Measurements and Error Analysis: Dimensions, Least Count, Error Propagation, Four-Probe Resistance, Grounding, DC Supplies and Lock-in Detection
1. Units, dimensions and dimensional analysis
The SI has seven base quantities — length (m), mass (kg), time (s), current (A), temperature (K), amount (mol) and luminous intensity (cd) — and every other unit is a product of their powers. Since 2019 they are fixed by exact values of constants: h, e, k_B and N_A are defined numbers. A dimensional formula records the powers of M, L, T and A: force is [MLT⁻²], Planck’s constant (energy × time) is [ML²T⁻¹], and ε₀, from F = q²/(4πε₀r²), is [M⁻¹L⁻³T⁴A²].
Dimensional analysis rests on homogeneity: every term of a physical equation has the same dimensions. It checks equations and derives power-law dependences. For a pendulum, suppose T = k mᵃ lᵇ gᶜ: matching M⁰L⁰T¹ = Mᵃ Lᵇ⁺ᶜ T⁻²ᶜ gives a = 0, c = −½, b = ½, so T = k√(l/g). The method cannot give the constant k (it is 2π), cannot separate a sum of terms, and fails when two quantities with the same dimensions can combine into a dimensionless ratio.
2. Least count and significant figures
- Vernier calliper. Least count = 1 MSD − 1 VSD. If 10 vernier divisions equal 9 main-scale divisions of 1 mm, LC = 1 − 0.9 = 0.1 mm.
- Screw gauge. LC = pitch / number of circular divisions: 0.5 mm / 50 = 0.01 mm. Reading = main scale + (circular division × LC) − zero error, where a positive zero error (the zero of the circular scale below the line when the jaws are closed) is subtracted.
- Significant figures. All non-zero digits count; zeros between them count; leading zeros never count; trailing zeros count after a decimal point. So 0.004020 has four. In a product or quotient keep the fewest significant figures of the inputs; in a sum or difference keep the fewest decimal places.
3. Errors, the standard error and the propagation of uncertainty
Systematic errors shift every reading the same way — a miscalibrated scale, a zero error, a lead resistance — and are not reduced by repetition. Random errors scatter readings about the mean and are reduced by averaging. For N readings with standard deviation σ, the best estimate is the mean and its standard error is σ/√N: sixteen readings with σ = 0.8 give a standard error of 0.2. Independent results xᵢ ± σᵢ are combined by the weighted mean Σ(xᵢ/σᵢ²)/Σ(1/σᵢ²), with uncertainty 1/√Σ(1/σᵢ²); 10.0 ± 0.1 and 10.3 ± 0.2 combine to 10.06 ± 0.09.
| Result | Uncertainty (in quadrature) | Maximum (worst-case) estimate |
|---|---|---|
| q = x ± y | δq = √(δx² + δy²) | δq = δx + δy |
| q = c·x | δq = |c| δx | δq = |c| δx |
| q = xᵃ yᵇ / zᶜ | δq/q = √[(a δx/x)² + (b δy/y)² + (c δz/z)²] | δq/q = a δx/x + b δy/y + c δz/z |
| q = f(x) | δq = |df/dx| δx | δq = |df/dx| δx |
Worked: g = 4π²L/T² from a pendulum, with L uncertain by 0.5% and T by 0.2%. The power of T doubles its contribution, so δg/g = √[(0.5)² + (2 × 0.2)²] % = √0.41 % = 0.64%. The worst-case estimate adds them, 0.5 + 0.4 = 0.9%. For q = x²y/z³ with 1%, 2% and 1%, the contributions are 2%, 2% and 3%, and δq/q = √17 % = 4.12%.
4. Two-probe and four-probe resistance measurement
In a two-probe measurement the same pair of leads carries the current and senses the voltage, so the measured resistance is R_sample + 2R_lead + 2R_contact. That is harmless for a 10 kΩ resistor and fatal for a 10 mΩ one, or for a superconductor whose true resistance is zero. In a four-probe measurement a current source drives I through the outer pair, and a high-impedance voltmeter reads V across the inner pair. Almost no current flows in the voltage leads, so their lead and contact resistances drop no voltage, and V/I is the sample’s resistance between the inner contacts alone.
- Collinear four-point probe on a thick sample (spacing s, far from edges): ρ = 2πs (V/I).
- On a thin film (thickness t ≪ s): the sheet resistance is R_s = (π/ln 2)(V/I) ≈ 4.532 V/I, in ohms per square, and ρ = R_s t. A film giving V/I = 2.0 Ω has R_s ≈ 9.06 Ω/□.
- Thermoelectric offsets at the contacts are not removed by four probes; they are removed by reversing the current and averaging V(+I) and −V(−I).
5. Grounding, ground loops, DC power supplies and lock-in amplifiers
A ground loop forms when two instruments are grounded at different points and also joined by a signal cable’s shield: the two "grounds" differ by a small mains-frequency voltage, current circulates round the loop, and a hum at 50 Hz and its harmonics appears on the signal. The cures break the loop: a single-point (star) ground, one shield grounded at one end only, differential or isolated inputs, and keeping loop areas small against magnetic pick-up. The safety earth is never removed to cure hum.
A DC power supply is four stages. A transformer sets the voltage and isolates the circuit from the mains. A rectifier — half-wave, centre-tapped full-wave or bridge — makes it unidirectional; the bridge needs diodes of peak inverse voltage V_m and gives a ripple at twice the mains frequency. A capacitor filter holds the peak between charging pulses, leaving a peak-to-peak ripple V_r ≈ I/(fC) for half-wave and I/(2fC) for full-wave rectification. A regulator — a Zener diode with a series resistor, or an IC regulator — removes the remaining ripple and the dependence on load. Worked: full-wave at 50 Hz with C = 1000 μF and 100 mA load gives V_r = 0.1/(2 × 50 × 10⁻³) = 1.0 V.
A lock-in amplifier recovers a signal buried in noise by modulating the experiment at a reference frequency f_r — chopping a light beam, or driving a current at f_r — and then detecting only at f_r. The phase-sensitive detector multiplies the input V_s sin(ω_r t + φ) by the reference sin(ω_r t); the product has a DC term ½V_s cos φ and a term at 2f_r, and a low-pass filter keeps only the DC. Noise at other frequencies averages away, so the equivalent bandwidth is set by the filter time constant τ (about 1/(4τ) for a single-pole filter): a longer τ means a narrower bandwidth, a cleaner output and a slower response. A dual-phase lock-in also forms the quadrature output and reports R = √(X² + Y²) and φ independently.
Key takeaways
- Dimensional analysis fixes power laws — T = k√(l/g) — but never the constant, and cannot handle a sum of terms or a dimensionless ratio.
- Screw gauge: reading = main scale + divisions × LC − (signed) zero error; LC = pitch/divisions.
- For q = xᵃyᵇ, independent relative uncertainties add in quadrature weighted by the powers; the worst case adds them linearly. The standard error of a mean is σ/√N.
- Four probes remove lead and contact resistance because the voltage leads carry no current; current reversal removes thermal EMFs. Thin-film R_s = 4.532 V/I.
- Ripple of a capacitor filter is I/(fC) half-wave and I/(2fC) full-wave; a lock-in outputs ½V_s cos φ with a bandwidth set by its time constant.
Practice questions (14)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
g is found from g = 4π²L/T². The relative uncertainty in L is 0.5% and in T is 0.2%, independent and random. The relative uncertainty in g, in percent, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.64
δg/g = √[(δL/L)² + (2δT/T)²] = √(0.25 + 0.16) = √0.41 = 0.64%. Adding linearly gives 0.9%, the worst-case bound rather than the independent-error estimate, and forgetting the power 2 on T gives √0.29 = 0.54%.q = x²y/z³, where x, y and z have independent random relative uncertainties of 1%, 2% and 1%. The relative uncertainty in q, in percent, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 4.12
The contributions are 2 × 1 = 2%, 1 × 2 = 2% and 3 × 1 = 3%, so δq/q = √(4 + 4 + 9) = √17 = 4.12%. The division by z³ does not reduce its contribution — powers enter as magnitudes. The linear sum is 7%, and ignoring the powers gives √6 = 2.45%.A screw gauge has a pitch of 0.5 mm and 50 circular divisions. With the jaws closed the circular-scale zero lies 3 divisions below the reference line (a positive zero error). Measuring a wire, the main scale reads 3.5 mm and the circular scale 27 divisions. The corrected diameter, in mm, is ____.
Numerical answer — type the value.
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Answer: 3.74
LC = 0.5/50 = 0.01 mm. Observed = 3.5 + 27 × 0.01 = 3.77 mm; a positive zero error of 3 × 0.01 = 0.03 mm is subtracted, giving 3.74 mm. Adding it gives 3.80 mm, and ignoring it gives 3.77 mm.How many significant figures does 0.004020 have?
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Answer: A — 4
Leading zeros only locate the decimal point and are not significant; 4, the zero between 4 and 2, the 2 and the trailing zero after the decimal point are — four in all. Three drops the trailing zero, which counts because it was deliberately recorded.The dimensional formula of the permittivity of free space ε₀ is:
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Answer: A — [M⁻¹L⁻³T⁴A²]
From F = q²/(4πε₀r²), ε₀ = q²/(Fr²) = [A²T²]/([MLT⁻²][L²]) = [M⁻¹L⁻³T⁴A²]. [MLT⁻²A⁻²] is μ₀, from F/l = μ₀I²/(2πd); [MLT⁻³A⁻¹] is the electric field.The speed v of ripples on a liquid surface is assumed to depend on the surface tension S, the density ρ and the wavelength λ. Dimensional analysis gives:
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Answer: A — v ∝ √(S/(ρλ))
S is [MT⁻²], ρ is [ML⁻³]. With v = Sᵃρᵇλᶜ: M: a + b = 0; T: −2a = −1, so a = ½ and b = −½; L: −3b + c = 1, so c = −½. Hence v ∝ √(S/(ρλ)). The option with λ in the numerator is the gravity-wave form √(gλ), which needs g.Sixteen repeated readings of a length have a standard deviation of 0.8 mm. The standard error of their mean, in mm, is ____.
Numerical answer — type the value.
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Answer: 0.2
The standard error is σ/√N = 0.8/√16 = 0.8/4 = 0.2 mm. The standard deviation itself (0.8 mm) is the spread of single readings, and σ/N = 0.05 mm understates the uncertainty of the mean.Two independent measurements of a length are 10.0 ± 0.1 cm and 10.3 ± 0.2 cm. Their weighted mean, in cm, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 10.06
Weights are 1/σ²: 100 and 25. Mean = (100 × 10.0 + 25 × 10.3)/125 = (1000 + 257.5)/125 = 10.06 cm, with uncertainty 1/√125 = 0.09 cm. The unweighted average 10.15 cm gives the poorer measurement equal say.Compared with a two-probe measurement, a four-probe measurement of a small resistance:
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Answer: A — Removes the resistance of the leads from the result; B — Removes the contact resistance at the voltage probes; C — Relies on the voltmeter drawing negligible current
With separate current and voltage leads, and a voltmeter that draws almost no current, no IR drop appears across the voltage leads or their contacts, so V/I is the sample alone. Thermal EMFs are DC offsets that appear in the voltage circuit regardless; they are removed by reversing the current and averaging.A collinear four-point probe on a thin film (thickness much smaller than the probe spacing) gives V/I = 2.0 Ω. Using R_s = (π/ln 2)(V/I), the sheet resistance in Ω per square, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 9.06
π/ln 2 = 3.1416/0.6931 = 4.532, so R_s = 4.532 × 2.0 = 9.06 Ω/□. Using the thick-sample factor 2πs instead requires the spacing and gives a resistivity, not a sheet resistance; reporting V/I = 2.0 Ω itself omits the geometric factor.A full-wave rectifier on 50 Hz mains feeds a 1000 μF reservoir capacitor and a load drawing a steady 100 mA. The peak-to-peak ripple voltage, in V, is approximately ____.
Numerical answer — type the value.
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Answer: 1
Full-wave rectification recharges the capacitor twice per mains cycle, at 100 Hz, so V_r = I/(2fC) = 0.1/(2 × 50 × 1000 × 10⁻⁶) = 1.0 V. Using the half-wave form I/(fC) gives 2 V.A 50 Hz hum appears on a small signal when a sensor and an amplifier, grounded at different mains sockets, are joined by a shielded cable grounded at both ends. The most appropriate remedy is to:
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Answer: A — Break the ground loop — single-point grounding, or ground the shield at one end only
The hum is driven by the potential difference between the two grounds circulating current through the loop the shield closes. Breaking the loop removes the current. Lifting the safety earth is dangerous and forbidden, raising the gain amplifies the hum with the signal, and a longer cable enlarges the loop.A lock-in amplifier receives V_s sin(ωt + φ) and a reference at the same ω. After phase-sensitive detection and low-pass filtering, its in-phase output is proportional to:
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Answer: A — V_s cos φ
sin(ωt + φ) sin ωt = ½[cos φ − cos(2ωt + φ)]; the filter removes the 2ω term and leaves ½V_s cos φ. The sin φ term is the quadrature (Y) output of a dual-phase instrument, and noise away from ω averages to zero rather than appearing at the output.Which statements about a lock-in amplifier are correct?
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Answer: A — The experiment must be modulated at the reference frequency; B — Increasing the output time constant narrows the detection bandwidth; C — It can recover a signal whose amplitude is smaller than the broadband noise
The lock-in detects only at the reference frequency, so the signal must be put there by modulation. Its equivalent bandwidth is about 1/(4τ), so a longer τ rejects more noise, which is how a signal far below the broadband noise is recovered. It is a narrow-band detector — the opposite of measuring all frequencies.