Potential Dividers

A potential divider (also called a voltage divider) is the most elegant little circuit in introductory electronics. It uses two resistors in series to produce an output voltage that is a fixed fraction of the input voltage — or, if one of the resistors is a sensor (thermistor, LDR, strain gauge), to produce an output voltage that varies with some physical quantity.

This is the last topic in our OCR H556 Module 4.3.1 course. It draws on everything we have learned — series resistance, current, voltage, Ohm's law, non-ohmic components, and Kirchhoff's laws — and shows how they combine into one of the most useful real-world circuits. Understanding potential dividers is critical both for the A-Level exam and for practical electronics.

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The Basic Potential Divider

Take a single supply voltage V_in (say, from a battery or lab PSU) and connect two resistors R₁ and R₂ in series across it. Call the junction between them the "tap point". The output voltage is measured between the tap point and the negative terminal of the supply, i.e. across R₂.

flowchart LR
  VIN[V in] --> R1[R1]
  R1 --> T((Tap point))
  T --> R2[R2]
  R2 --> GND[Ground]
  T --> VOUT[V out]
  VOUT --> GND

What is V_out?

Assume no current is drawn from the tap point (high-impedance load). Then the same current I flows through both resistors:

• I = V_in / (R₁ + R₂)
• V_out = I × R₂ = V_in × R₂ / (R₁ + R₂)

Hence the famous potential divider equation:

V_out = V_in × R₂ / (R₁ + R₂)

Some textbooks put R₁ across the output instead, giving V_out = V_in × R₁/(R₁+R₂) — the form depends on which resistor is at the "bottom" of the divider. Always write the equation in words first, and identify which resistor is across the output.

Quick Sanity Checks

• If R₂ >> R₁: V_out ≈ V_in (almost all the voltage drops across R₂).
• If R₂ << R₁: V_out ≈ 0 (almost all the voltage drops across R₁).
• If R₁ = R₂: V_out = V_in / 2 (half the supply).

Worked Example 1 — Simple Ratio

A 12 V supply feeds a potential divider made of a 1000 Ω and a 2000 Ω resistor. What is the output voltage across the 2000 Ω?

• V_out = 12 × 2000 / (1000 + 2000) = 12 × 2000 / 3000 = 12 × (2/3) = 8.0 V

And across the 1000 Ω?

• V₁ = 12 × 1000 / 3000 = 4.0 V

Note V_out + V₁ = 8 + 4 = 12 V. As in any series circuit, the two voltages add to the total.

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Why Bother?

Why not just use a battery of the right voltage in the first place? Several reasons:

1. Fractional voltages on demand. You may only have a 12 V supply but need 5 V, or 3.3 V, for logic circuits.
2. Adjustable output. If R₁ or R₂ is a potentiometer, the output is continuously variable.
3. Sensor readouts. If R₁ or R₂ depends on a physical quantity (temperature, light, pressure), V_out becomes a sensor signal proportional to that quantity.
4. Simple and reliable. No transistors, no ICs — just two resistors.

In professional designs, potential dividers are the "front-end" of almost every measurement system. They convert a varying resistance into a varying voltage that an ADC or op-amp can read.

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Worked Example 2 — Designing a 5 V Rail

You have a 9 V battery and need to feed 5 V to a logic chip that draws a negligible current. Design a suitable potential divider.

Design equation: 5 = 9 × R₂ / (R₁ + R₂) → R₂ / (R₁ + R₂) = 5/9 → 9 R₂ = 5(R₁ + R₂) → 4 R₂ = 5 R₁ → R₂ / R₁ = 5/4

Pick R₁ = 4 kΩ, R₂ = 5 kΩ (or any ratio 4:5). For example 4 kΩ / 5 kΩ gives V_out = 9 × 5/9 = 5.0 V exactly.

Watch out: this only works if the logic chip really draws negligible current. Otherwise it "loads" the divider and reduces the output (see "Loading Effect" below).

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The Loading Effect

The potential divider equation assumes no current is drawn from the tap point. In practice, whatever device you connect to V_out has some resistance R_load, and that acts in parallel with R₂.

V_out (loaded) = V_in × (R₂ ∥ R_load) / (R₁ + R₂ ∥ R_load)

where R₂ ∥ R_load = R₂ R_load / (R₂ + R_load).

Worked Example 3 — Loading

Using the 12 V, 1000 Ω/2000 Ω divider from Example 1, V_out is 8 V with nothing connected. Now connect a 2000 Ω load across the output. What is the new V_out?

• R₂ ∥ R_load = 2000 × 2000 / (2000 + 2000) = 1000 Ω
• V_out = 12 × 1000 / (1000 + 1000) = 6.0 V

The output has dropped from 8 V to 6 V — a 25% error. Loading is significant.

Rule of thumb: R_load should be at least 10 × R₂ for the unloaded formula to be accurate to ~10%. For precision work, use a buffer (a high-input-impedance op-amp) between the divider and the load, or make the divider resistances small compared with the load.

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The Potentiometer

A potentiometer is a mechanical potential divider: a single resistive track with a sliding contact (the "wiper") that can be moved to give any desired ratio R₁ : R₂.

Rotating a volume knob on a radio moves the wiper. When the wiper is at one end, V_out = 0; at the other end, V_out = V_in. In between, V_out varies approximately linearly with position.

A three-terminal potentiometer gives a continuously variable voltage. A two-terminal variable resistor — called a rheostat — gives a variable resistance but not a variable voltage, and cannot reach 0 V.

Exam Tip: OCR is strict about the distinction. A potential divider (three-terminal) gives a voltage that can go from 0 to V_in. A rheostat (two-terminal variable resistor) cannot reach zero output and is not a potential divider.

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Sensor Circuits

The real magic of potential dividers is pairing them with a sensor resistor.

Temperature Sensor with a Thermistor

Replace R₁ (say) with an NTC thermistor. Its resistance drops as temperature rises. So as T rises:

• R_thermistor falls.
• V across R_thermistor = V_in × R_therm / (R_therm + R₂) falls.
• V_out across R₂ rises.

Conversely, if you put the thermistor in the R₂ position, V_out falls as T rises.

Design choice: which way round you put the thermistor determines whether V_out rises or falls with temperature. Choose the one that makes sense for your application.

Worked Example 4 — Thermistor Alarm