Internal Resistance with a Non-Ideal Ammeter
A student is asked to determine the internal resistance of a dry cell marked 1.50 V using a voltmeter, an ammeter, and a 10.0 Ω resistor. The ammeter has a resistance of 0.50 Ω and the digital voltmeter has a resistance of 10 MΩ. The circuit has the cell, ammeter, and 10.0 Ω resistor in series; the voltmeter is connected across the series combination of the cell and ammeter, which is also across the resistor. The ammeter reads 0.120 A and the voltmeter reads 1.20 V. The student uses r = (1.50 − 1.20)/0.120 = 2.5 Ω and reports this as the internal resistance. A second student says the true value is smaller and that the 10.0 Ω resistor is not the only resistance outside the cell. The teacher asks the pair to identify the systematic error, recalculate, and explain why changing the resistor to 100 Ω would not remove the error.
1. Identify the systematic error in the student’s calculation and recalculate the internal resistance.
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The student has assumed that the whole drop from the marked emf to the voltmeter reading occurs across the internal resistance. In fact the ammeter is inside the part of the circuit across which the voltmeter is connected, so the drop is shared: E = V + I(r + R_A). Here E = 1.50 V, V = 1.20 V, I = 0.120 A, so r + R_A = (1.50 − 1.20)/0.120 = 2.5 Ω. Since R_A = 0.50 Ω, the true internal resistance is r = 2.0 Ω. The check V/I = 1.20/0.120 = 10.0 Ω confirms that the 10.0 Ω load is consistent; it is the extra 0.50 Ω of the ammeter that the student has wrongly assigned to the cell.
2. Show quantitatively whether the 10 MΩ voltmeter can be blamed for the discrepancy.
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The voltmeter is in parallel with the 10.0 Ω resistor. Its equivalent resistance is (10.0 × 10 × 106)/(10.0 + 10 × 106) ≈ 9.99999 Ω, a change of about one part in a million from 10.0 Ω. The current through the voltmeter is about 1.20 V / 10 × 106 Ω = 1.2 × 10−7 A, compared with 0.120 A through the load, so it is negligible. The voltmeter loading therefore cannot explain a 0.5 Ω discrepancy; the ammeter resistance does.
3. Why would replacing the 10.0 Ω resistor by 100 Ω not remove the systematic error? What is a better procedure?
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If the cell emf is still taken as exactly 1.50 V, the student’s formula always gives (E − V)/I = r + R_A, so the same +0.50 Ω offset remains regardless of the load. With 100 Ω the current would be only about 0.0146 A and E − V only about 0.036 V, so meter resolution, cell polarisation, and any error in the assumed emf would become relatively more important. A better method is to measure V and I for several different loads and plot V against I: the intercept gives E and the slope gives −(r + R_A), allowing R_A to be subtracted. Alternatively, use a potentiometer or a high-impedance null method to measure the emf directly and connect the ammeter outside the voltmeter’s terminals.