Potential Divider Circuits: GCSE Physics

Predict first: if two resistors of 300 Ω and 700 Ω are connected in series across 10 V, the 700 Ω resistor will take 7 V and the 300 Ω resistor will take 3 V. A potential divider shares a fixed supply voltage between two components in proportion to their resistances, and GCSE examiners use it constantly.

What is a potential divider?

A potential divider (also called a voltage divider) is a circuit consisting of two resistors connected in series across a fixed voltage supply. The supply voltage is shared between the two resistors: each takes a share of the total voltage in proportion to its resistance.

This circuit is fundamental to electronics because it allows you to produce any output voltage between 0 V and the supply voltage, and it forms the basis of many sensor circuits.

Why does voltage split in proportion to resistance?

In a series circuit, the same current flows through both resistors. By Ohm's law:

V = IR

So for resistor R₁: V₁ = I × R₁
For resistor R₂: V₂ = I × R₂

Since I is the same in both:

V₁ / V₂ = R₁ / R₂

The voltage across each resistor is proportional to its resistance. The larger the resistance, the larger the share of the total voltage it takes.

What is the potential divider equation?

If the output voltage V_out is measured across resistor R₂:

$$V_{\text{out}} = V_{\text{in}} \times \frac{R_2}{R_1 + R_2}$$

Where:

• V_in = total supply voltage (V)
• R₁ = the first resistor in series (Ω)
• R₂ = the second resistor in series, across which V_out is measured (Ω)
• V_out = the output voltage (V)

This equation is not given on the AQA equation sheet, so you must recall it — or derive it from Ohm's law.

Worked examples

Example 1 — basic calculation:

Two resistors, R₁ = 300 Ω and R₂ = 700 Ω, are in series across a 10 V supply. What is the voltage across R₂?

• V_out = 10 × 700 / (300 + 700)
• V_out = 10 × 700 / 1,000
• V_out = 10 × 0.7 = 7 V

The voltage across R₁ = 10 − 7 = 3 V. Check: 3/7 = 300/700 ✓

Example 2 — finding resistance:

A 12 V supply is connected across two resistors in series. The output voltage across R₂ is 4 V. R₁ = 600 Ω. What is R₂?

• V_out/V_in = R₂ / (R₁ + R₂)
• 4/12 = R₂ / (600 + R₂)
• 1/3 = R₂ / (600 + R₂)
• 600 + R₂ = 3R₂
• 600 = 2R₂
• R₂ = 300 Ω

How do sensors use potential dividers?

The most important GCSE application of the potential divider is using a variable resistor (LDR or thermistor) as one of the two resistors. Because the variable resistor's resistance changes with a physical quantity (light or temperature), the output voltage changes with that quantity too — creating a sensor circuit.

Light-dependent resistor (LDR) in a potential divider

An LDR has high resistance in the dark and low resistance in bright light.

Circuit: LDR (R₁) in series with fixed resistor R₂; V_out measured across R₂.

In bright light: LDR resistance falls → the LDR takes a smaller share of voltage → more voltage appears across R₂ (V_out rises). This rising V_out could switch on a circuit (e.g. turn off a streetlight when dawn breaks).

Swap the positions of LDR and R₂ (measure V_out across the LDR instead) and the output does the opposite:

Now V_out is HIGH in the dark — useful for triggering a streetlight to switch ON when it gets dark.

Thermistor (NTC) in a potential divider

A negative temperature coefficient (NTC) thermistor has high resistance when cold and low resistance when hot.

Circuit: thermistor (R₁) in series with fixed resistor R₂; V_out measured across R₂.

A rising V_out could trigger a fan to switch on when a circuit gets hot — for example, in a computer's cooling system.

How do you remember which way the voltage changes?

Use the voltage divider ratio: the component with the bigger resistance always takes the bigger share of the voltage. If R₁ gets smaller (e.g. LDR in bright light, or thermistor getting hot), V₁ falls, and since V₁ + V₂ = V_in, V₂ must rise.

Sketch this as a "seesaw": when one resistance goes up, the voltage across it goes up; the voltage across the other component goes down.

Frequently asked questions

Do I need to know the potential divider equation for GCSE?

It is expected at GCSE Higher tier (AQA, Edexcel, OCR all include it). The equation V_out = V_in × R₂/(R₁ + R₂) is not printed on the AQA equation sheet, so you should be confident deriving it from V = IR and the fact that current is the same throughout a series circuit. Always check your specification and past paper mark schemes to see the exact level of detail required.

What happens to the output voltage if R₁ increases while R₂ stays the same?

If R₁ increases, a greater fraction of V_in appears across R₁, so less is left for R₂. V_out (across R₂) decreases. You can verify this with the formula: V_out = V_in × R₂/(R₁ + R₂). As R₁ increases, the denominator (R₁ + R₂) increases while the numerator (R₂) stays fixed, so the fraction and therefore V_out decreases.

Can I use a potential divider with more than two resistors?

Yes — in principle any number of resistors in series divides the supply voltage, each taking a share proportional to its resistance. However, GCSE exam questions use exactly two resistors. The same logic applies: identify which resistor's voltage you need, note the total resistance, and apply the ratio.

Why are potential dividers used in sensor circuits rather than just the sensor alone?

A single LDR or thermistor connected across a voltage supply would vary its current but not produce a useful output voltage on its own. Pairing it with a fixed resistor in series creates a circuit where the voltage can switch between near-zero and near-V_in as the sensor's resistance changes dramatically. This large voltage swing is used to trigger switches (transistors or relay circuits) that control other components such as lamps, buzzers, or fans.

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