Choosing the Better Ammeter–Voltmeter Circuit
During a physics practical examination, a candidate must determine the resistance of a coil expected to be about 25 Ω at room temperature. The available elements are a 6 V battery, a rheostat, an ammeter of resistance 0.40 Ω, a voltmeter of resistance 4.0 kΩ, and the coil. Two wiring plans are proposed. In Plan A, the ammeter is in series with the coil and the voltmeter is connected only across the coil. In Plan B, the ammeter is still in series, but the voltmeter is connected across the series combination of the ammeter and the coil. With the rheostat adjusted, Plan A gives 2.50 V and 0.101 A; Plan B gives 2.54 V and 0.100 A. The candidate must report a single resistance value with justification. The examiner also asks whether each raw ratio V/I is an overestimate or underestimate, and whether the two corrected values agree within the meter least counts of 0.01 V and 0.001 A.
1. Calculate the raw resistance from each plan. Using the meter resistances, determine the corrected coil resistance and decide which plan gives the smaller systematic error.
Read the worked answer
Plan A raw ratio: R_A(raw) = 2.50/0.101 = 24.75 Ω. The voltmeter draws a small current, so the ammeter reads more than the coil current. I_V = 2.50/4000 = 0.000625 A, so I_coil = 0.101 − 0.000625 = 0.100375 A. Corrected R = 2.50/0.100375 = 24.91 Ω.
Plan B raw ratio: R_B(raw) = 2.54/0.100 = 25.40 Ω. The voltmeter includes the ammeter drop, so V_coil = 2.54 − (0.100)(0.40) = 2.50 V. Corrected R = 2.50/0.100 = 25.00 Ω.
Plan A's raw error is about −0.25 Ω (−1.0%), while Plan B's raw error is about +0.40 Ω (+1.6%). Plan A has the smaller raw systematic error because the voltmeter's loading is negligible for a 25 Ω coil, whereas the ammeter drop is not negligible relative to the coil voltage. After correction, the two values agree within about 0.1 Ω, consistent with the stated least counts. Either can be reported as approximately 25 Ω, but Plan A is the better default choice.
2. The candidate reports 25.0 Ω. A classmate objects that Plan A must be wrong because the ammeter reads more current than actually flows through the coil. Does this objection change the reported value? Explain how the correction should be applied.
Read the worked answer
The objection is physically correct but numerically small. In Plan A the ammeter measures I_coil + I_voltmeter, not I_coil alone, so the raw ratio V/I is slightly low. The correction is R = V/(I − V/R_V). Here the correction is only about 0.16 Ω, so the reported value remains approximately 25 Ω.
The key judgement is whether the correction is larger than the reading uncertainty. The current correction, 0.000625 A, is smaller than the 0.001 A least count of the ammeter; the resulting resistance change is below the 0.1 Ω level implied by the voltage least count. Therefore the candidate need not change the reported value, but should state that the raw Plan A value is a slight underestimate and that the corrected value is about 24.9 Ω.
3. Midway through the experiment the coil warms up. If the rheostat is not adjusted, state the direction of drift in the Plan A ammeter and voltmeter readings and in the calculated R.
Read the worked answer
For a metallic coil, resistance increases with temperature. With the battery and rheostat unchanged, the total circuit resistance rises, so the current falls: the ammeter reading decreases.
The voltmeter reading is V_coil = V_battery × R_coil/(R_coil + R_other). As R_coil increases, the coil takes a larger fraction of the battery voltage, so the voltmeter reading increases even though the current decreases. Therefore the ratio V/I increases, and the calculated resistance drifts upward. This is not a circuit fault; it is the expected temperature coefficient of resistance. To compare with the room-temperature value, the candidate should take readings quickly, keep the current low, or correct for the measured temperature rise.