Physics › Electricity › EMF and internal resistance
EMF and internal resistance
Every real cell resists its own current, and that hidden resistance is why the voltage on the label is not the voltage you get. One straight-line graph exposes both of a cell's secrets at once.
Builds on Circuits and Kirchhoff's laws.
IN THIS TOPIC
- Define emf as the energy given to each unit of charge, ε = E/Q.
- Use ε = I(R + r) and the terminal pd V = ε − Ir in circuits where r is not negligible.
- Extract ε and r from a graph of terminal pd against current.
WHAT YOU PROBABLY THINK
A 1.5 V battery gives you 1.5 volts.
What emf actually is
The electromotive force of a source is the energy it gives to each coulomb of charge passing through it:
measured in volts, joules per coulomb, exactly like pd. The name is a historical accident: it is an energy per charge, and no part of it is a force. It describes what the chemistry supplies; the terminal pd, coming next, describes what the outside world receives.
The resistance inside
Real cells are built of chemicals and electrodes that resist the very current they drive. Model this as a perfect source of emf in series with a small internal resistance r, both sealed inside the case.
The emf must pay for both resistances in the loop:
and what appears across the cell's terminals, the terminal pd, is the emf minus the part spent inside: V = ε − Ir, which is also just IR, the pd across the external circuit.
Lost volts
The difference Ir is nicknamed the lost volts: energy per coulomb spent crossing the cell's own innards, warming the battery rather than the circuit. It grows with the current, so the harder a cell works, the further its terminal pd sags below the emf. That is why headlights dim while the starter motor runs: the starter's huge current inflates Ir, and every other component feels the drop.
Only when no current flows does the sag vanish. A voltmeter across an unused cell reads the full emf, because with I = 0 there are no lost volts to subtract.
One graph, both secrets
Vary the external resistance, record the terminal pd and the current, and plot V against I. The model equation V = ε − Ir is a straight line: intercept ε, gradient −r.
The graph earns its place because it measures the emf properly. You cannot reach I = 0 by simply disconnecting things and hoping; the intercept extrapolates to it, giving ε uncontaminated by lost volts, and the slope hands you r in the same breath.
THE EXAM BIT
- From a V–I graph: the intercept is ε and the gradient is minus r. Quote r as a positive resistance and say the gradient is negative.
- The terminal pd equals the emf only when I = 0. A high-resistance voltmeter across an isolated cell reads ε for exactly that reason.
- In any calculation the cell's r sits in series with everything external: the loop equation is ε = I(R + r), never ε = IR.
- “Lost volts” = Ir. It is not a fixed number for a cell; double the current and you double the loss.
- Despite the name, emf is an energy per unit charge, in volts. Calling it a force costs the definition mark.
CHECK YOURSELF
A battery of emf 9.0 V and internal resistance 0.60 Ω drives a 4.4 Ω resistor. Find the current, the terminal pd, and the lost volts.
Show a hint
The emf pays for both resistances; the terminals only show the external share.
Show the answer
Current: I = εR + r = 9.0 / (4.4 + 0.60) = 1.8 A.
Terminal pd: V = IR = 1.8 × 4.4 = 7.9 V (equivalently ε − Ir).
Lost volts: Ir = 1.8 × 0.60 = 1.1 V. The 9.0 V splits as 7.9 outside plus 1.1 inside, and the books balance.
The emf is what the cell promises.
The terminal pd is what you get, minus Ir.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.