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AP Physics · enrichment

Charge, Fields & Circuits — Visual Lab

Module 5. Two ideas from the same chapter, side by side: a Circuit Bench where you wire three resistors and read every current and voltage, and a two-charge field sub-view where you add electric fields as vectors. Change a slider; the physics updates instantly and exactly.

Interactive Lab Module 05 · Circuits & Fields Ohm's Law · Series/Parallel · E-Fields

A circuit is just a set of rules about current and voltage. In series, one current threads through everything and the voltage splits up. In parallel, everyone shares the full voltage and the current splits up. A combination nests one inside the other. The same charges that move through wires also reach out into empty space as an electric field \(\big(E = \tfrac{kq}{r^2}\big)\) — and when two charges act at once, you simply add their field vectors. Drive the lab below and watch both pictures stay numerically honest.


Circuit Bench & Field Lab

Pick a topology (series, parallel, or combination), set the source voltage and three resistors, and read the equivalent resistance, total current, and every branch. Then drag the test point in the field sub-view and read the net \(\vec{E}\) from two point charges by superposition.


Orientation

What you're seeing

  • Ohm's law is the engine. \(V = IR\) connects the three quantities; the bench solves the whole network and reports the total current \(I = \tfrac{V}{R_{eq}}\).
  • Series adds resistance. \(R_{eq} = R_1 + R_2 + R_3\). One current flows through all three; the source voltage divides among them.
  • Parallel lowers resistance. \(\tfrac{1}{R_{eq}} = \sum \tfrac{1}{R_i}\). Every branch sees the full voltage and the branch currents add up.
  • The moving dots are charge flow. Their speed tracks the current — more current, faster flow. Power \(P = VI\) is the rate energy leaves the source.
  • The field sub-view adds vectors. Each point charge contributes \(E = \tfrac{kq}{r^2}\); the readout is the signed sum at the test point \(P\).
Investigation

Try this

  1. Switch series → parallel with the same three resistors. Watch \(R_{eq}\) drop below the smallest resistor and the total current jump.
  2. In combination mode, make \(R_2 = R_3\). The parallel pair becomes exactly half of one of them — check \(R_p = \tfrac{R}{2}\) in the readout.
  3. Double the source voltage. Every current doubles and the power quadruples (\(P = \tfrac{V^2}{R}\)). The flow dots visibly speed up.
  4. In the field view, drag \(P\) between two equal positive charges. Find the spot where the net field reads \(\approx 0\) — the fields exactly cancel there.

Worked Examples

Two of the moves this module asks for most: reducing a combination network to a single current, and adding two electric fields as vectors.

Example 1 — Solve a combination circuit

R₁ in series with R₂ ∥ R₃

A \(12\,\text{V}\) battery drives \(R_1 = 4\,\Omega\) in series with the parallel pair \(R_2 = 6\,\Omega\) and \(R_3 = 12\,\Omega\). Find the total current and the voltage across \(R_1\).

  1. Collapse the parallel pair. \(R_p = \dfrac{R_2 R_3}{R_2 + R_3} = \dfrac{6\cdot 12}{6 + 12} = \dfrac{72}{18} = 4\,\Omega\).
  2. Add the series part. \(R_{eq} = R_1 + R_p = 4 + 4 = 8\,\Omega\).
  3. Total current (Ohm's law). \(I = \dfrac{V}{R_{eq}} = \dfrac{12}{8} = 1.5\,\text{A}\). This is the current through \(R_1\).
  4. Voltage across \(R_1\). \(V_1 = I R_1 = 1.5 \cdot 4 = 6\,\text{V}\); the other \(6\,\text{V}\) sits across the parallel block.
Answer: \(I = 1.5\,\text{A}\), \(V_1 = 6\,\text{V}\). Verify it in the lab by choosing Combination with \(V=12\), \(R_1=4\), \(R_2=6\), \(R_3=12\).

Example 2 — Net field of two point charges

Superposition along a line

Two charges sit on the x-axis: \(+4\,\mu C\) at \(x=1\,\text{m}\) and \(+4\,\mu C\) at \(x=5\,\text{m}\). Find the net electric field at the midpoint \(P\) at \(x=3\,\text{m}\). Use \(k = 8.99\times10^{9}\).

  1. Distances to \(P\). Each charge is \(r = 2\,\text{m}\) from the midpoint.
  2. Each magnitude. \(E = \dfrac{kq}{r^2} = \dfrac{(8.99\times10^{9})(4\times10^{-6})}{2^2} = 8{,}990\ \text{N/C}\) (same for both).
  3. Directions. A positive charge pushes a positive test charge away. The left charge pushes \(P\) to the right \((+x)\); the right charge pushes \(P\) to the left \((-x)\).
  4. Add as vectors. Equal magnitudes, opposite directions: \(E_{net} = 8990 - 8990 = 0\ \text{N/C}\).
Answer: \(E_{net} = 0\) at the midpoint of two equal like charges. Drag \(P\) to \(x = 3\) in the lab to see the field vanish — then move it off-center to watch it return.
Why it matters Every wall outlet, phone charger, LED, and microcontroller is a network of resistances obeying these exact rules — engineers size wires and choose resistor values with Ohm's law and the series/parallel reductions you just drove. The field idea scales the same way: from the spark of static electricity to the capacitor that smooths a power supply to the deflection plates in an old CRT, charges shape the space around them. Master both halves here and the bridge to magnetism, induction, and circuit design opens up.

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Key Vocabulary

The precise words a physicist uses to describe what the lab is doing.

Ohm's law

The relationship \(V = IR\) between voltage, current, and resistance for an ohmic conductor.

Equivalent resistance

The single resistor \(R_{eq}\) that would draw the same total current from the source as the whole network.

Series vs. parallel

Series shares one current and adds resistances; parallel shares one voltage and adds reciprocals \(\big(\tfrac{1}{R_{eq}} = \sum \tfrac{1}{R_i}\big)\).

Electric field

The force per unit charge, \(E = \tfrac{F}{q}\). A point charge produces \(E = \tfrac{kq}{r^2}\), pointing away from \(+q\) and toward \(-q\).

Superposition

The total field at a point is the vector sum of the fields from every charge, each computed as if alone.

Electrical power

The rate energy is delivered: \(P = VI = I^2R = \tfrac{V^2}{R}\), measured in watts.

About this course AP Physics · enrichment — not a TEKS-graded course. This module is offered as supplemental, college-level practice in algebra-based DC circuits and electrostatics, mirroring the AP Physics 1/2 treatment. It is not part of a Texas Essential Knowledge and Skills (TEKS) sequence and carries no TEKS grade; the on-grade, standards-aligned coursework lives in the Mathematical Architects math pathway. Use this lab to stretch, explore, and build college readiness.

Ready for the full course map? Head back to AP Physics, or read the Student Support page to see how enrichment fits into the studio.

AP Physics · enrichment — a supplemental, college-level course offered alongside the Mathematical Architects math pathway. Not a TEKS-graded course; "AP" refers to the College Board Advanced Placement framework, not a state standard.