EMF, Internal Resistance and Terminal Voltage

A cell’s emf is what its chemistry supplies; its terminal voltage is what the outside world gets. This returns both, the lost volts in between, the share of the power that never leaves the battery, the short-circuit current, and the load that would draw the most power — for any number of identical cells in series and in parallel.

Last updated 11 Nov 2026

Terminal voltage, lost volts and where the power goes

Cells, internal resistance → V and η

EMF of one cell (V)The open-circuit voltage: what a perfect voltmeter drawing no current would read across one cell on its own. Printed on the cell — 1.5 V for a zinc-carbon or alkaline AA, 1.2 V for NiMH, about 3.7 V for a lithium-ion cell. The emf is energy per unit charge supplied by the chemistry, so its unit is the volt, and calling it a force is an accident of the nineteenth century that the name has never shaken off.

Internal resistance of one cell (Ω)A fresh AA alkaline cell is of the order of 0.1 to 0.3 Ω; a tired one can be several ohms, which is why a nearly dead battery reads 1.5 V on a meter and still will not turn a motor. Zero is allowed here and is worth trying: it is the ideal cell, the terminal voltage then equals the emf at every load, and the short-circuit rows vanish because an ideal cell’s short-circuit current is infinite.

Number of identical cells in seriesIn series the emfs add and so do the internal resistances: four 1.5 V cells give 6 V with four times the internal resistance. This is why a four-cell torch is brighter than one cell but not four times brighter, and why the gain falls away as the cells age and their internal resistance climbs.

Number of such series strings in parallelPutting m identical strings in parallel keeps the emf the same and DIVIDES the internal resistance by m. That is the whole reason for wiring cells in parallel: not more voltage, but a lower source resistance and so a stiffer terminal voltage under load. Set it above 1 and watch the lost volts fall while the total emf does not move.

External load resistance (Ω)Everything outside the battery, lumped into one resistance. Zero means a short circuit, and the page will still answer: the current is then the short-circuit current, the terminal voltage is zero and all the power is dissipated inside the battery. That is a real and dangerous condition, not a mathematical curiosity.

4.4118VExample

Four 1.5 V cells in series, each of internal resistance 0.45 Ω, driving a 5 Ω load

One loop, one subtraction, and the trade-off between power and efficiency

I = E / (R + r) · V_terminal = E − Ir = IR · η = R / (R + r) · P_load = I²R, maximum at R = r

E: the emf of the whole battery: the emf of one cell times the number in SERIES. Energy supplied per unit charge by whatever is driving the circuit. Measured in volts, and equal to the terminal voltage only when no current flows
r: the total internal resistance: one cell’s resistance times the number in series, divided by the number of parallel strings. It is inside the battery and you cannot get at it, which is why it is only ever inferred — from the slope of a V against I graph, which is the standard experiment
Ir: the LOST VOLTS. Not lost in any mysterious sense: this is the potential difference that appears across r, and the energy goes into heating the battery. It rises with current, which is the reason a terminal voltage sags under load
V = E − Ir: the terminal voltage, and the equation of a straight line if you plot V against I: intercept E, gradient −r. That is how the practical is done, and the gradient being NEGATIVE is the part students lose marks on
η = R/(R+r): the fraction of the chemistry’s power that reaches the load. It rises towards 1 as the load gets large, which is the opposite of the condition for maximum POWER. Both statements are true at once and that is not a contradiction: a big load gets a high fraction of a small total

Worked example

Four 1.5 V cells in series, each of internal resistance 0.45 Ω, driving a 5 Ω load
BUILD THE BATTERY FIRST. In series the emfs add: E = 4 × 1.5 = 6.0 V. So do the internal resistances: r = 4 × 0.45 = 1.8 Ω. Both of them, not just the emf — forgetting that r adds as well is the standard mistake with a series battery
ONE LOOP, ONE CURRENT. I = E/(R + r) = 6.0/(5 + 1.8) = 6.0/6.8 = 0.882353 A. The internal resistance is in the denominator beside the load because the current has to go through both
THE TERMINAL VOLTAGE, TWO WAYS. Either V = IR = 0.882353 × 5 = 4.4118 V, or V = E − Ir = 6.0 − 0.882353 × 1.8 = 6.0 − 1.5882 = 4.4118 V. They must agree, and computing it both ways is the quickest check there is on this topic
WHERE THE POWER WENT. The load gets I²R = 3.8927 W. The battery wastes I²r = 1.4014 W heating itself. Total 5.2941 W, which is also E×I — so 26.5 per cent of this battery's output never leaves the battery. That is what a 1.8 Ω source resistance costs against a 5 Ω load
AND THE TWO EXTREMES. Short the terminals and the current would be E/r = 3.3333 A, with the whole 20 W going into the cells. Match the load to the source instead, R = r = 1.8 Ω, and the load would take its maximum of E²/4r = 5.0 W — more than the 3.89 W it gets now, but at 50 per cent efficiency rather than 73.5 per cent. Maximum power and maximum efficiency are different targets and you cannot have both

What the same four cells do to the same load, wired three ways

Computed on this page’s own arithmetic with emf 1.5 V, r 0.45 Ω per cell and a 5 Ω load. Into a load this much larger than the internal resistance, series wins easily — voltage is what you are short of, not source stiffness. The ranking reverses when the load is SMALL: put 0.1 Ω across the same cells and the all-parallel arrangement delivers more power than the series one, because then it is the internal resistance that is limiting the current rather than the emf that is limiting the voltage. Which arrangement is better is a question about the load, and has no answer without one.
Wiring	Total emf	Total r	Current	Terminal voltage	Load power
4 in series	6.00 V	1.80 Ω	0.8824 A	4.4118 V	3.8927 W
2 series × 2 parallel	3.00 V	0.45 Ω	0.5505 A	2.7523 V	1.5150 W
4 in parallel	1.50 V	0.1125 Ω	0.2934 A	1.4670 V	0.4304 W

Why a voltmeter on a dead battery still reads 1.5 volts

A battery has two voltages and the whole topic is the gap between them. The emf is what the chemistry supplies per unit charge. The terminal voltage is what is left by the time the charge has fought its way out through the cell’s own resistance. They differ by Ir, the lost volts, and they are equal only when I is zero.

That single fact explains the dead-battery puzzle. Put a digital voltmeter across a nearly exhausted AA and it reads 1.5 V, because a digital voltmeter draws almost no current and so measures something very close to the emf. Put the same cell in a torch and nothing happens, because the internal resistance has climbed to several ohms and at the torch’s current the lost volts eat the lot. Ageing destroys a cell’s internal resistance long before it destroys its emf, which is why the voltage test everyone performs is the one test that cannot detect it.

The practical that measures r is a straight-line graph, and its gradient is negative. Vary the load, plot terminal voltage V on the y-axis against current I on the x-axis, and you get V = E − Ir: a straight line whose intercept is the emf and whose GRADIENT is −r. Students routinely report r as a negative number or quote the gradient’s magnitude as the emf; the intercept is the emf and the magnitude of the gradient is the internal resistance. The CBSE 2023-24 document puts internal resistance in the Unit II theory list, and does not list it among the Section A experiments — it appears instead as the first suggested investigatory project, “to study various factors on which the internal resistance/EMF of a cell depends”.

And the maximum-power result is a trap worth understanding rather than memorising. The load that extracts the most watts from a given source is the one matched to the source, R = r. At that point exactly half the power is burnt inside the source, so the efficiency is 50 per cent. Audio amplifiers and radio transmitters are matched because getting the power out is what matters; the electricity grid is emphatically not matched, because there the 50 per cent would be the point. If you want the circuit arithmetic that sits behind the load side of this, the metre bridge is the standard way a school lab measures an unknown resistance in the first place.

Frequently asked questions

What is the difference between emf and potential difference?

An emf is energy GIVEN to charge by a source — chemistry in a cell, a changing flux in a dynamo. A potential difference is energy TAKEN from charge by a component. Both are joules per coulomb and both read in volts. The terminal voltage of a cell is a potential difference; the cell’s emf is not. Cambridge 9702 makes this an explicit outcome, 10.1.4: “distinguish between e.m.f. and potential difference (p.d.) in terms of energy considerations”.

How do I find the internal resistance from a graph?

Plot terminal voltage against current and take the magnitude of the gradient. V = E − Ir is a straight line with intercept E and gradient −r, so r is the gradient’s size, in ohms, and the emf is where the line meets the voltage axis. Two cautions: the line is only straight while r is constant, which is not true at high current or for a cell close to flat; and you must plot V against I, not I against V, or the gradient is −1/r.

Why do the internal resistances add when cells are in series?

Because the current goes through all of them in turn, which is what series means. Four cells of 0.45 Ω present 1.8 Ω to the circuit, so you gain four times the emf and pay four times the lost volts per amp. Wiring the same four cells in parallel does the reverse: one cell’s emf, a quarter of one cell’s internal resistance. Series buys voltage, parallel buys a stiff terminal voltage, and the CBSE Unit II list asks for both.

Why does my terminal voltage go up when I use a bigger resistor?

A bigger load means less current, less current means fewer lost volts, and fewer lost volts means the terminal voltage sits closer to the emf. In the limit of an infinite load — an open circuit — no current flows, there are no lost volts at all, and the terminal voltage IS the emf. That limit is exactly what a high-resistance voltmeter is exploiting.

Is maximum power the same as maximum efficiency?

No, and they are at opposite ends. Maximum POWER to the load happens at R = r, where the efficiency is exactly 50 per cent. Maximum EFFICIENCY happens as R becomes large, where the load’s share of the power approaches 100 per cent but the total power approaches zero. Any real design picks a point between them, and which way it leans says what the designer was short of.

What happens if I set the internal resistance to zero?

You get the ideal cell. The terminal voltage then equals the emf at every load, the lost volts are zero, and the efficiency is 100 per cent whatever the load. The short-circuit and maximum-power rows disappear from the output, which is correct rather than a fault: an ideal cell’s short-circuit current is infinite and its maximum power transfer unbounded, so the page drops them rather than printing an infinity. The page refuses altogether only if the load is zero as well, which is an ideal cell short-circuited — a case with no finite answer of any kind.

Related calculators

References

Annual Syllabus Session 2023-24, Class XII Physics (Code 042), Central Board of Secondary Education, edustud.nic.in/edu/SYLLABUS_2023_24/12/12_physics_eng_2023_24.pdf (read 11 November 2026). The board’s own document, cited and not endorsed by it. Unit I Chapter 1 asks for “Coulomb’s law-force between two point charges; forces between multiple charges, superposition principle”; Chapter 2 for “electrical potential energy of a system of two point charges” and “capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor.(No derivation, formulae only)”. Unit II asks for “drift velocity, mobility and their relation with electric current”, “Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, kirchhoff’s rules, Wheatstone bridge”. Unit III asks for “Biot – Savart law and its application to the current carrying circular loop”, “Ampere’s law and its applications to infinitely long straight wire. Straight solenoid (only qualitative treatment)” and “force between two parallel current-carrying conductors-definition of ampere”.
Syllabus: Cambridge International AS & A Level Physics 9702, for examination in 2025, 2026 and 2027 (Version 1), Cambridge Assessment International Education, cambridgeinternational.org/Images/664565-2025-2027-syllabus.pdf (read 11 November 2026). Cited for the outcomes a Cambridge candidate is examined on, which are worded differently from the Indian board’s: 18.3.2 “recall and use Coulomb’s law F = Q1Q2 /(4πε0r2) for the force between two point charges in free space”, 18.5.3 “use V = Q/(4πε0r) for the electric potential in the field due to a point charge”, 9.1.4 “use, for a current-carrying conductor, the expression I = Anvq, where n is the number density of charge carriers”, 10.1.5 “understand the effects of the internal resistance of a source of e.m.f. on the terminal potential difference”, 19.1.3 “derive, using C = Q/V, formulae for the combined capacitance of capacitors in series and in parallel”, 20.2.2 “recall and use the equation F = BIL sin θ, with directions as interpreted by Fleming’s left-hand rule” and 20.3.2 “recall and use F = BQv sin θ”. The copy read here was truncated before the data-and-formulae sheet, so what is on THAT sheet could not be established from it.
2022 CODATA Recommended Values of the Fundamental Physical Constants (wallet card), NIST Committee on Data for Science and Technology, physics.nist.gov/cuu/pdf/wallet_2022.pdf (read 11 November 2026). Every constant on this page is taken from here, and the reason to read the card rather than a textbook is the bracket. The elementary charge is printed as 1.602 176 634 × 10⁻¹⁹ C with no bracket at all, because since the 2019 revision of the SI it is exact by definition — the coulomb is now defined from it, not the other way round. The same is true of c, h, k and N_A. But the vacuum electric permittivity is printed 8.854 187 8188(14) × 10⁻¹² F/m and the vacuum magnetic permeability 1.256 637 061 27(20) × 10⁻⁶ N/A²: both now have an uncertainty, about 1.6 parts in 10¹⁰, where before 2019 μ₀ was exactly 4π × 10⁻⁷ and ε₀ exactly 1/μ₀c². Almost every school book still prints the old exact forms. At school precision nothing changes; the logic of which constants are definitions changed completely.