Internal Resistance & EMF

Pick a source and a load. The page works out the current, what actually appears at the terminals, how the power splits between the load and the inside of the source, and how close you are to the most that source could ever give.

Worked example — Internal Resistance & EMF

Current round the loop	0.2992 A
Voltage at the terminals	1.4961 V
Power into the load	0.4477 W
Power wasted inside the source	0.0161 W
Share that reaches the load	96.53%
The most this source could ever deliver	3.3368 W
How much of that you are getting	13.42%
What a dead short would draw	8.611 A

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Lesson

The theory — Internal Resistance & EMF

A real source is two things in series: an EMF E, the energy it gives each coulomb, and an internal resistance r, which is everything inside it that opposes the current — electrolyte, electrodes, the connections to the terminals. What you can measure from outside is the terminal voltage, V = E − Ir, and it equals the EMF only when no current flows. The current round the loop is I = E/(R+r), the load receives I²R, the source itself dissipates I²r, and the fraction that reaches the load is η = R/(R+r).

What each symbol means

E: the EMF, in volts. The energy per coulomb the source supplies, and the reading you get with nothing connected
r: internal resistance, in ohms. Not a component you can point at — it is the whole of the source's own opposition to current, gathered into one number
R: the load, in ohms. The one quantity in this list that belongs to you rather than to the source
I: the current, in amperes. The same everywhere in the loop, because the source and the load are in series
η: efficiency, R/(R+r). The share of the supplied energy that reaches the load rather than heating the source

Assumes: A constant EMF, a constant internal resistance and steady DC — the Thévenin equivalent of the source, which is exact for a linear source and a decent first approximation for a chemical one. A real cell breaks the first two slowly: r climbs as it discharges and climbs sharply in the cold, and E falls with state of charge, which is why a cold car on a winter morning turns over slowly for two reasons at once rather than one. The figures for each source in the list are representative rather than measured from a particular cell.
Breaks when: On any source whose characteristic is not a straight line, and the solar cell in the list is the worst offender on this page. A photovoltaic cell's terminal voltage against current is a curve, so a single r is only a local fit and its maximum-power point is nowhere near R = r — which is exactly why solar inverters carry maximum-power-point tracking rather than a fixed matched load. A real chemical cell bends too, falling away faster than the line predicts once the current is large enough that the electrolyte cannot keep up. And the 50% result is about steady power, not about energy stored: running a battery at its matched load gets the most watts out of it and less than half of its charge, which is why nothing that has to last is ever run there.

The most power you can get arrives at 50% efficiency 🖖

Press Matched load. The cell is driving 0.18 Ω, the same as its own internal resistance, and the load takes 3.3368 W — the most this cell can deliver to anything at all. Now read the row beneath it: the cell is dissipating 3.3368 W as well. Exactly half of everything the chemistry supplies is being turned into heat inside the battery, and that is not a fact about this cell: η = R/(R+r) is one half whenever R = r, for any source, at any voltage. The first panel draws the two powers together — one humps, one falls, and they cross at the peak. Getting the most out of a source and wasting the least of it are different settings, and you have to pick one.

Nine times the load, a ninth of the load, the same watts 🖖

Press Nine times r, then A ninth of r. Both deliver 1.2013 W to the load, 36.00% of the peak. What differs is the price. At 1.62 Ω the cell runs at 90.00% and warms by 0.1335 W. At 0.02 Ω it runs at 10.00% and warms by 10.8113 W, which is eighty-one times as much heat for the same power delivered. The table below is that symmetry laid out: the power column reads the same downwards and upwards, and the efficiency column does not. Which side of the peak you sit on is the entire question — a radio receiver is matched because the signal is all there is, and a power station never is, because the fuel is.

Both of these batteries read fine on a meter 🖖

A voltmeter across an AA cell says 1.55 V, because a voltmeter draws almost nothing: R is enormous, I is nearly zero, and V = E − Ir has nothing to subtract. Connect something real and the reading becomes a statement about the load — 1.4961 V into a torch bulb, 0.7750 V into 0.18 Ω. That is why a battery tester puts a load across the cell instead of just measuring it. The same arithmetic says what a source can do: a dead short draws E/r, which is 8.611 A for this cell and 2,520 A for a car battery. That is 293 times the current from 8 times the voltage, and it is the internal resistance doing all of it.

Problem solved in full

Finding a battery's internal resistance without opening it 6 steps
You have a cell, a voltmeter, an ammeter and two resistors, 10 Ω and 1 Ω. Across the 10 Ω you read 0.152259 A at 1.522593 V. Across the 1 Ω you read 1.313559 A at 1.313559 V. Find the EMF and the internal resistance — and then say what the cell would do into a dead short, without shorting it.

Open this problem in the tool
1. Every measurement you can take sits on the same straight line, V = E − Ir. Two points fix a line, so two resistors are enough.
  
  $V = E - I r$
2. Take the slope between the two readings. It comes out at −0.180000 V per ampere.
  
  $\frac{Δ V}{Δ I} = \frac{1 . 313559 - 1 . 522593}{1 . 313559 - 0 . 152259} = \frac{- 0 . 209034}{1 . 161300} = - 0.180000$
3. That slope is −r, so the internal resistance is 0.1800 Ω. Nothing had to be taken apart to get it.
  
  $r = - \frac{Δ V}{Δ I} = 0.1800 Ω$
4. The intercept is the EMF: add back the volts the internal resistance was eating at either reading.
  
  $E = V + I r = 1.522593 + 0.152259 \cdot 0.18 = 1.550000 V$
5. Check it on the other point, which is the whole reason for measuring twice. 1.313559 + 1.313559 × 0.18 gives 1.550000 V again.
  
  $1.313559 + 1.313559 \cdot 0.18 = 1.550000 V$
6. Now both ends of the line are known, and so is everything between them: the peak the cell can deliver, where it sits, and the current a dead short would draw.
  
  $P_{m a x} = \frac{E ^{2}}{4 r} R = r \frac{E}{r} 3.3368 W 0.1800 Ω 8.611 A$
Answer

E = 1.550000 V and r = 0.1800 Ω, from two voltmeter readings. The cell's peak output is E²/4r = 3.3368 W into a 0.1800 Ω load, and a dead short would pull E/r = 8.611 A. The last figure is the point of the method: you can state what a cell does in the one condition you must never put it in, because the line goes all the way to the axis whether or not you follow it there. It is also how a battery management system knows a pack is ageing — r climbs first, well before the open-circuit voltage gives anything away.

$E = 1.550000 V, r = 0.1800 Ω \Rightarrow \frac{E}{r} = 8.611 A$

Learning path

What Ohm’s law leaves out

References (2)

Insight block 1 is one reading of the maximum power transfer theorem and there are several, which is what this survey is for — the load that takes the most power, the source resistance that does, and the conjugate match are three different questions with three different answers: S. C. Dutta Roy, "Many Faces of the Maximum Power Transfer Theorem." International Journal of Electrical Engineering & Education 40(2), 2003, pp. 103-111. DOI 10.7227/ijeee.40.2.1
The Thevenin equivalent the whole page rests on, and the three forms of resistive power the results table uses. Cited across this chapter on purpose: one circuits text. J. W. Nilsson and S. A. Riedel, Electric Circuits, 10th ed. Pearson/Prentice Hall, 2015. ISBN 978-0-13-376003-3.

By the 42Math team · Reviewed against published sources. Corrections welcome.