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 \$R1\$ and \$R2\$ 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{R2}{R1+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, \$E1\$ and \$E2\$, 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 \$E1\$ across a galvanometer and a sliding contact. Move the contact until the galvanometer reads zero (null condition). Record the balance length \$l1\$.
4. Repeat the procedure for \$E2\$, obtaining balance length \$l2\$.
5. Since the potential gradient is uniform, the emf’s are proportional to their balance lengths:

   \$\frac{E1}{E2} = \frac{l1}{l2}\$

4. Symbol Table

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, \$EA\$ and \$EB\$, give balance lengths \$lA=0.250\,\$m and \$lB=0.375\,\$m on the same potentiometer. Determine the ratio \$EA:EB\$.
3. In a potential divider, \$R1=2.0\,\$kΩ and \$R2=3.0\,\$kΩ are connected across \$12\,\$V. Calculate \$V{\text{out}}\$ across \$R2\$.
4. Explain why a potentiometer is preferred over a simple voltmeter when measuring a very small emf.