Current, power, and component limits

Current needs a closed path

Current is the rate at which charge passes a cross-section: $I=\Delta Q/\Delta t$. One ampere is one coulomb per second. Conventional current points in the direction a positive charge would move. Electron motion in a metal points the other way; circuit equations consistently use conventional current.

A source connected to only one resistor terminal does not establish a sustained DC current through the resistor. A complete return path is needed. The same current passes every element of a series path when no branch splits off. Voltage is measured across an element; current is measured through a branch. An ammeter must be inserted in series. Putting a current-input meter directly across a supply creates an unintended low-resistance branch.

Energy budget

With current entering the positive voltage terminal, absorbed power is $P=VI$. A resistor dissipates $I^2R=V^2/R$. Suppose 5 V is applied to 1 kΩ. Current is 5 mA and power is 25 mW. A hypothetical 0.125 W rating exceeds this calculated dissipation, but actual permitted power depends on temperature and the manufacturer's derating curve. A label alone is not a thermal design.

Now reduce resistance to 100 Ω. Current becomes 50 mA and power 250 mW. The voltage stayed the same while dissipation increased tenfold. Selecting a lower resistance can solve loading but cost current, heat, and battery life. A source supplying power has negative absorbed power under the same sign convention; this makes the total power sum a useful check.

A practical record

Create a table containing voltage, resistance, current, calculated dissipation, and an explicit power limit. Note which numbers are assumptions. For a 3.3 V rail and 330 Ω, the ideal current is 10 mA and resistor power is 33 mW. A measured current significantly above that prediction suggests a different resistance, extra parallel path, or wrong supply setting. Do not hide a mismatch by changing units until the number looks plausible.