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Electricity & Circuits — Visual Lab
Module 5. A battery pushes charge; resistors push back. Wire two or three resistors in series or in parallel and watch the equivalent resistance, the total current, and the voltage and current in each resistor stay in perfect agreement — Ohm's law made visible.
Every circuit obeys one tiny rule — Ohm's law, \(V = IR\) — and two wiring choices. In series, resistors sit end-to-end: one current threads through all of them and the source voltage splits between them. In parallel, they sit side-by-side: every resistor feels the full voltage and the current divides among the branches. Drive the lab below and keep the schematic, the equivalent resistance, and the per-resistor numbers locked together.
Circuit Builder Lab
Set the source voltage and each resistance, then flip the series / parallel switch (and optionally add a third resistor). The schematic redraws, charge drifts around the live loop, and the readout reports the equivalent resistance, the total current \(I = V/R_{eq}\), and the exact voltage and current in every resistor.
What you're seeing
- The battery is the source. Its voltage \(V\) is the "push." The drifting dots picture the current; the readout gives the exact amps — the dots are only a feel for it.
- Each zig-zag is a resistor. Resistance \(R\) (ohms) opposes current. Ohm's law ties them together: \(V = IR\) across any resistor.
- Series = one path. The same current flows through every resistor; resistances add, so \(R_{eq}\) is bigger than any single one.
- Parallel = many paths. Every resistor feels the full source voltage; the reciprocals add, so \(R_{eq}\) is smaller than any single one.
- Power follows. The source delivers \(P = IV\) watts; each resistor dissipates its own \(P = I^2R\). The readout shows all of it.
Try this
- Set both resistors to 10 Ω in series with 12 V. Read \(R_{eq} = 20\,\Omega\), \(I = 0.6\,\text{A}\), and \(6\,\text{V}\) across each — the voltage splits evenly.
- Flip the same circuit to parallel. Now \(R_{eq} = 5\,\Omega\), each resistor still sees \(12\,\text{V}\), each draws \(1.2\,\text{A}\), and the battery supplies \(2.4\,\text{A}\) total.
- Make \(R_2\) huge in parallel. Its branch current shrinks toward zero — current prefers the path of least resistance. The smallest resistor hogs the current.
- Add a third resistor in series. Watch \(R_{eq}\) climb and the current fall — more resistance in one path means less current everywhere on it.
Worked Examples
Two of the moves this module asks for most: combining resistors and pushing Ohm's law all the way to the current in a single resistor.
Example 1 — A 12 V battery with two series resistors
A \(12\,\text{V}\) battery drives a \(4\,\Omega\) resistor and an \(8\,\Omega\) resistor wired in series. Find the equivalent resistance, the current, and the voltage across each resistor.
- Series resistances add. \(R_{eq} = R_1 + R_2 = 4 + 8 = 12\,\Omega\).
- Ohm's law for the whole loop. \(I = \dfrac{V}{R_{eq}} = \dfrac{12}{12} = 1\,\text{A}\) — this one current flows through both resistors.
- Voltage across each (the divider). \(V_1 = I R_1 = 1 \cdot 4 = 4\,\text{V}\); \(V_2 = I R_2 = 1 \cdot 8 = 8\,\text{V}\).
- Check. The drops sum to the source: \(4 + 8 = 12\,\text{V}\). \(\checkmark\)
Example 2 — Two resistors in parallel
A \(12\,\text{V}\) battery drives a \(6\,\Omega\) resistor and a \(12\,\Omega\) resistor in parallel. Find the equivalent resistance, the current in each branch, the total current, and the power delivered.
- Parallel reciprocals add. \(\dfrac{1}{R_{eq}} = \dfrac{1}{6} + \dfrac{1}{12} = \dfrac{2}{12} + \dfrac{1}{12} = \dfrac{3}{12} = \dfrac{1}{4}\), so \(R_{eq} = 4\,\Omega\) — smaller than either resistor.
- Each branch feels the full 12 V. \(I_1 = \dfrac{12}{6} = 2\,\text{A}\); \(I_2 = \dfrac{12}{12} = 1\,\text{A}\) — the smaller resistor draws more.
- Total current. \(I = I_1 + I_2 = 2 + 1 = 3\,\text{A}\), which also equals \(\dfrac{V}{R_{eq}} = \dfrac{12}{4} = 3\,\text{A}\). \(\checkmark\)
- Power delivered. \(P = IV = 3 \cdot 12 = 36\,\text{W}\).
Key Vocabulary
The precise words a physicist uses to describe what the lab is doing.
The energy per unit charge a source supplies, measured in volts (V). It is the "push" that drives current around the circuit.
The rate of flow of charge, measured in amperes (A). One ampere is one coulomb of charge passing a point each second.
How strongly a component opposes current, measured in ohms (Ω). Ohm's law links them: \(V = IR\).
Components on a single path. The same current passes through each; resistances add: \(R_{eq} = R_1 + R_2 + \cdots\).
Components on separate branches sharing two nodes. Each feels the same voltage; reciprocals add: \(\tfrac{1}{R_{eq}} = \sum \tfrac{1}{R_i}\).
Energy delivered per second, measured in watts (W). In a circuit \(P = IV = I^2R = \dfrac{V^2}{R}\).
Where this fits
This is a Physics enrichment lab — not a TEKS-graded course. It is offered to apply and extend the algebra you build in the math pathway (solving \(V = IR\) for any variable, adding fractions and reciprocals for parallel resistors, and working in scientific notation). The physics content here is the standard introductory, algebra-based treatment of DC circuits.
Ready for more? Head back to Physics or visit Student Support for office hours and help.
Physics · enrichment. Offered alongside the math pathway as enrichment; not part of the tested TEKS sequence and not graded against TEKS standards. Circuit relationships follow the standard algebra-based physics treatment of DC circuits.