Emf & Internal Resistance

Connect a resistor across a cell and positive charge drifts through the external circuit from the + terminal to the −. Inside the cell it must go the other way, from − to +, uphill against the very electric field that the separated charge has built up. Something has to push it, and it is not that field: in a cell it is chemistry at the electrodes, in a dynamo the engine that drags a conductor through a magnetic field, in a photocell the energy of an absorbed photon. Define ε = dW/dq, the work that agency does per coulomb carried through. The unit is the joule per coulomb, which is the volt, and that is where the name misleads — electromotive force is an energy per charge, not a force at all. It is not a potential difference either, because the work is not done by an electrostatic field. Keep the two apart and the rest follows: ε belongs to the source, while the p.d. at the terminals belongs to whatever you attached.

Go once round a single-loop circuit adding potential changes; energy conservation requires the total to be zero. Start at the − terminal and climb through the source: the chemistry lifts you by ε, but the same current I that flows in the wires also crosses the electrolyte and electrodes, dropping I r on the way out. Cross the load and you fall by I R, arriving back where you began. So ε − I r − I R = 0, and that rearranges two ways. Solve for current, I = ε/(R + r): the internal resistance simply joins the loop in series. Or group it as V = ε − I r = I R, where V is the terminal voltage, the p.d. a meter across the source actually reads. At I = 0 the internal drop vanishes and V = ε, which is why the open-circuit reading is the one that reports the emf. Under load V is always smaller, and the missing I r volts are not destroyed — they appear as heat inside the source.

Neither ε nor r is printed on a cell; you measure them. Vary the load, record current and terminal voltage together, and plot V against I. The model says the points fall on the straight line V = ε − I r, so the intercept is the emf and the magnitude of the gradient is the internal resistance. A 9 V battery driving two different loads gives (0.10 A, 8.85 V) and (0.50 A, 8.25 V). The gradient is (8.25 − 8.85)/(0.50 − 0.10) = −0.60/0.40 = −1.50 Ω, so r = 1.50 Ω, and extrapolating back, ε = 8.85 + 0.10 × 1.50 = 9.00 V. Use more than two points and fit a line: the scatter about it is your evidence that ε and r really were constant over the range tested. One caution about that intercept — a voltmeter drawing 1 μA leaves an internal drop of only 1.5 μV here, utterly negligible, but the same meter across a source whose r is megohms would read far below ε.

Push R towards zero and the current does not run away: I = ε/(R + r) climbs only to ε/r, the short-circuit current. Every source has one, and its size separates sources that look alike on paper. The 9 V battery above has r = 1.50 Ω, so it can never deliver more than 9.00/1.50 = 6.00 A; a wire laid across it merely gets warm while the battery gets hot. A 12.6 V lead-acid car battery has r near 0.024 Ω, giving a short-circuit current of about 5.3 × 10² A — that is why it can crank a starter motor, and why a spanner dropped across its terminals welds itself in place. The ceiling also says where the energy goes at that moment: with V = 0 the load takes nothing, so the whole ε²/r, some 6.6 × 10³ W for the car battery, is dissipated inside the battery itself.

Multiply ε = I r + I R through by I and the loop equation becomes a power statement: ε I = I²r + I²R. The left side is the rate at which the chemistry releases energy; the right side splits it into heating inside the source and useful power in the load. The useful fraction is η = I²R/(ε I) = R/(R + r), which equals V/ε. Now watch P_R = ε²R/(R + r)² as R varies. It is zero at R = 0 (all current, no voltage), rises to a maximum, and falls back towards zero for very large R (all voltage, no current). The maximum sits at R = r and equals ε²/(4r). Take ε = 12.0 V and r = 0.50 Ω: at R = 0.50 Ω the load takes 72.0 W at η = 50%, while at R = 11.5 Ω it takes only 11.5 W but at η = 95.8%. Maximum power and maximum efficiency are different design targets.

Constant ε and constant r are a first-order description, good inside a limited window. r is a lumped stand-in for electrolyte conductivity, electrode area and contact resistance, and none of those stay put: a cell's internal resistance typically doubles or worse by the end of its discharge, and it rises sharply in the cold, which is why a car that starts in July hesitates in January. ε drifts too, sagging with state of charge, so the open-circuit voltage is only a rough fuel gauge. At large currents an extra non-ohmic drop appears at the electrodes — polarisation — and the V–I line bends downward, so an r fitted at 0.1 A underestimates the sag at 5 A. Some sources are not linear at all: a photovoltaic cell holds nearly constant current up to a knee and then collapses, so it is described by its whole I–V curve, not by a pair of numbers.