recall and use the principle of the potentiometer as a means of comparing potential differences

Potential Dividers

Learning Objective

Recall and use the principle of the potentiometer as a means of comparing potential differences.

1. The Potentiometer Principle

A potentiometer is a length of uniform resistive wire of total resistance $R_{\text{wire}}$ and length $L$. When a constant current $I$ flows through it, the potential drop is linear along the wire:

$$V(x)=\frac{V_{\text{total}}}{L}\,x$$

where $x$ is the distance measured from the end at zero potential and $V_{\text{total}}$ is the total potential difference across the wire.

2. Potential Divider Circuit

A potential divider consists of two series resistors $R_1$ and $R_2$ connected across a supply voltage $V_{\text{s}}$. The junction between them provides a fraction of the supply voltage:

$$V_{\text{out}} = V_{\text{s}}\frac{R_2}{R_1+R_2}$$

This is the same relationship that underlies the potentiometer when the resistive wire is treated as a continuous series of infinitesimal resistors.

3. Using a Potentiometer to Compare Two Unknown Potentials

To compare two unknown emf’s, $E_1$ and $E_2$, the following steps are used:

1. Connect the potentiometer wire to a stable reference voltage source (the driver).
2. Adjust the driver so that a known voltage $V_{\text{ref}}$ appears across a calibrated length $l_{\text{ref}}$ of the wire.
3. Place the unknown emf $E_1$ across a galvanometer and a sliding contact. Move the contact until the galvanometer reads zero (null condition). Record the balance length $l_1$.
4. Repeat the procedure for $E_2$, obtaining balance length $l_2$.
5. Since the potential gradient is uniform, the emf’s are proportional to their balance lengths: $$\frac{E_1}{E_2} = \frac{l_1}{l_2}$$

4. Symbol Table

Symbol	Quantity	Unit
$V_{\text{total}}$	Total potential across the potentiometer wire	V
$L$	Length of the potentiometer wire	m
$x$	Distance from zero‑potential end	m
$R_1, R_2$	Resistances in a potential divider	Ω
$V_{\text{s}}$	Supply voltage	V
$V_{\text{out}}$	Output voltage of the divider	V
$E_1, E_2$	Unknown emf’s being compared	V
$l_1, l_2$	Balance lengths for $E_1$ and $E_2$	m

5. Example Calculation

Given a potentiometer wire of length $L=1.00\,$m with a driver set so that $V_{\text{total}}=5.00\,$V, the potential gradient is $5.00\,$V m\(^{-1}\).

If an unknown emf $E$ balances at $l=0.320\,$m, its value is:

$$E = \left(\frac{5.00\ \text{V}}{1.00\ \text{m}}\right) \times 0.320\ \text{m}=1.60\ \text{V}$$

6. Common Pitfalls

• Assuming the driver voltage is perfectly stable; any drift changes the potential gradient.
• Neglecting contact resistance at the sliding contact, which can introduce a small error.
• Using a non‑uniform wire (e.g., temperature gradients) which invalidates the linear relationship.
• Reading the balance length from the wrong end of the wire.

7. Summary

• The potentiometer provides a direct, high‑precision method for comparing voltages without drawing current from the source.
• It works on the same principle as a potential divider: a uniform potential gradient along a resistive element.
• At the null point, the unknown emf equals the potential drop over the measured length of the wire.
• Because the galvanometer reads zero, the method eliminates loading errors.

8. Practice Questions

1. A potentiometer wire of length $1.20\,$m is driven by a $6.00\,$V source. An unknown emf balances at $0.450\,$m. Find the emf.
2. Two unknown emf’s, $E_A$ and $E_B$, give balance lengths $l_A=0.250\,$m and $l_B=0.375\,$m on the same potentiometer. Determine the ratio $E_A:E_B$.
3. In a potential divider, $R_1=2.0\,$kΩ and $R_2=3.0\,$kΩ are connected across $12\,$V. Calculate $V_{\text{out}}$ across $R_2$.
4. Explain why a potentiometer is preferred over a simple voltmeter when measuring a very small emf.