ε cancels — should be automatic; do not solve by substitution under exam time. The charging flip in (c) is a five-second decision worth a full mark: spot the word "charged", change the sign.

Calling emf a force. Emf is energy per unit charge, in volts — equation (1). The name is an eighteenth-century fossil. Boards from CBSE to A-Level explicitly award or dock the definitional mark on this point.

Using $V$ where $\varepsilon$ belongs (and vice versa). In $I = \varepsilon/(R+r)$, the numerator is the emf, never the loaded terminal voltage. Conversely the power delivered to the external circuit is $VI$, not $\varepsilon I$ (that is the total including internal heating). Place each symbol in its hardware before substituting numbers.

Assuming the voltmeter reads the emf exactly. It reads the emf only if it draws truly zero current. School voltmeters are high-resistance, so the reading is close to $\varepsilon$ — good enough at board level when stated, but "the voltmeter reads slightly less than the emf because it draws a small current" is the full-credit sentence. (The instrument that genuinely draws zero at balance is the potentiometer — a story for later in this chapter.)

The charging sign. "A battery is being charged at…" flips equation (2) to $V = \varepsilon + Ir$. Terminal voltage above emf is not an error; it is the signature of charging.

Graph-reading slips. On a $V$–$I$ graph the intercept is $\varepsilon$ and the slope is $-r$: a steeper line means a worse (higher-resistance) source. Do not report the slope with a positive sign, and do not confuse this graph with the $I$–$V$ characteristic of a resistor from Ohm's Law & Resistance — here the falling line belongs to the source.

Pairing errors in two-reading problems. Bigger $R$ goes with smaller $I$. A negative $r$ from your algebra is the standard symptom of mispairing — treat it as an alarm, not an answer.

Constants and conventions. All boards here work in SI volts, amperes, ohms. No $g$ is needed anywhere in this module. IB and AP-2 papers typically give $r$ explicitly; JEE and NEET expect you to extract it from data or graphs as in the competitive example.

Texts. NCERT Physics Class 12, Ch. 3 (Current Electricity, cells and emf sections) · HC Verma, Concepts of Physics Vol. 2, Ch. 32 · CIE 9702 Topic 10 (practical circuits: emf and internal resistance) · IB Physics guide, Theme B.5.

Provenance. All results are standard. A verifier should recompute: Example 1's $I = 6.0/12.0 = 0.50\ \mathrm{A}$, $V = 5.75\ \mathrm{V}$ both routes, $I_{sc} = 12\ \mathrm{A}$, efficiency $5.75/6.0 = 95.8\%$; Example 2's $r = 0.50\ \Omega$, $\varepsilon = 2.0\ \mathrm{V}$, $V = 1.8\ \mathrm{V}$, charging $V = 3.0\ \mathrm{V}$; quiz values $1.5 - 0.5 \times 0.3 = 1.35\ \mathrm{V}$, $12.0/0.05 = 240\ \mathrm{A}$, $12.6 + 10 \times 0.02 = 12.8\ \mathrm{V}$, $(1.50 - 1.26)/0.60 = 0.40\ \Omega$.

Prerequisites. Electric Potential & Potential Difference · Ohm's Law & Resistance. Leads to. Combining Cells: Series & Parallel · Kirchhoff's Rules · the battery-problems archetype (maximum power, efficiency — coming later in E3) · EMF and Internal Resistance (UG deep dive).