Recall and apply the principle of the potentiometer as a null‑method technique for comparing potential differences (emf’s) without drawing current from the source.
| Section | Syllabus Code | Topic Covered | Related Practical Skills (Paper 3/5) |
|---|---|---|---|
| 10.1 | Potential difference & e.m.f. | Definition of emf & p.d., internal resistance, null‑method principle | Null‑method measurement, error analysis, driver‑resistance determination |
| 10.2 | Potential dividers | Derivation of \(V{\text{out}} = Vs\frac{R2}{R1+R_2}\) using Kirchhoff’s laws; continuous‑divider view | Construction of a calibrated voltage divider, investigation of loading effects |
| 10.3 | Measuring emf & p.d. without drawing current | Potentiometer as a comparator, balance‑length method, uncertainty propagation | Driver set‑up, sliding‑contact technique, data‑record tables, repeatability checks |
| 18 | Electric fields | Potential gradient as a uniform electric field along the wire | Field‑mapping experiments (contextual background) |
| 19 | Capacitance | Often combined with potentiometer in RC‑timing investigations | RC‑circuit measurements (optional extension) |
| e.m.f. (electromotive force) | The energy supplied per unit charge by a source when no current flows; numerically equal to the open‑circuit potential difference between its terminals. |
| Potential difference (p.d.) | The energy change per unit charge between two points in a circuit while a current may be flowing. |
| Internal resistance (r) | The resistance inherent to a source; it causes a voltage drop \(I r\) when a current \(I\) flows, so the terminal p.d. is \(E - I r\) where \(E\) is the emf. |
A potentiometer consists of a straight length L of uniform resistive wire (resistance \(R_{\text{wire}}\)) connected to a stable driver (voltage source). When a current I flows, the voltage drop along the wire is linear:
\[
V(x)=k\,x,\qquad k=\frac{V{\text{total}}}{L}=\frac{I\,(R{\text{wire}}+r_d)}{L}
\]
\[
k = \frac{Vd}{L}\;\frac{R{\text{wire}}+rd}{R{\text{wire}}}
\]
The factor \(\frac{R{\text{wire}}+rd}{R_{\text{wire}}}\) corrects for the voltage drop across the driver’s internal resistance.
The uniform wire can be regarded as an infinite series of infinitesimal resistors \(dR\). Over a length \(x\) the equivalent resistance is \(\frac{R_{\text{wire}}}{L}\,x\). Substituting this into the divider formula gives
\[
V(x)=V_{\text{total}}\frac{x}{L}=k\,x,
\]
which is exactly the potentiometer relation derived above. The key advantage of the potentiometer is that the “load” (the unknown source) is connected only at the balance point, where the galvanometer reads zero, so the divider is never loaded – a point worth emphasising for AO3 (explain why the method is superior to a simple voltage‑divider measurement).
+———[Driver +]———+———[R_wire]———+———[Driver –]———-
| |
| |
| |
[Sliding] [Galvanometer]
Contact (zeroed)
| |
+———[Unknown emf]———+
When the sliding contact is at the balance point the galvanometer reads zero; the potential difference across the unknown emf equals the potential drop along the wire between the zero end and the contact.
\[
k = \frac{E{\text{ref}}}{l{\text{ref}}}\quad\text{or}\quad
k = \frac{Vd}{L}\frac{R{\text{wire}}+rd}{R{\text{wire}}}.
\]
\[
E1 = k\,l1,\qquad E2 = k\,l2,\qquad
\frac{E1}{E2}= \frac{l1}{l2}.
\]
For a single emf:
\[
\Delta E = \sqrt{(k\,\Delta l)^2 + (l\,\Delta k)^2}
\]
where
\[
\Delta k = k\sqrt{\left(\frac{\Delta Vd}{Vd}\right)^2
+\left(\frac{\Delta L}{L}\right)^2
+\left(\frac{\Delta rd}{R{\text{wire}}+r_d}\right)^2 }.
\]
For the ratio of two emfs:
\[
\frac{\Delta (E1/E2)}{E1/E2}= \sqrt{\left(\frac{\Delta l1}{l1}\right)^2+\left(\frac{\Delta l2}{l2}\right)^2}
\]
Typical classroom values:
Given:
Solution:
\[
k = \frac{V_d}{L}=5.00\ \text{V m}^{-1}
\]
\[
E = k\,l = 5.00 \times 0.320 = 1.60\ \text{V}
\]
\[
\Delta E = k\,\Delta l = 5.00 \times 0.00005 = 2.5\times10^{-4}\ \text{V}\approx\pm0.0003\ \text{V}
\]
Data:
Gradient (uncorrected): \(k0 = Vd/L = 5.00\ \text{V m}^{-1}\).
Corrected gradient:
\[
k = k0\frac{R{\text{wire}}+rd}{R{\text{wire}}}=5.00\frac{10.0+0.20}{10.0}=5.10\ \text{V m}^{-1}
\]
EMF:
\[
E = k\,l = 5.10 \times 0.450 = 2.30\ \text{V}
\]
Uncertainty:
\[
\Delta k = k\sqrt{\left(\frac{0.01}{6.00}\right)^2+\left(\frac{0.01}{10.20}\right)^2}=0.009\ \text{V m}^{-1}
\]
\[
\Delta E = \sqrt{(k\Delta l)^2+(l\Delta k)^2}
= \sqrt{(5.10\times0.00005)^2+(0.450\times0.009)^2}
\approx 0.004\ \text{V}
\]
\[
\boxed{E = 2.30 \pm 0.004\ \text{V}}
\]
Suppose a simple voltage divider is used to measure an unknown emf \(E\) of 2.00 V. The divider output is connected to a digital voltmeter of internal resistance 10 MΩ. If the divider has a total resistance of 1 MΩ, the voltmeter draws a current that reduces the measured voltage by about 0.2 %.
Using the potentiometer, the galvanometer is set to zero; no current flows from the unknown source, so the measured emf is free from this loading error. This conceptual point should be highlighted in the answer to an AO3 question.
| Trial | Source (Standard / \(E1\) / \(E2\)) | Balance length \(l\) (mm) | Driver voltage \(V_d\) (V) | Driver resistance \(r_d\) (Ω) | Calculated emf (V) | Uncertainty (V) |
|---|---|---|---|---|---|---|
| 1 | Standard (1.50 V) | 300.2 | 5.00 | 0.20 | 1.50 | ±0.001 |
| 2 | \(E_1\) | 320.5 | 5.00 | 0.20 | 1.60 | ±0.001 |
| 3 | \(E_2\) | 280.0 | 5.00 | 0.20 | 1.40 | ±0.001 |
Having mastered the potentiometer, you will encounter the following related topics in the remaining syllabus:
| Symbol | Quantity | Unit |
|---|---|---|
| V_total | Total potential across the potentiometer wire | V |
| L | Length of the potentiometer wire | m |
| k | Potential gradient (V m⁻¹) | V m⁻¹ |
| R_wire | Resistance of the wire | Ω |
| r_d | Internal resistance of the driver | Ω |
| I | Current through the wire | A |
| V_s | Supply voltage for a divider | V |
| R₁, R₂ | Resistances in a potential divider | Ω |
| V_out | Output voltage across R₂ | V |
| E₁, E₂ | Unknown emf’s being compared | V |
| l₁, l₂ | Balance lengths for E₁ and E₂ | m |
| Δ (uncertainty) | Absolute uncertainty | Same as quantity |
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