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LESSON 12 / 22 · TOPIC 11.5

A loaded battery’s terminal voltage can fall

You will be able to: Account for internal and wire resistance and separate their power losses.

Calculus-based circuit analysisFree study resourceReview editionTeacher review pending

Why is terminal voltage different from emf while a battery supplies current?

A battery might read 12 V with no load but less while supplying a device. Some voltage is dropped across resistance inside the battery, and connecting wires can lose additional voltage.

A useful starting point: Reduce a network one valid group at a time →

Words and symbols before equations

Emf ε
Energy supplied per charge by the ideal source part, measured in volts, not force units.
Internal resistance r
Series resistance inside the battery model.
Terminal voltage
Voltage across the battery’s external terminals.
Wire resistance R_w
Total resistance of the external connecting wires.
Battery and external resistances: lumped series model+ε = 12 Vr+R_w = 2 ΩLoad 4 ΩReference I →Schematic geometry; ideal wires unless labeled otherwise
Read this model snapshot. I = 2 A; battery terminal V = 10 V; load V = 8 V. Emf supplies 24 W: internal 4 + wire 4 + load 16 W.
What this picture assumes

Discharging battery with positive load; emf in series with internal r. External wire resistance is modeled separately. Terminal voltage = emf − Ir. Charging polarity is outside this model.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. I = 2 A; battery terminal V = 10 V; load V = 8 V. Emf supplies 24 W: internal 4 + wire 4 + load 16 W.
  3. Check what the picture assumes below. Use the Explore task to predict one change before moving a control.

Connect the picture to the physics

For a discharging battery feeding R_load and R_w, I = ε/(r+R_w+R_load). The terminal voltage is ε−Ir.

The load receives V_load = I R_load. This can be lower than terminal voltage because wires drop I R_w. Ideal-wire approximations must be reconsidered when wire resistance is significant.

Power supplied by the emf is εI. Internal, wire and load resistor powers sum to I²(r+R_w+R_load). The relation ε−Ir here assumes discharge; charging the battery reverses the relevant current direction.

A worked example, step by step

A 12 V emf has r = 1 Ω, wires total 1 Ω, and a 4 Ω load. Find current and voltages.

  1. Total resistance is 6 Ω, so I = 2 A.
  2. Battery terminal voltage is 12−2(1) = 10 V.
  3. Wire drop is 2 V and load voltage is 8 V.
  4. Power balance: 24 W supplied = 4 W internal + 4 W wires + 16 W load.
Common mix-up

Battery terminal voltage and load voltage differ when external wires have appreciable resistance.

CHECK THE IDEA

What does a nearly open-circuit voltmeter read in this model?

Compare with an explanation

With negligible current, terminal voltage approaches the emf.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Increase load resistance and compare current, terminal voltage and useful load power. Then vary wire resistance separately from internal resistance.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

Battery and external resistances: lumped series model+ε = 12 Vr+R_w = 2 ΩLoad 4 ΩReference I →Schematic geometry; ideal wires unless labeled otherwise

I = 2 A; battery terminal V = 10 V; load V = 8 V. Emf supplies 24 W: internal 4 + wire 4 + load 16 W.

Power destinationW · same scale for all bars0Inside battery4External wires4Load16

Discharging battery with positive load; emf in series with internal r. External wire resistance is modeled separately. Terminal voltage = emf − Ir. Charging polarity is outside this model.

Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant current, circuit topology, charge or energy conservation, or RC relationship to justify your prediction.

Apply the idea to a fresh problem Practice →

Show what you understand.

Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.

1. Emf is measured in…

Show answer and reasoning

volts. It describes energy supplied per charge.

2. With ε = 9 V, r = 1 Ω and I = 2 A during discharge, terminal V is…

Show answer and reasoning

7 V. V_terminal = ε−Ir = 7 V.

Original written challenge

4 points · self-check · not an official AP question

A 6 V battery with r = 0.5 Ω supplies a 2 Ω load through 0.5 Ω of wire. Find current, terminal/load voltages and internal power.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: I = 6/(0.5+0.5+2) = 2 A.
  2. 1 point: Terminal voltage is 6−2(0.5) = 5 V.
  3. 1 point: Load voltage is 2(2) = 4 V.
  4. 1 point: Internal power is I²r = 2 W.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1When do terminal voltage and emf match?

When current is zero or internal resistance is negligible.

RECALL 2Which wire resistance is used?

The total resistance of the outgoing and return paths.

RECALL 3Why can a short be dangerous in real circuits?

Small total resistance can permit a large current and power dissipation.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

A loaded battery’s terminal voltage can fall

  • I = ε/(r+R_w+R_load).
  • Discharge: V_terminal = ε−Ir.
  • P_emf = P_internal+P_wire+P_load.

Remember: Battery terminal voltage and load voltage differ when external wires have appreciable resistance.

Conditions: Discharging battery with positive load; emf in series with internal r. External wire resistance is modeled separately. Terminal voltage = emf − Ir. Charging polarity is outside this model.

Refresh Kid · AP Physics C: Electricity and Magnetism Unit 4 (official Unit 11) · Objectives 11.5.B · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 11.5, objectives 11.5.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is E&M Unit 4: Electric Circuits, numbered Unit 11 in the official combined Physics C sequence. Topics 11.1–11.8 retain their official identifiers. Models include signed charge flow, prescribed current-density and resistivity integrals, DC resistor networks, nonideal batteries and meters, capacitor combinations and finite-resistance RC transients. Circuit schematics use conventional-current references and explicit node connectivity. Initial capacitor voltage and positive time constants are stated. Unequal ideal sources are never directly wired in parallel. The optional spatial wire view supplements a complete 2D current-density explanation. Checked with the Fall 2026 clarifications. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.

Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.

Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.

Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.

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