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What makes a circuit complete, and what does a battery actually supply to drive a current?

Topic 11.2 Simple Circuits: interpret circuit schematics and explain the role of emf, the complete circuit and the conventions for open and short circuits.

A focused answer to AP Physics 2 Topic 11.2, covering circuit schematics and their symbols, the complete (closed) circuit, the role of electromotive force as energy per charge supplied by a source, internal resistance and terminal voltage, and open and short circuits, with full worked examples.

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  1. What this topic is asking
  2. Schematics and the complete circuit
  3. Electromotive force and terminal voltage
  4. Open and short circuits
  5. Try this

What this topic is asking

The College Board (Topic 11.2) wants you to interpret circuit schematics, explain the role of electromotive force (emf) as the energy per charge a source supplies, and understand complete, open and short circuits, including the effect of internal resistance on terminal voltage.

Schematics and the complete circuit

A circuit only carries current when the loop is unbroken: charge must have a continuous path back to the source. A schematic is the map of that path, stripped of physical detail down to which components connect to which. The first step in any circuit problem is to read the schematic, identify the loop, and see which components are in the current's path.

Electromotive force and terminal voltage

Despite its name, emf is not a force; it is energy per charge, the same kind of quantity as voltage. It measures the "push" a source gives each coulomb. Internal resistance is the small resistance inside the source itself, and some of the emf is spent driving current through it, which is why the terminal voltage drops below the emf as the current rises. This is why a car's headlights dim when the starter motor (a heavy load drawing large current) cranks: the large current causes a big IrI r drop inside the battery.

Open and short circuits

Two extreme cases are tested directly. An open circuit has a break in the loop (a switch off, a cut wire), so the resistance is effectively infinite and no current flows. A short circuit provides a path of near-zero resistance across the source, so by I=V/RI = V/R a very large current flows, limited mainly by the internal resistance; this can overheat wires and is what fuses protect against. The strategic role of this topic is to set up the analysis tools for the rest of the unit: you must read a schematic, recognize the source's emf as the energy supply, account for internal resistance when present, and identify whether the circuit is complete, open or shorted. With the current found from I=E/(R+r)I = \mathcal{E}/(R + r), the resistance rules (Topic 11.3), power (Topic 11.4) and Kirchhoff's laws (Topics 11.6, 11.7) all follow.

Try this

Q1. State what emf measures and give its unit. [2 points]

  • Cue. The energy supplied per unit charge by a source, measured in volts.

Q2. State what happens to the current in a circuit if the loop is broken (open circuit). [1 point]

  • Cue. The current falls to zero (no complete path for charge to flow).

Exam-style practice questions

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

AP 2024 (style)6 marksSection II (short FRQ). A battery has an emf of 1212 V and an internal resistance of 0.500.50 ohms. It is connected to an external resistor of 5.55.5 ohms. (a) Calculate the current in the circuit. (b) Calculate the terminal voltage of the battery. (c) Explain why the terminal voltage is less than the emf.
Show worked answer β†’

A 6-point FRQ on emf and internal resistance.

(a) Current (2 points): the total resistance is 5.5+0.50=6.05.5 + 0.50 = 6.0 ohms, so I=ER+r=126.0=2.0I = \dfrac{\mathcal{E}}{R + r} = \dfrac{12}{6.0} = 2.0 A.
(b) Terminal voltage (2 points): Vterm=Eβˆ’Ir=12βˆ’(2.0)(0.50)=11V_{term} = \mathcal{E} - I r = 12 - (2.0)(0.50) = 11 V (or Vterm=IR=(2.0)(5.5)=11V_{term} = IR = (2.0)(5.5) = 11 V).
(c) Explanation (2 points): some of the emf is "used up" driving the current through the internal resistance, so the voltage available at the terminals is less than the full emf.

Markers reward the current with total resistance, the terminal voltage, and explaining the drop across the internal resistance.

AP 2023 (style)1 marksSection I (multiple choice). A circuit has a battery and a bulb. If a thick, low-resistance wire is connected directly across the battery's terminals (a short circuit), what happens? (A) no current flows (B) a very large current flows (C) the current is unchanged (D) the bulb glows brighter. Justify your reasoning.
Show worked answer β†’

A 1-point MCQ on short circuits. The answer is (B).

A short circuit provides a path of almost zero resistance across the source. By I=V/RI = V/R, a very small resistance gives a very large current, limited mainly by the internal resistance. The trap is (A): a short circuit gives a large current, not zero.

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