Real Batteries, Internal Resistance & Meters
- Distinguish emf from terminal voltage and compute the difference
- Explain why terminal voltage falls as current increases
- State the ideal resistance of an ammeter and a voltmeter and explain why
A real battery has resistance of its own
A battery's emf ε is the energy it supplies per unit charge — a fixed property. But current flowing through the battery's own internal resistance r dissipates some of that energy inside the battery, so what appears across its terminals is less. Terminal voltage = ε − Ir. At zero current the two are equal, which is why a voltmeter on a disconnected battery reads the emf.
Why headlights dim when the engine cranks
A starter motor has very low resistance and draws enormous current. From V = ε − Ir, a large I produces a large Ir drop inside the battery, so terminal voltage falls sharply and everything else on the circuit dims. As the battery ages its internal resistance rises, which is why an old battery can read a healthy voltage with no load and still fail to start the car — the no-load reading measures emf and says nothing about r.
What meters should ideally be
An ammeter measures current and is connected in series, so it must have near-zero resistance — otherwise it changes the current it is measuring. A voltmeter measures potential difference and is connected in parallel, so it must have near-infinite resistance — otherwise it draws current through itself and reduces the voltage it is measuring. Both requirements come from the same principle: a good measurement disturbs the system as little as possible.
What goes wrong when they are swapped
Connecting an ammeter in parallel with a component places a near-zero resistance across it — a short circuit, drawing very large current and typically destroying the meter. Connecting a voltmeter in series inserts a near-infinite resistance, which essentially stops the current and makes the circuit read almost nothing. Both mistakes are diagnosable from the reading, and the exam asks about them because they test whether you understand why the ideal resistances are what they are.
A battery of emf 12.0 V and internal resistance 0.50 Ω is connected to a 5.5 Ω resistor. Find the current, the terminal voltage, and the power dissipated inside the battery.
- 1.Total resistance = R + r = 5.5 + 0.50 = 6.0 Ω.
- 2.Current: I = ε/(R + r) = 12.0/6.0 = 2.0 A.
- 3.Terminal voltage: V = ε − Ir = 12.0 − (2.0)(0.50) = 12.0 − 1.0 = 11.0 V.
- 4.Power inside the battery: P = I²r = (2.0)²(0.50) = 2.0 W, dissipated as heat rather than delivered.
Internal resistance is in series with the external circuit, so add it to the total before computing current — do not compute the current from R alone and then subtract. Order matters here, and doing it the wrong way round gives a current that is too large.
As the current drawn from a real battery increases, its terminal voltage:
An ideal voltmeter has:
An ammeter is mistakenly connected in parallel with a resistor. The likely result is:
When a question distinguishes "emf" from "terminal voltage", it is testing internal resistance. Write V = ε − Ir explicitly rather than treating the two as the same number — the distinction is usually the point of the question.
Answer the 3 checkpoints as you read.
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