Potentiometer With No Balance Point
A potentiometer is wired with a 10.0 m uniform resistance wire, a driver cell of emf 2.0 V and negligible internal resistance, and a standard cell of emf 1.20 V used for calibration. With the standard cell in the circuit and the galvanometer key closed, the balance point is at 6.00 m measured from the end where the driver cell is connected. The standard cell is replaced by a cell of unknown emf E, and the balance point moves to 7.20 m. The same unknown cell is then shunted by a 2.0 Ω resistor, and the balance point returns to 6.00 m. A second student repeats the calibration with longer leads, finds no balance point anywhere on the wire, and concludes that the driver cell is defective. The demonstrator asks for a reasoned analysis of that conclusion before accepting it, and for the values of E and of the internal resistance of the unknown cell.
1. Is the second student's conclusion that the driver cell is defective justified? Give the reasoning and the checks that would distinguish the possible causes.
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Not justified on the evidence available - it is one hypothesis among several, and the cheapest checks should come first.
The driver establishes a potential drop of at most 2.0 V across the whole wire. A cell of emf below 2.0 V, connected so that it opposes the driver, must therefore have a balance point somewhere on the wire. Since 1.20 V < 2.0 V, a null point is guaranteed if the circuit is correct.
Possible causes of no null point:
- Reversed standard cell - the two emfs then add, the galvanometer deflects the same way everywhere, and no null exists.
- Open circuit - a loose terminal, a broken joint or a jockey not making contact means no current flows in the wire, so there is no potential gradient at all and no point can balance 1.20 V.
- Driver unable to deliver current - a run-down cell can read about 2 V on a high-resistance voltmeter yet have a terminal p.d. that collapses on load, lowering the potential gradient so much that the null point lies beyond the wire.
- Faulty galvanometer or key.
Discriminating checks: reverse the standard cell and see whether the deflection direction changes; measure the driver's terminal p.d. while it is in the circuit; test continuity of the wire and leads; move the jockey slowly over the whole length and watch for any deflection minimum. Only if the driver holds its terminal p.d. under load, the polarity is correct and the wire is continuous should a defective driver be blamed.
2. Calculate the emf E of the unknown cell and its internal resistance r.
Read the worked answer
Potential gradient. The standard cell (1.20 V) balances at 6.00 m:
$$k = \frac{1.20}{6.00} = 0.200\ \text{V m}^{-1}.$$
Emf of the unknown cell. The balance length is 7.20 m:
$$E = k,l = 0.200\times7.20 = 1.44\ \text{V}.$$
Internal resistance. With the 2.0 Ω shunt connected, the balance length is again 6.00 m, which measures the terminal potential difference of the cell while it supplies current:
$$V = 0.200\times6.00 = 1.20\ \text{V}.$$
Shunt current: $I = V/R = 1.20/2.0 = 0.60$ A.
$$r = \frac{E - V}{I} = \frac{1.44-1.20}{0.60} = 0.40\ \Omega.$$
Equivalently $r = R\left(\frac{E}{V}-1\right) = 2.0\left(\frac{1.44}{1.20}-1\right) = 2.0\times0.20 = 0.40\ \Omega$, which confirms the result. Note that the potentiometer measures the emf only when the cell draws no current; the shunt deliberately makes it draw current so that the internal drop becomes measurable.
3. Suppose the unknown cell had an emf of 2.4 V. What would happen, and what could be changed to make the measurement possible?
Read the worked answer
No balance point would be found anywhere on the wire. The driver can produce a maximum potential difference of 2.0 V between the ends of the 10.0 m wire, and the balance condition requires the potential difference tapped off to equal 2.4 V. Since $2.4 > 2.0$, the required length would be
$$l = \frac{2.4}{0.200} = 12.0\ \text{m},$$
which is longer than the wire, so the galvanometer would deflect one way over the entire length.
Remedies, in order of practicality:
- use a driver cell of emf comfortably above 2.4 V (say 3–4 V), or
- keep the driver but lengthen the wire / add series resistance so that the potential gradient falls: with the same 2.0 V driver and a 12 m effective length the gradient becomes 0.167 V/m and 2.4 V would balance at 14.4 m - so in practice a larger driver emf is the real fix;
- alternatively, measure the cell against a known fraction using a potential divider.
Simply pushing the jockey harder or changing the galvanometer cannot help, because the limitation is the total potential drop available along the wire, not the sensitivity of the detector.