Potential dividersEdexcel A-Level Physics: Revision notes
Section 1
The potential divider principle
A potential divider is two or more resistors in series across a supply. The same current flows through each, so the supply p.d. is shared in proportion to their resistances:
V_out = V_in × R₂/(R₁ + R₂)
where V_out is the p.d. across R₂. The larger the share of the total resistance, the larger the share of the p.d.
Worked example: 12 V across R₁ = 4.0 kΩ and R₂ = 2.0 kΩ. The current is 12/6000 = 2.0 mA, and V_out across R₂ = 12 × 2.0/6.0 = 4.0 V (and 8.0 V across R₁).
The resistances can be left in kΩ in the ratio, but convert to Ω when finding a current.
Check you are using the resistor the output is taken across in the numerator. The p.d. across R₁ is V × R₁/(R₁ + R₂).
Section 2
Uniform wire and sliding contact
A uniform wire connected across a cell is a continuously variable potential divider. Resistance and p.d. are both proportional to length, so the p.d. between one end and a sliding contact is
V = V_cell × (length to contact / total length)
Worked example: a 1.00 m wire of 20 Ω across 6.0 V has current 0.30 A. A contact 0.35 m from Q gives V = 6.0 × 0.35 = 2.1 V.
The potential falls uniformly along the wire from one end to the other when the cell has negligible internal resistance.
Section 3
Effect of a load
A potential divider only gives the calculated output when nothing is drawing current from it. If a load is connected across R₂, the load and R₂ are in parallel. Their combined resistance is smaller than R₂, so the output p.d. falls.
Worked example: a 2.0 kΩ load across the 2.0 kΩ R₂ in the 12 V divider gives a parallel resistance of 1.0 kΩ, so V_out = 12 × 1.0/(4.0 + 1.0) = 2.4 V (not 4.0 V).
The same applies to a voltmeter, which is why it must have a very high resistance.
Section 4
Dividers with thermistors and LDRs
Replacing one resistor with a sensor makes the output vary with the surroundings.
- NTC thermistor: resistance falls as temperature rises (more conduction electrons released, outweighing lattice vibrations).
- LDR: resistance falls as light intensity rises (light releases more conduction electrons).
With a sensor and fixed resistor R in series, the p.d. across R is V × R/(R + R_sensor). If the sensor's resistance falls, the sensor takes a smaller share of the supply, so the p.d. across the fixed resistor rises and the p.d. across the sensor falls.
To switch at a chosen output p.d., choose R. For a switch-on at half the supply, set R equal to the sensor's resistance at the switching condition.
Say which component the output is across. Swapping the output from the sensor to the fixed resistor reverses the effect of temperature or light on the output.
Must Know
- V_out = V_in × R₂/(R₁ + R₂); the same current flows through each resistor
- Along a uniform wire, p.d. is proportional to length
- A load in parallel lowers the output p.d.
- NTC thermistor and LDR: resistance falls as temperature or light rises
- The sensor with the lower resistance takes the smaller share of the p.d.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Potential dividers
- A potential divider is made from a 12 V battery of negligible internal resistance, a 4.0 kΩ resistor R₁ and a 2.0 kΩ resistor R₂ connected in series. R₁ is joined to the positive terminal and R₂ to the negative terminal. The output p.d. is taken across R₂.A load of resistance 2.0 kΩ is now connected across R₂. Calculate the new output p.d.2 marks
- A garden light switches on automatically at dusk. A 5.0 V supply of negligible internal resistance is connected across a 10 kΩ fixed resistor in series with a light-dependent resistor (LDR). The output p.d. is taken across the fixed resistor, and the lamp is switched on when this output falls below 2.0 V. The LDR has a resistance of 2.0 kΩ in daylight and 40 kΩ in the dark.Explain why the output p.d. falls when it gets dark.2 marks
- A uniform resistance wire PQ of length 1.00 m and resistance 20 Ω is connected across a 6.0 V cell of negligible internal resistance. A sliding contact J is placed on the wire at a distance of 0.35 m from Q, and the output p.d. is taken between Q and J.Calculate the output p.d. between Q and J when nothing else is connected to the circuit.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).