Mini Physics
Part of Mini EducationSwitch subject
Search

Internal resistance, terminal p.d. and output power

Key idea: H2 Physics lessons on circuit representation, material behaviour, networks and RC transients.

  • GCE A-Level H2 Physics 2027

Build the idea

Learn the idea

Big question: Why does a real source's terminal voltage fall under load?

A source with e.m.f. ε and internal resistance r delivering current I has terminal p.d. V = ε − Ir. The lost volts Ir represent energy per coulomb transferred inside the source. Output power is IV; increasing current can raise internal heating I²r and reduce terminal p.d.

Follow energy per charge through a real source

E.m.f. ε is energy supplied per coulomb by the source. While the source delivers current I, energy per charge Ir is transferred inside its internal resistance r, leaving terminal p.d. V = ε − Ir for the external circuit.

At open circuit I = 0 and V = ε. A falling terminal voltage under load is therefore not a used-up e.m.f.; it reflects a larger internal energy transfer per charge.

Check your understanding: A 1.5 V cell has r = 0.20 Ω and supplies 2.0 A. Find terminal p.d.

V = 1.5 − 2.0(0.20) = 1.1 V.

Check the source power account

The source power is εI. External output power is VI and internal heating is I²r, so εI = VI + I²r while delivering current.

A graph of V against I has intercept ε and gradient −r. When an external supply charges the cell, current direction changes and the terminal p.d. can exceed ε; decide the operating mode before choosing signs.

Check your understanding: What do intercept and gradient give on a terminal-V against I graph?

Intercept ε and gradient −r.

Real source with internal resistance and external loadA circuit model with ideal electromotive force epsilon and internal resistance r inside the source boundary, connected in series to an external load R. Current leaves the positive terminal and voltage labels show epsilon equals V plus Ir during discharge.real source+−ideal e.m.f. εinternal rlost p.d. = IrRterminal p.d. VIdischarging: ε = V + Irsource energy = load transfer + internal heating
Discharging-source model: the ideal e.m.f. and internal resistance are in series. The terminal p.d. across the load is V = ε − Ir.

Key ideas to keep

  • With no current, terminal p.d. equals e.m.f. in this model.
  • Use ε + Ir when an external source is charging the cell.
  • Lost volts are a potential difference, not energy unless multiplied by charge.

See the reasoning

Worked example

Derive the terminal behaviour of a loaded source

Question: For a source of e.m.f. E and internal resistance r supplying load R, derive terminal p.d. and load power.

  1. Step 1: Apply the loop equation

    Why: The load and internal resistance carry the same current.

    Working: ε = I(R + r), so I = ε/(R + r).

  2. Step 2: Find terminal p.d.

    Why: It is the load drop after the internal lost volts.

    Working: V = IR = ε − Ir.

  3. Step 3: Separate useful and internal powers

    Why: Energy supplied by the source has two transfer routes.

    Working: Pload = I²R and Pinternal = I²r; together they equal εI.

Answer: Kirchhoff's loop relation gives E = I(R + r), so I = E/(R + r). Terminal p.d. is V = IR = E − Ir. Load power is P = I²R = E²R/(R + r)²; internal heating is I²r.

Check: At I = 0, lost volts vanish and terminal p.d. returns to ε.

Use a hint if needed

Practise with support

Try this

A 2.0 V source with r = 0.50 Ω supplies 1.0 A. Find terminal p.d. and internal power.

Hint: Lost volts are Ir, not r/I.

Check your answer

V = E − Ir = 1.5 V. Internal power is I²r = 0.50 W.

Now work without the hint

Practise independently

Your turn

Explain how increasing load current affects terminal p.d., useful output power and internal dissipation.

Check your answer

The lost volts Ir grow, so terminal p.d. E − Ir falls. Useful power is IV = I(E − Ir), while internal dissipation I²r grows. Output power is not simply proportional to current because terminal p.d. changes.

Avoid these traps

Common mistakes

Common mistake

Terminal p.d. always equals e.m.f.

What is wrong with this reasoning?

Show better thinking

When current is supplied, Vterminal = E − Ir; Ir is the internal lost volts.

Write for the examiner

Exam guidance

State whether the source is delivering or being charged before choosing the voltage sign.

Exam-style practice [6 marks]

A 12 V source with r = 1.0 Ω supplies a 5.0 Ω load. Find I, terminal p.d., useful power and internal power.

Plan before you answer

  • Include r in the total resistance.
  • Use V = ε − Ir.
  • Account for both output powers.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

Model answer

I = 12/6 = 2.0 A. V = 10 V. Useful power = 20 W; internal power = I²r = 4.0 W.

Come back in three days

Check what stayed with you

Recall question

A 1.8 V cell supplies 0.60 A and has terminal p.d. 1.5 V. Find r.

Check the answer

Lost volts = 1.8 − 1.5 = 0.30 V, so r = 0.30/0.60 = 0.50 Ω.

Keep moving

Try this next

Continue to the next lesson in this topic.

Series, parallel and potential-divider networks

Syllabus and review details

This lesson covers the listed H2 Physics 9478 outcomes. Topic 16 states no explicit exclusions. I–V graph gradient language always names the plotted axes; resistance is V/I at an operating point rather than automatically a graph gradient. Metal temperature dependence is explained through drift velocity and NTC behaviour through carrier number density. Terminal p.d. is E − Ir while a source supplies current. Capacitor rules are not copied from resistor rules. The exponential forms assume constant R and C, an ideal source or isolated discharge loop, and τ = RC.

  • GCE A-Level H2 PhysicsTopic 16(g) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027