A real battery loses some voltage internally
You will be able to: Model a delivering battery with emf and series internal resistance.
Why can battery voltage sag under load?
A battery reads 12 V with almost no current, but its terminal voltage drops when a device draws more current. Model that behavior with an ideal 12 V source and a small resistor inside the battery boundary.
A useful starting point: Reduce a network, then work backward →
Words and symbols before equations
- Emf ε
- Open-circuit voltage of the source model, in V; despite its name, not a force in newtons.
- Internal resistance r
- Resistance inside the source model, in Ω.
- Terminal voltage V_t
- Voltage across the battery’s external terminals.
- Load R
- External device resistance receiving energy.
What this picture assumes
Battery delivers current with fixed emf 12 V. Internal resistance and load are constant; wire resistance is neglected. This is a discharge model, not a charging relation.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- I=2 A; V_terminal=12−Ir=10 V. Power: 4 W internal + 20 W load = 24 W supplied.
- 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 battery delivering current, the loop contains r and R in series. Therefore I=ε/(R+r), and terminal voltage is V_t=ε−Ir=IR. This sign assumes discharge: a charging battery can have a different terminal relation.
The source supplies chemical-to-electrical power εI. The external load receives I²R, while I²r heats the battery internally. Those two powers sum to εI.
Lower load resistance draws more current and produces a larger internal drop. Real wires can also add series resistance. Ideal-wire approximations are justified only when wire resistance is small compared with relevant component resistances; they fail for an otherwise shorted source.
A worked example, step by step
A 12 V emf battery has r=1 Ω and drives R=5 Ω. Find I, terminal voltage and internal power.
- Total resistance=5+1=6 Ω.
- I=12/6=2 A.
- V_t=12−(2)(1)=10 V, matching IR=(2)(5).
- Internal heating is I²r=4 W; load gets 20 W, totaling εI=24 W.
Emf and terminal voltage are equal only when internal drop is negligible, such as the open-circuit limit.
What does the terminal voltage approach as current approaches zero?
Compare with an explanation
The emf ε, because Ir approaches zero.
Predict. Change one thing. Explain.
Reduce load resistance and compare source current, terminal voltage and internal heating. Keep the battery emf fixed.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
I=2 A; V_terminal=12−Ir=10 V. Power: 4 W internal + 20 W load = 24 W supplied.
Battery delivers current with fixed emf 12 V. Internal resistance and load are constant; wire resistance is neglected. This is a discharge model, not a charging relation.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant current, voltage or energy 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.
Original written challenge
4 points · self-check · not an official AP questionA battery with ε=6 V and r=0.5 Ω drives a 2.5 Ω load. (a) Find I. (b) Find V_t. (c) Find internal heating power. (d) Check total power.
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Compare with the answer and four-point rubric
- 1 point: I=6/3=2 A.
- 1 point: V_t=6−1=5 V.
- 1 point: I²r=2 W.
- 1 point: Load gets 10 W; 10+2=12 W=εI.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Is emf a mechanical force?
No; its unit is volts.
RECALL 2Which way does V_t change under heavier discharge?
It falls as internal voltage drop grows.
RECALL 3Where does internal electrical dissipation go?
Into thermal energy within the battery model.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A real battery loses some voltage internally
- Delivering battery: I=ε/(R+r).
- V_t=ε−Ir.
- εI=I²R+I²r.
Remember: Emf and terminal voltage are equal only when internal drop is negligible, such as the open-circuit limit.
Conditions: Battery delivers current with fixed emf 12 V. Internal resistance and load are constant; wire resistance is neglected. This is a discharge model, not a charging relation.
Refresh Kid · AP Physics 2 Unit 3 (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. Refresh Kid calls this the third AP Physics 2 unit; College Board numbers it Unit 11; the first unit in this course is official Unit 9. 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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