Internal Resistance with the Voltmeter in the Wrong Place
A student determines the emf and internal resistance of a cell. The circuit contains the cell, an ammeter, and a variable resistor in series. The voltmeter is connected across the variable resistor, not across the cell, and the student does not notice. The voltmeter is assumed ideal. The readings are: I/A: 0.100, 0.200, 0.300, 0.400, 0.500 V/V: 1.42, 1.34, 1.26, 1.18, 1.10 The student plots V against I, draws a best-fit straight line, and finds intercept 1.50 V and gradient −0.80 Ω. The ammeter has a resistance of 0.15 Ω. The cell is rated 1.5 V. A classmate says the internal resistance is 0.80 Ω and the cell is therefore 'in poor condition' because it should be much lower.
1. Use the first and last data points to confirm the gradient and intercept. State the emf and the apparent internal resistance, and show that the intermediate points are consistent with a straight line.
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Using the first and last points: gradient = (1.10 − 1.42)/(0.500 − 0.100) = −0.32/0.400 = −0.80 Ω. The intercept is 1.50 V, since V = 1.50 − 0.80I.
Check an intermediate point: at I = 0.300 A, V = 1.50 − 0.80 × 0.300 = 1.26 V, which matches the table. At I = 0.200 A, V = 1.34 V, also matches. The data are consistent with a straight line of intercept 1.50 V and gradient −0.80 Ω.
The student therefore reports emf = 1.50 V and apparent internal resistance = 0.80 Ω.
2. Explain why the ammeter resistance is included in the gradient in this circuit, but would not be included if the voltmeter were connected across the cell. Calculate the corrected internal resistance and comment on the classmate's conclusion.
Read the worked answer
Because the voltmeter is across the variable resistor, V = IR. The cell sees the variable resistor, the ammeter and its own internal resistance in series, so E = I(R + R_A + r). Substituting R = V/I gives E = V + I(R_A + r), or V = E − I(r + R_A).
Thus the gradient is −(r + R_A), not −r. The ammeter resistance is included in the measured slope. The corrected internal resistance is r = 0.80 − 0.15 = 0.65 Ω.
If the voltmeter were connected across the cell instead, it would read the cell terminal voltage V = E − Ir. The ammeter resistance would still affect the current for a given setting of the variable resistor, but it would not appear in the V–I line of the cell; the gradient would be −r directly. The classmate's conclusion that the cell is in 'poor condition' is not supported: 0.65 Ω is a plausible internal resistance, and the 0.80 Ω value was an artefact of the circuit connection.
3. Explain why a best-fit line through all five points is more reliable than a calculation from the first and last points. Identify one further systematic error that could affect the result even if the line is fitted correctly.
Read the worked answer
A best-fit line through all five points uses the whole data set. It averages the random reading uncertainties, checks that the relationship is actually linear, and reduces the influence of any one poor reading. A two-point calculation uses only the extremes and gives no check on curvature or scatter; a single misread value at either end would shift the result.
A further systematic error: the voltmeter is assumed ideal, but a real voltmeter has finite resistance and draws current. That current is not measured by the ammeter, so the cell current is slightly larger than I. The effect is largest at low currents and would make the intercept slightly too low and the gradient slightly wrong. Other acceptable answers: the cell's emf or internal resistance changes as it warms during the run, or contact resistance in the variable resistor changes.