Skip to main content
EnglandPhysicsSyllabus dot point

How do Kirchhoff's laws, internal resistance and potential dividers determine currents and voltages in a circuit?

Electrical circuits: Kirchhoff's two laws, series and parallel combinations of resistors, electromotive force and internal resistance, and the potential divider including sensor circuits.

A focused answer to the OCR H556 content on electrical circuits, covering Kirchhoff's two laws, series and parallel resistor combinations, electromotive force and internal resistance with terminal potential difference, and the potential divider used in sensor circuits.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

OCR wants you to apply Kirchhoff's two laws, combine resistors in series and parallel, account for electromotive force and internal resistance with the terminal potential difference, and analyse a potential divider including sensor circuits using a thermistor or LDR.

The answer

Kirchhoff's laws

Series and parallel resistors

Electromotive force and internal resistance

A graph of terminal p.d. against current is a straight line with intercept ε\varepsilon and gradient −r-r, which is how internal resistance is measured experimentally.

Potential dividers

Examples in context

Potential dividers with thermistors form the temperature-sensing input to thermostats and fire alarms; with LDRs they switch streetlights at dusk. Internal resistance limits the maximum current a battery can deliver and explains why a car's headlights dim when the starter motor draws a huge current. Kirchhoff's laws are the foundation of all circuit analysis, from simple resistor networks to the design of measuring instruments. Multi-cell battery packs trade off emf (cells in series) against current capacity (cells in parallel).

Try this

Q1. State Kirchhoff's first law and the conservation principle behind it. [2 marks]

  • Cue. The total current into a junction equals the total current out; it follows from conservation of charge.

Q2. Two resistors of 6.0 Ω6.0\ \Omega and 3.0 Ω3.0\ \Omega are connected in parallel. Find their combined resistance. [2 marks]

  • Cue. 1R=16.0+13.0=12.0\frac{1}{R} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{1}{2.0}, so R=2.0 ΩR = 2.0\ \Omega.

Q3. A cell of emf 9.0 V9.0\ \text{V} and internal resistance 1.0 Ω1.0\ \Omega drives a current of 1.5 A1.5\ \text{A}. Find the terminal p.d. [2 marks]

  • Cue. V=ε−Ir=9.0−1.5×1.0=7.5 VV = \varepsilon - Ir = 9.0 - 1.5 \times 1.0 = 7.5\ \text{V}.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20184 marksA cell of emf 1.5 V1.5\ \text{V} and internal resistance 0.50 Ω0.50\ \Omega is connected to a 4.0 Ω4.0\ \Omega resistor. Calculate the current in the circuit and the terminal potential difference across the cell.
Show worked answer →

Total resistance is the external resistor plus the internal resistance: Rtotal=4.0+0.50=4.5 ΩR_{\text{total}} = 4.0 + 0.50 = 4.5\ \Omega.

Current: I=εR+r=1.54.5=0.333 AI = \frac{\varepsilon}{R + r} = \frac{1.5}{4.5} = 0.333\ \text{A}.

Terminal p.d. is the voltage across the external resistor: V=IR=0.333×4.0=1.33 VV = IR = 0.333 \times 4.0 = 1.33\ \text{V}. (Equivalently V=ε−Ir=1.5−0.333×0.50=1.33 VV = \varepsilon - Ir = 1.5 - 0.333 \times 0.50 = 1.33\ \text{V}.)

Markers reward including the internal resistance, the current about 0.33 A0.33\ \text{A}, and the terminal p.d. about 1.3 V1.3\ \text{V}.

OCR 20214 marksA potential divider consists of a 12 V12\ \text{V} supply across two resistors in series: a fixed 2.0 kΩ2.0\ \text{k}\Omega resistor and a light-dependent resistor (LDR). In darkness the LDR has a resistance of 10 kΩ10\ \text{k}\Omega. Calculate the output voltage taken across the LDR in darkness.
Show worked answer →

The potential divider equation gives the voltage across the LDR: Vout=Vin×RLDRRLDR+RfixedV_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R_{\text{LDR}} + R_{\text{fixed}}}.

Substitute: Vout=12×1010+2.0=12×1012=10 VV_{\text{out}} = 12 \times \frac{10}{10 + 2.0} = 12 \times \frac{10}{12} = 10\ \text{V}.

Markers reward the potential divider equation, putting the LDR resistance on top (because the output is across the LDR), and the value 10 V10\ \text{V}. In light the LDR resistance falls, so the output voltage falls, which is how the circuit senses light.

Related dot points

Sources & how we know this