understand the principle of a potential divider circuit

Potential Dividers – Cambridge AS & A Level Physics (9702)

Learning Objectives

  • Explain the principle of a potential (voltage) divider and why the two terms are synonymous in the syllabus.

  • Derive the output‑voltage expression using Kirchhoff’s loop rule, and show how source internal resistance modifies it.

  • Design a divider for a required output voltage; assess the effect of resistor tolerances and temperature coefficients.

  • Analyse the loading effect, introduce the Thevenin‑equivalent view, and describe how to minimise loading (high‑impedance design or buffering).

  • Apply the divider concept to potentiometers, galvanometer null‑balance measurements, and sensor circuits (thermistors, LDRs).

1. What Is a Potential Divider?

A potential divider (also called a voltage divider) is a linear circuit that provides a known fraction of an input emf. In the Cambridge syllabus the term “potential” is preferred, but both describe the same device.

2. Two‑Resistor Divider – Theory

2.1 Circuit diagram

Two resistors, R1 and R2, in series across a source of emf Vs. The output voltage Vout is taken across R2.

2.2 Derivation (ideal source – Kirchhoff’s loop rule)

Applying Kirchhoff’s loop rule to the series loop gives

\$Vs - I R1 - I R2 = 0 \;\;\Longrightarrow\;\; I = \frac{Vs}{R1+R2}\$

The voltage across the lower resistor is

\$V{\text{out}} = V{R2}= I R2 = Vs\frac{R2}{R1+R2}\tag{1}\$

If the output is taken across R1, replace R2 by R1 in (1).

2.3 Including source internal resistance

Real batteries or supplies have an internal resistance r (shown in series with the divider). The loop equation becomes

\$Vs - I r - I R1 - I R2 = 0 \;\;\Longrightarrow\;\; I = \frac{Vs}{r+R1+R2}\$

Hence

\$V{\text{out}} = Vs\frac{R2}{r+R1+R_2}\tag{2}\$

Equation (2) shows that a large r reduces the obtainable fraction of the emf.

2.4 Numerical illustration of internal resistance

Source emf VsInternal r (Ω)R₁ (Ω)R₂ (Ω)Ideal Vout (V)With r (V)
1201 k7145.005.00
12100 Ω1 k7145.004.74
12500 Ω1 k7145.004.23

The table shows how a 100 Ω internal resistance already drops the output by 5 %.

3. Loading Effect and Thevenin Equivalent

3.1 Load resistance across the output

Connecting a load RL in parallel with R2 gives an equivalent resistance

\$R{\text{eq}}=\frac{R2RL}{R2+R_L}\$

The loaded output voltage is

\$V{\text{out}} = Vs\frac{R{\text{eq}}}{R1+R_{\text{eq}}}\tag{3}\$

3.2 Rule of thumb

For negligible loading, design the divider so that RL ≥ 10 × R2.

3.3 Thevenin‑equivalent view

  • The divider (including source internal resistance) can be replaced by a Thevenin source Vth = Vout(open‑circuit) in series with a Thevenin resistance Rth = R1‖(R2+r).

  • The load sees a simple series circuit, making it easy to calculate the loaded voltage with the voltage‑divider rule again.

  • Buffering with an op‑amp voltage follower makes Rth appear infinite to the load, eliminating loading errors.

4. Resistor Tolerance and Temperature Coefficient

4.1 Propagation of tolerance

\$\$\frac{\Delta V{\text{out}}}{V{\text{out}}}\approx

\sqrt{\left(\frac{\Delta R1}{R1}\right)^2+

\left(\frac{\Delta R2}{R2}\right)^2}\$\$

Two 1 % resistors give a worst‑case voltage error of ≈ ±1.4 %.

4.2 Temperature coefficient

For precision references use resistors with ≤ 50 ppm / °C. The change in output due to temperature is

\$\Delta V{\text{out}} \approx V{\text{out}}\;\alpha_{\text{eff}}\;\Delta T\$

where αeff is the effective coefficient of the resistor pair (often the difference of their individual coefficients).

5. Practical Applications Required by the Syllabus

5.1 Potentiometer (null‑balance method)

A potentiometer is a three‑terminal uniform‑resistance wire of length L driven by a stable source. The potential gradient is

\$\frac{dV}{dx}= \frac{V_s}{L}\$

When a jockey connected to a galvanometer touches the wire at position ℓ, the voltage between the ends of the galvanometer is

\$V = \frac{ℓ}{L}\,V_s\$

At the balance point the galvanometer reads zero, so the unknown voltage Vx equals the known reference voltage Vref multiplied by the ratio of the two balance lengths. Because the galvanometer carries no current at balance, the measurement is free from loading errors – a direct illustration of the potential‑divider principle in a null method.

5.2 Galvanometer in a potentiometer set‑up

  • Detects any residual current between the two voltage sources.

  • Zero deflection indicates equal potentials at the two galvanometer terminals.

  • Used in the syllabus to measure emf of cells with high accuracy.

5.3 Sensor Dividers – Thermistor and Light‑Dependent Resistor (LDR)

  • Thermistor (NTC example): \$RT(T)=R0\,e^{\beta\left(\frac{1}{T}-\frac{1}{T_0}\right)}\$

    Output voltage: \$V{\text{out}}=Vs\frac{RT(T)}{R1+R_T(T)}\$ – voltage falls as temperature rises.

  • LDR: \$R_{LDR}=k\,I^{-\alpha}\$ (where \$I\$ is illumination).

    Output voltage: \$V{\text{out}}=Vs\frac{R{LDR}}{R1+R_{LDR}}\$ – voltage rises in darkness.

These circuits convert a physical quantity into a proportional voltage, satisfying the syllabus requirement for “thermistor & LDR applications”.

6. Example Calculations

6.1 Designing a 5 V Output from a 12 V Supply (ideal source)

  1. Choose \$R_1=1.0\,\$kΩ (standard value).

  2. Use (1): \$5 = 12\frac{R2}{1.0\text{k}+R2}\$ → \$R_2 = \frac{5\,000}{7}\approx714\;\Omega\$.

  3. Standard E12 values: 680 Ω gives \$V_{\text{out}}=4.90\,\$V; 720 Ω gives \$5.07\,\$V.

  4. Check tolerance: with ±1 % resistors, worst‑case \$V_{\text{out}}\$ lies between 4.85 V and 5.15 V.

6.2 Loading Example (same as syllabus)

Divider: \$R1=2.2\,\$kΩ, \$R2=1.0\,\$kΩ, \$V_s=9\,\$V.

  • Open‑circuit: \$V_{\text{out}}=9\frac{1.0}{2.2+1.0}=2.78\,\$V.

  • With \$RL=5\,\$kΩ: \$R{\text{eq}}=0.833\,\$kΩ → \$V_{\text{out}}=9\frac{0.833}{2.2+0.833}=2.51\,\$V.

  • Improvement: add a unity‑gain op‑amp buffer (voltage follower) after the divider; the load then sees a very high input impedance, so \$V_{\text{out}}\$ remains ≈ 2.78 V.

6.3 Thermistor Temperature Sensor

Given \$R0=10\,\$kΩ at \$T0=298\,\$K, \$\beta=3500\,\$K, \$R1=5\,\$kΩ, \$Vs=12\,\$V.

  1. At \$25^{\circ}\mathrm{C}\$ (\$T=298\,\$K): \$R_T=10\,\$kΩ.

    \$V_{\text{out}}=12\frac{10}{5+10}=8.0\,\$V.

  2. At \$75^{\circ}\mathrm{C}\$ (\$T=348\,\$K): \$R_T=10\,e^{3500\left(\frac{1}{348}-\frac{1}{298}\right)}\approx3.3\,\$kΩ.

    \$V_{\text{out}}=12\frac{3.3}{5+3.3}=5.0\,\$V.

  3. Output changes ≈ 3 V over a 50 °C range → sensitivity ≈ 60 mV / °C, suitable for many laboratory temperature‑measurement tasks.

7. Practical Considerations

  • Resistor selection: tolerance ≤ 1 % (0.5 % for high‑precision references); power rating \$P=I^2R\$ – calculate using the divider current.

  • Power example (12 V → 5 V): \$I=12/(1.0\text{k}+0.714\text{k})=7.0\,\$mA.

    \$P{R1}=I^2R1=0.049\,\$W, \$P_{R2}=0.025\,\$W → a 0.25 W resistor is adequate.

  • High‑impedance loads: keep \$RL\ge10R2\$ or buffer with an op‑amp voltage follower.

  • Temperature stability: use low‑TC (≤ 50 ppm/°C) resistors when the divider supplies a reference voltage for instrumentation.

  • Buffering (Thevenin view): The voltage follower makes the Thevenin resistance appear infinite to the load, preserving the designed \$V_{\text{out}}\$.

8. Common Mistakes to Avoid

  • Assuming the output voltage is unchanged by a load – always verify \$RL\gg R2\$ or use a buffer.

  • Choosing resistor values that are too low, causing unnecessary power dissipation and temperature drift.

  • Confusing which resistor the output is taken across; remember \$V_{\text{out}}\$ is across the resistor *below* the output node.

  • Neglecting the source’s internal resistance, especially with batteries that have a noticeable voltage drop under load.

9. Summary

A potential divider creates a predictable fraction of an input voltage. The ideal formula is

\$V{\text{out}}=Vs\frac{R2}{R1+R_2}\$

Including source internal resistance r gives

\$V{\text{out}}=Vs\frac{R2}{r+R1+R_2}\$

Accuracy depends on resistor values, tolerances, loading, and temperature effects. The Thevenin‑equivalent model and buffering are useful tools for analysing and improving performance. The concept underpins potentiometer null‑balance measurements, galvanometer methods, and sensor circuits (thermistors, LDRs), fulfilling all Cambridge AS & A Level requirements for the topic.

10. Practice Questions

  1. Design a divider that gives \$3.3\,\$V from a \$9\,\$V supply using standard E12 resistor values. Show your calculations and include the effect of ±1 % tolerance on the final voltage.

  2. A divider uses \$R1=2.2\,\$kΩ and \$R2=1.0\,\$kΩ.

    • What is the open‑circuit output voltage?

    • If a \$5\,\$kΩ load is connected, what is the new output voltage?

    • Explain how you could improve the accuracy without changing the resistor values.

  3. Explain why a voltage follower (buffer) is often placed after a potential divider in precision circuits. Include a diagram showing the buffer’s effect on loading.

  4. In a potentiometer experiment a galvanometer reads zero when the jockey is at 42 cm on a 1 m wire supplied by a 6 V source. What is the unknown voltage being compared to the 6 V reference?

  5. A thermistor with \$R0=10\,\$kΩ at \$25^{\circ}\mathrm{C}\$ and \$\beta=3500\,\$K is placed in the lower leg of a divider with \$R1=5\,\$kΩ and \$V_s=12\,\$V. Calculate the output voltage at \$25^{\circ}\mathrm{C}\$ and at \$75^{\circ}\mathrm{C}\$ (use Kelvin). Discuss the suitability of this circuit as a temperature sensor.