Work on paper or with supported browser models first. For physical experiments use an appropriate isolated low-voltage source and suitably rated parts. Power off before rewiring; never connect a current-mode meter across a source.
1. Map the nodes
Choose one circuit below. Draw its full return path and label each shared connection with one node name. Mark both terminals of every voltage measurement; draw the break where a correctly configured series ammeter would go.
My circuit / source / component values: ____________________
Source positive node: __________ Return node: __________
Voltage measurement points: __________ and __________
2. Make a prediction
| Circuit and assumptions | Predict before checking |
|---|---|
| Regulated 5 V across one ideal 1 kΩ resistor | Resistor current: ______ mA Resistor dissipation: ______ mW |
| 5 V, assumed 2 V LED drop, 330 Ω and switch in series | Closed-switch current: ______ mA Open-switch steady branch current: ______ mA |
| 5 V divider: upper 10 kΩ, lower 10 kΩ; no output load | Output to return: ______ V Series current: ______ mA |
| Same divider with 10 kΩ from output to return | Lower equivalent resistance: ______ kΩ Loaded output: ______ V |
3. Record observations without rewriting the prediction
Record whether the values came from arithmetic, an educational model or a physical instrument. A browser result is not a measurement of physical hardware.
Method and meter mode: ____________________
Expected: __________ Observed: __________ Units: __________
One plausible reason for the difference: ____________________
If a result surprises you, power off and check connections, breadboard strips, rail continuity, component value and polarity before changing formulas. Use voltage mode across a resistor to estimate its current from its known resistance when appropriate.
Worked answers and checks
Resistor: I = 5 / 1000 = 0.005 A = 5 mA; P = 5 × 0.005 = 0.025 W = 25 mW.
LED: I ≈ (5 − 2) / 330 = 9.09 mA with the switch closed. With an ideal open switch the steady branch current is zero, but not every voltage in the circuit is necessarily zero. The LED forward voltage is an assumption; actual current and ratings need separate checks.
Unloaded divider: Vout = 5 × 10 / (10 + 10) = 2.5 V, current = 5 / 20000 = 0.25 mA. Each resistor dissipates 0.625 mW.
Loaded divider: 10 kΩ ∥ 10 kΩ = 5 kΩ; Vout = 5 × 5 / (10 + 5) ≈ 1.67 V. This is why the unloaded formula cannot be used unchanged after attaching a load.
Follow the worked builds
Check a resistor network with the series and parallel resistance calculator, or return to the learning roadmap.