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EMF and internal resistance

Curricula: A-level RP6, IB B.5, 1re (source réelle), Sek II

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Aim

Determine the EMF and internal resistance of a battery from how the voltage across the external circuit depends on the current.

The bench

A 9 V (PP3) battery; its internal resistance is unknown. The load is a 50 Ω rheostat (slider in the middle, 25 Ω) through a switch (closed) and an ammeter. The voltmeter is across the whole external circuit: from the switch input to the rheostat slider.

Procedure

  1. Open the switch and note the voltage with no current — this is the EMF.
  2. Close the switch; moving the rheostat slider, take 6–8 "current–voltage" pairs.
  3. Plot the graph \(U(I)\): the intercept with the \(U\) axis gives the EMF, the magnitude of the slope gives the internal resistance.
  4. Find the slider position at which the power in the external circuit, \(P = U \cdot I\), is greatest.
Expected result

With no current — 9.00 V. At 25 Ω: about 337 mA and 8.48 V. The points lie on the straight line \(U = E - I \cdot r\) with a slope of about −1.55 Ω: 1.5 Ω inside the battery and 0.04 Ω from the two cables between it and the voltmeter. The power in the external circuit is greatest when its resistance equals the internal resistance: about 2.9 A and 4.5 V — half the EMF, 13 W. At the left end of the rheostat the current reaches 5 A — the battery has no protection, and the current is limited only by its internal resistance and the cables. On a real bench this must not be done: a 9 V (PP3) battery is rated for tens of milliamps, at amps it heats up and its voltage drifts, and the rheostat winding overheats. Real measurements are taken at currents up to 0.5 A, and the greatest power is found from the graph.

Questions

  1. Why does the terminal voltage drop as the current increases?
  2. How can the short-circuit current be found from the graph?
Answers
  1. Part of the EMF is spent driving the current through the battery itself: its internal resistance takes a voltage \(I \cdot r\) that grows with the current, and the external circuit gets the rest, \(U = E - I \cdot r\). At 337 mA the loss is about 0.337 A · 1.55 Ω ≈ 0.52 V, so the voltmeter shows 8.48 V instead of 9.00 V.
  2. Extend the straight line to the \(I\) axis, where \(U = 0\): the intercept is the short-circuit current \(I = E / r\) ≈ 9.00 V / 1.55 Ω ≈ 5.8 A, with \(r\) taken from the slope. On the bench the current stops at about 5.2 A at the left end of the rheostat: the ammeter's shunt and the cables keep about 0.2 Ω in the external circuit, and the voltmeter still shows about 1 V.