AeroLab Reading room

Physics / AS level

D.C. circuits

Apply charge and energy conservation to networks and sources.

Learning objectives

The big idea

At a junction, total current entering equals total current leaving because charge does not accumulate steadily. Around a closed loop, the sum of emfs equals the sum of potential drops when signs are chosen consistently.

A real source has internal resistance, so some energy is transferred inside it when current flows. A potential divider shares voltage between series resistances; loading its output can change the voltage it supplies.

1. Draw connections rather than trusting appearance

Circuit behaviour depends on which points share connections, not on where components appear on a page. Mark junctions and follow each conducting branch. Series components share current; components across the same two nodes share potential difference. Replacing a complicated resistor network with an equivalent resistance can simplify total-current calculations, but individual branch currents still require returning to the original connections.

2. Kirchhoff’s rules express conservation

At a junction, total current entering equals total current leaving for steady operation. Around a closed loop, directed emf and potential-drop sums balance. Choose a loop direction and current directions consistently; a negative calculated current can indicate that the actual direction is opposite your initial choice. These are not separate rules about how much current a battery chooses to send: the source and entire connected load determine the result.

3. A real source uses energy internally

An emf ε supplies energy per unit charge; internal resistance r transfers some energy within the source. For a source delivering current I, terminal voltage is ε − Ir. A 1.5 V cell delivering 0.50 A with internal resistance 0.20 Ω has terminal voltage 1.40 V. The model assumes a suitable operating range and does not make emf and internal resistance immutable under every temperature, charge state or current.

4. Potential dividers respond to their loads

Two series resistors across a source share voltage in proportion to resistance. Measuring an unloaded output across one resistor gives the familiar divider ratio. Connecting a load places another resistance across that section, reducing its equivalent resistance and usually changing the output. A thermistor or light-dependent resistor can turn a divider into a sensor, but the output must be analysed for the actual component placement and loading.

A method you can reuse

  1. Redraw nodes clearly and simplify only genuine series or parallel groups.
  2. Apply current conservation at relevant junctions.
  3. Write one consistently signed loop equation for each needed path.
  4. Include internal resistance when relating emf and terminal voltage.
  5. Recalculate the divider network when a load or sensor changes.
Why does a low-resistance voltmeter disturb a potential-divider measurement?

It acts as a parallel load across the measured section, changing that section’s equivalent resistance and therefore the voltage division. A suitably high input resistance reduces this effect.

Equations and relationships

Vterminal=ε−IrV_{\text{terminal}}=\varepsilon-Ir

A source delivering current I with internal resistance r.

Vout=VinR2R1+R2V_{\text{out}}=V_{\text{in}}\frac{R_2}{R_1+R_2}

Unloaded two-resistor potential divider with output across R₂.

Worked example

A 12 V source supplies 2 A through an internal resistance of 1 Ω. What is its terminal voltage?

Follow the solution

V = 12 − 2 × 1 = 10 V.

Investigate and explain

Build a simulated potential divider, measure its unloaded output, then add a load and explain the change.

Use separate source and load resistors to model internal resistance explicitly. Compare an unloaded divider with a loaded one; ideal meters and sources can conceal real measurement effects.

Open the interactive investigation

Check your understanding

1. Kirchhoff’s junction rule expresses conservation of what?

  1. Resistance
  2. Charge
  3. Potential difference at every component
Check the answer and reasoning

Charge

Steady current cannot continually create charge at a junction.

2. A 12 V source supplies 2 A through an internal resistance of 1 Ω. What is its terminal voltage?

  1. 14 V
  2. 6 V
  3. 10 V
Check the answer and reasoning

10 V

V = 12 − 2 × 1 = 10 V.

3. Why can a divider’s output change when a load is connected?

  1. The load changes the equivalent lower-branch resistance.
  2. Kirchhoff’s laws stop applying.
  3. The supply always increases its emf.
Check the answer and reasoning

The load changes the equivalent lower-branch resistance.

The parallel load alters the resistance ratio.