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LNGR SPKRHIFI. TWO WAYS. HANDCRAFTED.
Knowledge / Understand components
LNGR · Learning workshop

The potentiometer

Three terminals turn a resistor into an adjustable voltage divider. The connected circuit always influences the result.

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Resistor · two terminals

Start without prior knowledgeUnderstand the model limitsDE / EN
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Circuit 04 · 5 small steps

Dividing voltage — with a load

Dividing voltage — with a loadThe two resistors represent a 10 kΩ potentiometer at its midpoint. M is the wiper. Initially, the M–B output is unloaded.+−5 VAB · 0 VR1 = 5 kΩR2 = 5 kΩR_L10 kΩM
The two resistors represent a 10 kΩ potentiometer at its midpoint. M is the wiper. Initially, the M–B output is unloaded.

  1. What is the resistance between A and B?

    Hint

    Add R1 and R2.

    Answer and working
    Calculation notes
    R_{\mathrm{total}}=R_1+R_2
    {R_{\mathrm{total}}=5\,\mathrm{k\Omega}+5\,\mathrm{k\Omega}=10\,\mathrm{k\Omega}}

  2. What current flows through the unloaded divider?

    Hint

    I = 5 V / 10,000 Ω.

    Answer and working
    Calculation notes
    I=\frac{U}{R_{\mathrm{total}}}
    I=\frac{5\,\mathrm{V}}{10000\,\Omega}
    {I=0.0005\,\mathrm{A}=0.5\,\mathrm{mA}}

  3. What is the output voltage between M and B?

    Hint

    UMB = I × R2.

    Answer and working
    Calculation notes
    U_{MB}=I\cdot R_2
    U_{MB}=0.0005\,\mathrm{A}\cdot5000\,\Omega
    {U_{MB}=2.5\,\mathrm{V}}

  4. Now connect RL = 10 kΩ between M and B. Find R2 parallel RL.

    Hint

    Rparallel = R2 × RL / (R2 + RL).

    Answer and working
    Calculation notes
    R_{2\parallel L}=\frac{R_2R_L}{R_2+R_L}
    R_{2\parallel L}=\frac{5\,\mathrm{k\Omega}\cdot10\,\mathrm{k\Omega}}{5\,\mathrm{k\Omega}+10\,\mathrm{k\Omega}}
    {R_{2\parallel L}\approx3.333\,\mathrm{k\Omega}}

    The dashed load is parallel to the lower resistor.

  5. Calculate the loaded output voltage and check the potentiometer model.

    Hint

    Uout = 5 V × Rlower / (R1 + Rlower).

    Answer and working
    Calculation notes
    U_{\mathrm{out}}=U\cdot\frac{R_{2\parallel L}}{R_1+R_{2\parallel L}}
    U_{\mathrm{out}}\approx5\,\mathrm{V}\cdot\frac{3.333\,\mathrm{k\Omega}}{5\,\mathrm{k\Omega}+3.333\,\mathrm{k\Omega}}
    {U_{\mathrm{out}}=2\,\mathrm{V}}

    The load reduces output voltage. Check the model: 10 kΩ, midpoint, 10 kΩ load.

Think it through before opening the answer. Always write the formula, substituted values and units.

01

The mental picture

Think of a movable tap on a resistive track. The wiper divides it into an upper and a lower section. Moving it changes their ratio. Resistance between the two outer terminals ideally stays constant.

02

The principle

An unloaded linear potentiometer produces voltage proportional to position. A load between wiper and ground sits in parallel with the lower section and changes the division. A logarithmic volume pot deliberately has another curve. This tool models a linear pot.

03

Try it yourself

Interactive experiment

The potentiometer

Set the wiper to 50% and compare a 100 kΩ load with 1 kΩ. The knob stays still, but output voltage falls as loading increases.

Model limits: Linear resistive track, ideal 5 V source and resistive load. No contact noise or intermittent wiper contact.

04

A worked example

A 10 kΩ pot across 5 V gives 2.5 V at its unloaded midpoint. A 10 kΩ load across the lower 5 kΩ section reduces this to 2 V. The circuit connected to the wiper belongs in the calculation.

A closer look · equations and assumptions
Formula notes
U_{\mathrm{out}}=U_{\mathrm{in}}\frac{R_2}{R_1+R_2}
R_{2\parallel L}=\frac{R_2R_L}{R_2+R_L}
R_{\mathrm{source}}=\frac{R_1R_2}{R_1+R_2}

The Thevenin equivalent is the unloaded divider voltage in series with R1 ∥ R2. A potentiometer’s source resistance is highest at the midpoint. A high-input-impedance buffer can reduce loading, but its input characteristics still require checking.

A common pitfall. An ordinary signal potentiometer is not a loudspeaker power attenuator. A speaker L-pad requires appropriate power ratings and a different resistor arrangement.

05

Workshop takeaway

The third terminal is no mystery: two ends and one wiper. Then check what the wiper is connected to.

Sources & transparency

Original illustrations and educational calculations. No measurements of a specific product. Edited 11 October 2026.

Work it out · answers unfold

Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.

14 · Divider: R1 = 2 kΩ upper, R2 = 3 kΩ lower, input 10 V. Unloaded output across R2. Find output.

Show worked answer
  1. Total resistance: 5 kΩ.
  2. Current: 2 mA.
  3. Voltage across R2: 6 V.
Calculation
U_o=10\frac{3}{2+3}=6\,V

15 · Load the divider in task 14 with 3 kΩ across R2. Find the new output.

Show worked answer
  1. R2 parallel load = 1.5 kΩ.
  2. Total resistance: 3.5 kΩ.
  3. Output = 10 · 1.5/3.5 = 4.286 V.
Calculation
U_o=10\frac{1.5}{2+1.5}\approx4.286\,V

16 · An unloaded divider should turn 12 V into 3 V. Choose R2 = 1 kΩ. Find R1.

Show worked answer
  1. Output is one quarter of input.
  2. Total resistance must be four times R2.
  3. R1 = 4 kΩ - 1 kΩ = 3 kΩ.
Calculation
R_1=R_2\left(\frac{U_i}{U_o}-1\right)=3\,k\Omega

18 · A 1 kΩ resistor has ±5% tolerance. Find minimum and maximum current at ideal 5 V.

Show worked answer
  1. Rmin = 950 Ω, Rmax = 1050 Ω.
  2. Minimum I uses maximum R: 4.762 mA.
  3. Maximum I uses minimum R: 5.263 mA.
Calculation
I_{min}=\frac{5}{1050};\quad I_{max}=\frac{5}{950}

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