Potentiometer internal-resistance discrepancy
A student uses a potentiometer with a 2.00 V driver cell and a uniform 10.00 m resistance wire. With the unknown dry cell on open circuit, the balance point is at 6.00 m. The cell is then shunted by a 10.0 Ω resistor and the balance point moves to 3.00 m. A second trial with a 5.0 Ω shunt gives 2.20 m. From the first shunt the student calculates an internal resistance of 10.0 Ω; from the second, 8.6 Ω. The student argues that the 5.0 Ω result is unreliable because the cell is 'being drained', while a partner says the driver cell must be running down. The laboratory has a standard cell, a high-impedance voltmeter, and a reversing switch available.
1. Compute the open-circuit emf and the internal resistance from each shunt trial. Are the two values consistent?
Read the worked answer
Open circuit: E = (6.00/10.00)(2.00 V) = 1.20 V. For R = 10.0 Ω: V = (3.00/10.00)(2.00) = 0.600 V, I = 0.600/10.0 = 0.0600 A, so r = (1.20 - 0.600)/0.0600 = 10.0 Ω. For R = 5.0 Ω: V = (2.20/10.00)(2.00) = 0.440 V, I = 0.440/5.0 = 0.0880 A, so r = (1.20 - 0.440)/0.0880 = 8.64 Ω. The two values differ by 1.36 Ω, about 14%, which is much larger than the reading uncertainty of a metre wire (typically 0.2 Ω here). They are not consistent.
2. Which explanation is quantitatively plausible? What check would you perform with the available equipment?
Read the worked answer
If r were really 10.0 Ω, the 5.0 Ω trial should give V = 1.20 × 5.0/(5.0 + 10.0) = 0.400 V, balancing at (0.400/2.00) × 10.00 = 2.00 m, not 2.20 m. To observe 2.20 m with V = 0.400 V, the driver emf would have to be Vd = (0.400/2.20) × 10.00 = 1.82 V, a 9% drop. A run-down driver cell therefore explains the longer balance point. If the dry cell polarized and its internal resistance increased, its terminal voltage would fall and the balance length would become shorter, not longer. A decrease of r to 8.6 Ω would also explain the result, but that requires a 14% change and is less likely. Check by re-measuring the open-circuit balance immediately after the 5 Ω trial; if it is no longer 6.00 m, the driver has changed. A standard cell can calibrate the driver, and a high-impedance voltmeter can check the shunt resistor value.
3. Could a 5% tolerance on the 5.0 Ω resistor explain the 2.20 m reading?
Read the worked answer
At E = 1.20 V and r = 10.0 Ω, a balance at 2.20 m requires V = 0.440 V. Then 0.440 = 1.20 R/(R + 10.0), giving 0.440R + 4.40 = 1.20R, so R = 5.79 Ω. That is 15.8% above 5.0 Ω, outside a ±5% tolerance. A 5% resistor could not explain the reading; a 15% error or a wrong resistor would be needed.