Topic 5 of 6

Potential dividers and the potentiometer

A potential divider provides an output voltage from part of a series resistance. Name the output terminals before deciding which resistance sets the output.

Series resistances carry the same current when no current branches off between them. Their voltage drops add to the supply p.d. Here C is labelled 0 V as a reference; this does not require a physical connection to Earth.

Derive the voltage share

Connect an upper resistance R_upper from A to B and a lower resistance R_lower from B to C. The supply is across A/C. Take the output V_BC across the lower resistance, with B measured relative to C.

If the output draws negligible current, the same current passes through both resistances:

I = V_s/(R_upper + R_lower)V_BC = IR_lower = V_s x R_lower/(R_upper + R_lower).

The numerator is the resistance across the chosen output. If output were measured across the upper resistance instead, its resistance would go in the numerator. The two drops add to V_s.

Move a potentiometer's wiper

A potentiometer has a resistive track with two end connections and a movable contact called the wiper. Connect the track ends A and C across the supply. The wiper is B, and output is taken between B and C.

Take the output between the wiper B and reference C

The uniform 10.0 kohm track is connected across 6.0 V at A/C. The output meter takes negligible current. The arrow touching the track is a connected wiper, not current flow.

Wiper one quarter of the track from C

V_BC = 1.50 V. The lower portion's resistance is 2.50 kohm. C is a chosen reference, not an Earth connection.

Wiper three quarters of the track from C

V_BC = 4.50 V. The lower portion's resistance is 7.50 kohm. C is a chosen reference, not an Earth connection.

A uniform 10.0 kΩ track spans A at 6.0 V to C at 0 V. With negligible output current, a wiper one quarter of the resistive length from C gives V_BC = 1.50 V; three quarters from C gives 4.50 V.

Moving the wiper changes how much of the track lies between B and C. The complete A/C track resistance stays the same. In the ideal unloaded model, output ranges from 0 V when B is at C to the full supply voltage when B is at A.

Use the named reference end

One quarter of a uniform track from 0 V

The track is 10.0 kΩ across 6.0 V. Here 1 kΩ = 1000 Ω.

One quarter from C gives R_lower = 2.50 kΩ and R_upper = 7.50 kΩ. Thus V_BC = 6.0 x 2.50/(7.50 + 2.50) = 1.50 V.

At three quarters from C, the lower share is 7.50/10.0, giving 4.50 V. The track current is 6.0/10 000 = 0.000600 A = 0.600 mA at either position in this unloaded model.

The position-to-voltage relationship is linear here because the track has uniform resistance per unit length. Do not assume that every type of potentiometer has the same relationship.

A two-terminal variable resistor instead includes a changing amount of resistance in one series path. A potentiometer used as a divider needs both track ends and the wiper connection; it changes the voltage share rather than the resistance of the whole connected track.

Keep the output-current condition

A high-resistance voltmeter or suitable input circuit draws negligible current, so it approximately preserves the simple divider ratio. A significant load connected across B/C becomes another parallel path and changes the circuit.

For example, two 2 kΩ arms across 6 V give an unloaded output of 3 V. Adding a 2 kΩ load across the lower arm makes the lower parallel equivalent 1 kΩ. The new output is 6 x 1/(2 + 1) = 2 V. Apply the whole-circuit method when the loading cannot be ignored.

Optional check A uniform 10 kilohm potentiometer track is connected across 6.0 V, from A at 6.0 V to C at 0 V. Its wiper B is 80% of the resistive length from C towards A. What is the output V_BC when the output draws negligible current?