One Potentiometer, Three Cells, One Driver
A Class XII student must measure two emfs on the same potentiometer. The slide wire is 10 m long with a resistance of 20 Ω, the driver is a 2.0 V lead–acid cell, and a 1.018 V standard cell is available for calibration. With the standard cell in circuit the balance point is found at 509 cm, so the potential gradient is 0.200 V m⁻¹. The two test objects are a fresh Daniell cell, expected to be about 1.10 V, and a Chromel–Alumel thermocouple with its cold junction in ice at 0 °C and its hot junction in boiling water, expected to give about 4.1 mV. The jockey position can be read to the nearest millimetre. The student also wants to check a nickel–cadmium pack labelled 2.4 V. A senior student warns him to take readings quickly and never to leave the driver circuit closed for long, because the wire and the driver cell both warm up.
1. From the calibration data, find the potential gradient and the balance length expected for the Daniell cell. Why must the driver current be identical during calibration and during measurement, and what error appears if the driver emf has fallen by 2 % since calibration while the student still uses the old gradient?
Read the worked answer
Gradient. k = E_standard / L_standard = 1.018 V / 5.09 m = 0.200 V m⁻¹ = 2.00 mV cm⁻¹.
Daniell cell. E = kL, so L = 1.10 / 0.200 = 5.50 m = 550 cm. This lies on the wire (550 cm < 1000 cm), so the measurement is possible.
Why the current must be constant. The wire is a uniform resistance; the p.d. per metre is k = Iρ/A. The current I is fixed by the driver emf and the whole circuit resistance (wire + rheostat + driver internal resistance + contact resistances). Any change in I changes k, and a length then no longer corresponds to a fixed voltage. Warming of the wire raises its resistance and lowers I; a falling driver emf does the same.
2 % drop. New gradient k′ = 0.98k. The same emf now balances at L′ = E/k′ = L/0.98. The student reports E_reported = kL′ = E/0.98 ≈ 1.02E, i.e. about 2 % too high (≈ 22 mV high on 1.10 V). Hence the rule: take the reading at once, do not touch the rheostat between calibration and measurement, and re-calibrate if the session is long.
2. Calculate the balance length for the thermocouple at the present gradient and estimate the percentage error from a ±1 mm reading. How could the student make the thermocouple measurement trustworthy, and what must be recalculated if he does so?
Read the worked answer
Balance length. L = 4.1 × 10⁻³ / 0.200 = 2.05 × 10⁻² m ≈ 2.05 cm = 20.5 mm.
Error. ±1 mm on 20.5 mm is 1/20.5 ≈ 4.9 %, i.e. ±0.20 mV on 4.1 mV. That is far too coarse for a thermocouple calibration.
Remedy - reduce the gradient. To reach about 1 % precision the balance length must be ≈ 100 mm, so k must become 4.1 × 10⁻³/0.100 = 0.041 V m⁻¹, one-fifth of the present value. Since k = IR_wire/L_wire with R_wire = 20 Ω over 10 m, the present current is I = kL/R = 0.200 × 10/20 = 0.10 A. The new current must be 0.0205 A, so the total circuit resistance must be 2.0/0.0205 ≈ 98 Ω, i.e. add about 78 Ω of stable series resistance.
What must be redone. The gradient, not just the length, has changed, so the standard-cell calibration must be repeated with the new setting; the galvanometer must be sensitive enough at the smaller current; and thermal emfs in the copper leads and the thermocouple's own resistance now matter at the 0.04 mV level. Also check the polarity - a reversed couple drives the jockey the wrong way.
3. Why can the 2.4 V nickel–cadmium pack not be measured on this arrangement? Explain the symptom the student would see, state the minimum change needed, and say why this change cannot be combined with the thermocouple setting.
Read the worked answer
Condition for balance. The p.d. available across the whole wire is kL_max = 0.200 × 10 = 2.0 V. A potentiometer can only measure emfs smaller than this, because the unknown must be balanced against a fraction of the wire's p.d. Since 2.4 V > 2.0 V, no balance point exists.
Symptom. The galvanometer deflects the same way for every jockey position, including at the far end of the wire - the classic signature of an unknown emf larger than the driver's.
Minimum change. Raise k above 2.4/10 = 0.24 V m⁻¹, so the driver must exceed 2.4 V; in practice a 3 V driver (two cells, or a 0–3 V stabilised supply) is used, keeping k ≈ 0.28–0.30 V m⁻¹ so the balance length (≈ 80–86 cm) stays on the wire and away from the ends.
Why not combined. The thermocouple needed k ≈ 0.041 V m⁻¹ - the opposite direction. Sensitivity (long balance length) and range (large emf measurable) pull against each other in one instrument. The student must re-calibrate against the standard cell for each setting, because a calibration made at one gradient is meaningless at another.