Potential divider circuit CIE physics

A potential divider circuit works by connecting two or more resistors in series across a supply voltage, so that the voltage is shared between them in proportion to their resistances. You can then tap off a fraction of the total voltage from across one resistor. According to expert tutors at My Physics Buddy, this is one of the most practically important circuits in A/AS Level Physics (9702).

How a Potential Divider Circuit Works — The Full Physics

Think of a potential divider like two people sharing a single pizza. If one person takes a bigger slice, the other gets less. The total pizza stays the same — just like the total supply voltage across the series resistors never changes. The fraction each resistor “takes” depends on how large its resistance is compared to the total.

The Core Idea: Voltage Sharing in a Series Circuit

When two resistors, R₁ and R₂, are connected in series across a supply voltage V_in, the same current I flows through both. By Ohm’s Law, the voltage across each resistor is proportional to its resistance. The output voltage V_out is taken across R₂ (the lower resistor).

Since the current through both resistors is the same:

I = V_in / (R₁ + R₂)

The voltage across R₂ is:

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

Where:

• V_in = supply (input) voltage in volts (V)
• R₁ = upper resistor in ohms (Ω)
• R₂ = lower resistor in ohms (Ω) — output is taken across this
• V_out = output voltage in volts (V)

This is the potential divider equation, and you must be completely comfortable with it for your CIE 9702 exam.

Worked Example

A potential divider has R₁ = 6 kΩ and R₂ = 4 kΩ connected in series across a 10 V supply. What is V_out across R₂?

Step 1 — Identify values:
V_in = 10 V, R₁ = 6000 Ω, R₂ = 4000 Ω

Step 2 — Apply the potential divider equation:
V_out = 10 × 4000 / (6000 + 4000)
V_out = 10 × 4000 / 10000
V_out = 10 × 0.4
V_out = 4 V

Notice that R₂ is 4/10 of the total resistance, so it takes 4/10 of the total voltage. This proportional logic is the fastest way to check your answer under exam conditions.

Using a Variable Resistor (Potentiometer)

In many CIE 9702 questions, R₂ is replaced by a light-dependent resistor (LDR), a thermistor, or a variable resistor. This is where potential dividers become genuinely powerful. As the sensor’s resistance changes — say, as light intensity increases and an LDR’s resistance falls — V_out changes too. This varying output voltage can then trigger switching circuits or feed into data loggers.

As an IBDP and A-Level Physics Specialist, I can tell you that the single most common mistake I see students make here is forgetting that connecting a load (like a voltmeter or a device) across R₂ changes the effective resistance of the lower branch, which changes V_out. This only becomes negligible when the load resistance is very much greater than R₂. For a high-resistance voltmeter, this is usually a safe assumption — but CIE exam questions will explicitly test whether you understand this condition.

Summary Table: Effect of Changing R₂ on V_out

Condition	Effect on Vout
R2 increases (e.g. LDR in dark)	Vout increases
R2 decreases (e.g. LDR in light)	Vout decreases
R2 = R1	Vout = Vin / 2
R2 very small compared to R1	Vout approaches 0 V

For a deeper look at how these circuits connect to broader electrical concepts, the CIE A-Level electricity revision resources at Physics and Maths Tutor are well worth bookmarking. You should also explore related topics through our Physics resources to strengthen your overall circuit analysis skills.

Common Mistakes with Potential Dividers

✗ Mistake: Using the potential divider formula with resistors in parallel instead of in series.
✓ Fix: The formula V_out = V_in × R₂/(R₁+R₂) only works when R₁ and R₂ are in series. Always confirm the circuit topology before applying the equation.

✗ Mistake: Forgetting that connecting a load across R₂ reduces the effective lower resistance and lowers V_out.
✓ Fix: When a load R_L is connected across R₂, calculate the parallel combination of R₂ and R_L first, then use that combined value as the lower resistance in the formula.

✗ Mistake: Assuming V_out is taken across R₁ (the upper resistor) rather than R₂ (the lower resistor).
✓ Fix: Always identify which resistor the output terminals are connected across in the circuit diagram. CIE questions deliberately vary this, so label both resistors and output terminals clearly before calculating.

Exam Relevance: Potential divider circuits appear in CIE A/AS Level Physics 9702, Edexcel A Level Physics, and OCR A Level Physics examinations. Questions test both the calculation of V_out and the qualitative reasoning about sensor-based divider circuits.

Pro Tip from Mamatha M: Always write the potential divider equation with R₂ in the numerator matching the resistor your output is taken across — this one habit eliminates the most common calculation error instantly.