Potentiometer Readings When the Driver Cell Is Changed
A Class XII group uses a 10.0 m uniform potentiometer wire with a 2.0 V accumulator as the driver cell. The rheostat is adjusted until the potential gradient along the wire is 0.200 V/m. A cell of unknown emf gives a balance point at 6.00 m from the end connected to the positive terminal. The group then connects a 5.0 Ω resistor across the cell and the balance point shifts to 5.50 m. From these two readings they calculate the internal resistance of the cell. Later, another group borrows the same wire and uses a 1.0 V driver cell without changing anything else; they cannot find any balance point for a 1.2 V cell and conclude that the cell is dead. The teacher asks them to check the driver-cell polarity and whether its emf is adequate, and to explain the role of the potential gradient. The galvanometer is ideal and the wire is uniform.
1. Calculate the emf and internal resistance of the cell from the two balance lengths, and explain physically why the balance length becomes shorter when the 5.0 Ω resistor is connected.
Read the worked answer
At balance the galvanometer carries no current, so the potentiometer measures the potential difference without drawing current from the cell under test.
Emf (open circuit): E = gradient × l₁ = 0.200 × 6.00 = 1.20 V.
With the 5.0 Ω resistor: the balancing p.d. is the terminal p.d. of the cell: V = 0.200 × 5.50 = 1.10 V. The current through the resistor is I = V/R = 1.10/5.0 = 0.220 A. Since V = E − Ir, r = (E − V)/I = (1.20 − 1.10)/0.220 = 0.455 Ω. Equivalently, r = R(l₁ − l₂)/l₂ = 5.0(0.50)/5.50 = 0.455 Ω.
Physically, with the resistor connected the cell drives current through its own internal resistance, so the terminal p.d. is less than the emf by Ir. A smaller p.d. needs a shorter length of wire to balance it, so the balance point moves from 6.00 m to 5.50 m. The difference l₁ − l₂ is proportional to the internal drop, which is why the formula works.
2. Explain why the second group cannot obtain a balance with the 1.0 V driver, even though the cell is not dead. What checks should they make?
Read the worked answer
A potentiometer can only balance a p.d. up to the total p.d. across the whole wire. With a 1.0 V driver and a 10 m wire, the maximum possible potential difference is 1.0 V, i.e. a gradient of 0.10 V/m. To balance 1.20 V would need l = 1.20/0.10 = 12.0 m, which is longer than the 10 m wire. No point on the wire can reach 1.20 V, so the galvanometer deflects one way everywhere.
Checks before blaming the cell:
- Driver emf must exceed the test emf by a comfortable margin; here 1.0 V < 1.2 V, so balance is impossible.
- Polarity: the driver cell and the cell under test must be connected so that their emfs oppose each other (positive terminals joined through the galvanometer circuit in the standard way). If they aid each other, the galvanometer deflects one way at every point.
- Continuity and contact: a broken wire or dirty contact can also prevent a balance, but that would usually give no deflection rather than a one-sided deflection.
- The rheostat must not be set so that the full wire p.d. is too small; even with a 2 V driver, a badly set rheostat can reduce the gradient and push the balance point off the wire.
Conclusion: the cell is not dead; the driver is inadequate. Replace it with a driver of emf greater than 1.2 V, e.g. 2 V, and adjust the rheostat.
3. Suppose the group uses a 1.5 V driver with the rheostat set so that the full 10 m wire carries the whole 1.5 V. Where is the balance point for the 1.20 V cell, and how does the sensitivity compare with the original 0.200 V/m setting? What is the maximum emf this arrangement can measure?
Read the worked answer
With 1.5 V across 10.0 m, the gradient is k = 1.5/10.0 = 0.150 V/m.
The balance length for 1.20 V is l = E/k = 1.20/0.150 = 8.0 m.
This is still on the wire, so the measurement is possible. Compared with the original 6.0 m at 0.200 V/m, the balance point is farther along the wire. A given small change ΔE in the cell's emf produces a shift Δl = ΔE/k; a smaller gradient gives a larger shift, so the 0.150 V/m setting is more sensitive (provided the balance point stays on the wire).
The maximum measurable emf is the full-wire p.d., 1.5 V. A driver of exactly 1.2 V would put the balance point at the far end (10.0 m), leaving no margin: any small drop in the driver emf, or any contact resistance, would make balance impossible. The practical rule is to choose a driver emf comfortably above the emf to be measured, but not so high that the balance length becomes short and sensitivity is lost.