Ammeter Placement in a Two-Resistor Circuit
A student builds a circuit with a 9.0 V battery (internal resistance 0.50 ohm), a 10 ohm resistor, and a 20 ohm resistor in series. They insert an ammeter (internal resistance 2.0 ohm) in series to measure current and read 0.27 A. They then place a voltmeter (internal resistance 1.0 M ohm) across the 20 ohm resistor and read 5.1 V. A classmate says the ammeter reading should be 9.0/30 = 0.30 A and that the voltmeter is 'loading' the circuit, so the readings are wrong.
1. Compute the expected ammeter reading including all resistances and compare with the measured 0.27 A.
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Total resistance = 10 + 20 + 2.0 + 0.50 = 32.5 ohm. Current = 9.0/32.5 = 0.277 A, which rounds to 0.28 A, not 0.30 A.
The measured 0.27 A is consistent within reading error. The classmate's 0.30 A ignores both the ammeter resistance and the battery internal resistance, so it is the wrong prediction. The ammeter does perturb the circuit, but the perturbation is small and predictable.
2. Compute the expected voltmeter reading across the 20 ohm resistor and assess whether the voltmeter is 'loading' the circuit significantly.
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With the ammeter in place, the current is 0.277 A through the 20 ohm resistor, giving V = 0.277 * 20 = 5.54 V.
With the voltmeter (1.0 M ohm) in parallel with 20 ohm, the parallel resistance is (20 * 1e6)/(20 + 1e6) ≈ 20.0 ohm (change < 0.002%). The loading is negligible: the reading should be about 5.5 V, and the measured 5.1 V differs by about 8%, which is more likely a meter calibration or reading error than loading.
So the classmate's 'loading' explanation is wrong here. A 1 M ohm voltmeter across a 20 ohm resistor barely changes anything. Loading would matter if the resistor were comparable to 1 M ohm.
3. The student wants to measure the 20 ohm resistor's value directly. Describe the correct method and the main pitfall.
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Use an ohmmeter with the resistor removed from the circuit (or at least isolated), because an ohmmeter applies its own small voltage and any parallel path or battery would corrupt the reading. Alternatively, measure V and I simultaneously with the voltmeter across the resistor and the ammeter in series, then compute R = V/I, correcting for meter resistances.
The main pitfall is leaving the battery connected: the ohmmeter would be damaged or would read the battery's EMF rather than the resistance. A second pitfall is that the ammeter's 2.0 ohm resistance is not negligible compared with 20 ohm, so the V/I method must account for it or use a four-terminal measurement.