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Internal resistance practical questions

Use terminal voltage and current data to find emf, internal resistance and load power.

Practise

Signed V–I gradient

Calculate the signed gradient of the supplied V–I line using the first and last readings. Give its value in Ω (V/A). Give 3 significant figures.

Practical setup and assumptions

A cell supplies a variable external resistor. An ideal ammeter measures the circuit current and an ideal voltmeter measures terminal p.d.

The graph follows V = ε − Ir. Assume the cell’s emf and internal resistance remain constant and the leads have negligible resistance.

Readings for this dataset
Same-cell readings
I / AV / V
0.11.46
0.21.42
0.31.38
0.41.34
0.51.3

The linked tasks use the same readings. These are generated practice data: treat them as exact unless an uncertainty is stated. Use the stated uncertainty rather than estimating it from scatter.

Terminal potential difference against currentPractice readings with a given straight line; use the line shown for calculations. Horizontal axis: Current / A, 0 to 0.8. Vertical axis: Terminal p.d. / V, 0 to 2.5. Exact values are also listed below the graph.00.10.20.30.40.50.60.70.800.250.50.7511.251.51.7522.252.5Current / ATerminal p.d. / V12345
Read the axes and units. The values are also available in the table below.
Larger graph
Terminal potential difference against currentPractice readings with a given straight line; use the line shown for calculations. Horizontal axis: Current / A, 0 to 0.8. Vertical axis: Terminal p.d. / V, 0 to 2.5. Exact values are also listed below the graph.00.10.20.30.40.50.60.70.800.250.50.7511.251.51.7522.252.5Current / ATerminal p.d. / V12345
Graph values
Coordinates
PointCurrent / ATerminal p.d. / V
10.11.46
20.21.42
30.31.38
40.41.34
50.51.3

Use decimals or e notation, e.g. 4.2e-3. Enter the unit separately.

Hint

Gradient = ΔV/ΔI. V decreases with I, so the gradient is negative.

Worked solution
  1. Gradient = (1.3 − 1.46)/(0.50 − 0.10) = −0.4 Ω.
  2. Answer: -0.400 Ω (3 significant figures).
Next task in this dataset
Work through this dataset
  1. Choose V–I axes
  2. Signed V–I gradient (current)
  3. EMF from the intercept
  4. Internal resistance
  5. Lost volts
  6. External resistance
  7. Current with a new load
  8. Power in the load
  9. Absolute uncertainty in r
  10. Voltmeter zero offset

These tasks keep the same readings. Another question starts a new dataset. Your answers are checked separately.

Another question
Question type: Signed V–I gradient
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Find emf and internal resistance from V = ε − Ir

A terminal-potential-difference–current line passes through (0.20 A, 1.44 V) and (0.60 A, 1.32 V). Gradient = −0.30 V A⁻¹, so internal resistance is 0.30 Ω. Extrapolating to zero current gives emf 1.50 V.

Internal resistance is the negative of this line’s gradient, so it is positive here. The intercept gives emf only for these axes and the stated source model. A terminal reading while current flows is generally below the emf for a source delivering energy.

Practise circuit current and potential difference

Read the line as V = ε − Ir

Plot terminal p.d. V vertically against current I. The intercept at zero current is emf ε, and the signed gradient is −r. Internal resistance r is positive.

Separate the cell and the external load

Lost volts inside the cell are Ir = ε − V. The external resistance is V/I; ε/I includes both external and internal resistance. With a new load R, current is ε/(R + r). Power delivered to the load is VI; I²r is dissipated internally.

Interpret a measurement offset

A constant additive voltmeter offset shifts the V–I intercept but leaves the gradient unchanged. Gain error, contact resistance and changing cell parameters need different treatment. The case assumes constant ε and r and ideal meters.

A-level relevance

AQA 7408: 3.5.1.6; required practical 6 (selected); OCR A H556: 4.3.2(c) (selected). Tasks cover the selected skills listed above.

AQA electricity specification · OCR A specification · AQA practical handbook · OCR practical skills handbook

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