Potentiometer That Cannot Balance a 1.5 V Cell
A student sets up a potentiometer to compare the emf of a dry cell with a standard cell. The slide wire is 4.00 m long and has resistance 8.0 Ω. The driver is a 2.0 V accumulator of negligible internal resistance, connected in series with a protective resistance box. With the resistance box set at 6.0 Ω, the standard cell of emf 1.018 V balances at 3.56 m. The student then tries to balance an unknown cell and finds no balance point anywhere on the wire, even after reversing the cell terminals. The unknown cell is known to be about 1.5 V. The connections are otherwise correct, and the galvanometer deflects in the same direction at both ends of the wire.
1. Calculate the potential gradient and the maximum emf measurable with this setting. Explain why the 1.5 V cell cannot be balanced.
Read the worked answer
Total resistance in the driver circuit = 8.0 + 6.0 = 14.0 Ω, so I = 2.0/14.0 = 0.143 A. Potential drop across the whole wire = I × 8.0 = 1.14 V. Potential gradient = 1.14/4.00 = 0.286 V m⁻¹. The maximum measurable emf is the full wire drop, about 1.14 V. For 1.5 V the balance length would need to be L = 1.5/0.286 = 5.25 m, longer than the 4.00 m wire, so no balance point exists. Reversing the unknown cell only reverses the galvanometer deflection; it cannot reduce the required length. The same-direction deflection at both ends confirms that the wire never provides a potential difference equal and opposite to the cell emf.
2. The student proposes increasing the protective resistance to 10 Ω to protect the galvanometer and obtain a balance. Evaluate this proposal and suggest one change that would actually work.
Read the worked answer
Increasing the protective resistance lowers the current and therefore lowers the potential gradient, making the problem worse. With 10 Ω, I = 2.0/18.0 = 0.111 A, the wire drop is only 0.889 V, and the gradient is 0.222 V m⁻¹. A 1.5 V cell would then need 6.75 m, even further beyond the wire. To bring the balance point onto the wire, the wire must have a larger total potential drop. Reducing the protective resistance to 2.0 Ω gives I = 2.0/10.0 = 0.200 A, a wire drop of 1.60 V and a gradient of 0.400 V m⁻¹; the balance length for 1.5 V becomes 3.75 m, which fits. Raising the driver emf instead, for example to 3.0 V with 6.0 Ω, gives a gradient of about 0.429 V m⁻¹ and a balance length of 3.5 m. Lowering the protective resistance is the simpler fix.
3. What does the galvanometer deflecting in the same direction at both ends tell you about the fault? If the unknown cell had been connected with reversed polarity, how would the observations have differed?
Read the worked answer
A potentiometer balance requires the unknown cell to oppose the potential drop along the wire. As the jockey moves, the potential difference between the ends changes sign relative to the cell, so a correct setup with a measurable emf must show a zero crossing, with the galvanometer deflecting in opposite directions near the two ends. A same-direction deflection at both ends means the wire voltage never matches the cell emf; either the cell emf exceeds the full wire drop or the cell is connected the wrong way round. If the polarity were reversed, the galvanometer would deflect strongly one way at one end and the opposite way at the other, and simply reversing the cell terminals would produce a balance. Since reversing the terminals changed nothing, the fault is the insufficient wire drop, not polarity.