Describe the action of a variable potential divider

4.3.3 Action and Use of Circuit Components

Objective

Describe the action of a variable potential divider and use the required equations in practical IGCSE situations.

1. Relationship between resistance, current and potential difference

• For a constant current I flowing through a conductor, the potential difference across it is
  

   V = I R

• Therefore, if the resistance of a part of a circuit is increased while the current remains the same, the voltage drop across that part also increases. This principle underlies the operation of a potential divider.

2. What is a variable potential divider?

A variable potential divider is a single resistive component that can be adjusted so that a chosen fraction of a fixed supply voltage appears across a part of the component. The adjustment is made by moving a wiper along a uniform resistive track.

3. Construction – potentiometer and rheostat

4. How it works – basic principle

• The resistive track has a total resistance Rtotal and total length L.
• The wiper is at a distance x from the left‑hand end.
• Resistances on either side of the wiper are

  \[

  R{1}=R{\text{total}}\frac{x}{L},\qquad

  R{2}=R{\text{total}}\Bigl(1-\frac{x}{L}\Bigr)

  \]

  where R₁ is between the left end and the wiper and R₂ is between the wiper and the right end.

5. Potential‑divider equations (required by the syllabus)

When the two fixed ends are connected to a stable supply voltage V₁, the voltage taken between the wiper and the nearer fixed end is the output voltage V₂:

\[

V{2}=V{1}\frac{R{2}}{R{1}+R{2}}=V{1}\frac{R{2}}{R{\text{total}}}

\]

Using the position ratio:

\[

V{2}=V{1}\Bigl(1-\frac{x}{L}\Bigr)

\]

5.1 Rearranged form (syllabus statement)

If V₁ (supply) and V₂ (required output) are known, the unknown resistance can be found from

\[

R{1}= \frac{V{1}\,R{2}}{V{2}}\qquad\text{or}\qquad

R{2}= \frac{V{2}\,R{1}}{V{1}}

\]

6. Using a variable potential divider – step‑by‑step

1. Connect the two fixed ends of the track across the supply voltage V₁.
2. Connect the load (or measuring instrument) between the wiper and the appropriate fixed end.
3. Adjust the wiper while monitoring the voltage with a voltmeter until the required V₂ is obtained.
4. For accurate division, ensure the load resistance RL is much larger than the resistance of the portion of the track it is connected to (recommended RL ≥ 10 R₂).

7. Load effect

If the load resistance is comparable to R₂, the effective resistance becomes the parallel combination

\[

R{2}^{\prime}=R{2}\parallel R{L}= \frac{R{2}R{L}}{R{2}+R_{L}}

\]

The output voltage then becomes

\[

V{2}=V{1}\frac{R{2}^{\prime}}{R{1}+R_{2}^{\prime}}

\]

8. Power‑rating considerations

The track must be able to dissipate the heat generated:

\[

P = I^{2}R \quad\text{or}\quad P = \frac{V^{2}}{R}

\]

Check the manufacturer’s rating; exceeding it can cause overheating and permanent damage.

9. Example 1 – solving for an unknown resistor

Given:

• Supply voltage V₁ = 12 V
• Required output voltage V₂ = 5 V
• Resistance of the lower part of the track R₂ = 4 kΩ

Find R₁:

\[

R{1}= \frac{V{1}\,R{2}}{V{2}}

= \frac{12\;\text{V}\times4\;\text{kΩ}}{5\;\text{V}}

= 9.6\;\text{kΩ}

\]

The wiper must be positioned so that the left‑hand resistance is 9.6 kΩ and the right‑hand resistance is 4 kΩ.

10. Example 2 – position‑based calculation

Data:

• Rtotal = 10 kΩ across a V₁ = 12 V supply.
• Wiper is at x/L = 30 % from the left end.

Find the voltage between the wiper and the left end (V₂).

\[

V{2}=V{1}\Bigl(1-\frac{x}{L}\Bigr)=12\;\text{V}\times(1-0.30)=12\;\text{V}\times0.70=8.4\;\text{V}

\]

11. Common IGCSE experimental applications

• Adjusting the brightness of a lamp (rheostat).
• Calibrating a voltmeter using a known voltage source (potentiometer).
• Providing a variable reference voltage for biasing a transistor.
• Testing the linearity of a sensor by supplying a known, adjustable voltage.

12. Summary

A variable potential divider splits a fixed supply voltage into a controllable output voltage by varying the ratio of two resistances within a single component. The key relationships are

\[

V{2}=V{1}\frac{R{2}}{R{1}+R_{2}},\qquad

R{1}= \frac{V{1}\,R{2}}{V{2}}

\]

Remember:

• For a constant current, the voltage drop across a resistor increases with its resistance (V = I R).
• Keep the load resistance much larger than the portion of the track it is connected to to avoid loading errors.
• Check the power rating of the track and avoid exceeding it.
• A uniform resistive track gives a linear relationship between wiper position and output voltage, but wear can introduce non‑linearity.