The short answer

Ohm’s law connects the voltage across an ohmic resistor to the current through it. First find the two nodes and the path; then do the arithmetic.

On this page
Source, resistor and return pathA closed loop joins the positive terminal of a low-voltage source to a resistor and back to the negative terminal. Voltage V is across the resistor and current I follows the loop.+−RV across R← I · return pathIllustrative low-voltage loop · not to scale
Source, resistor and return path

A relationship between two terminals

For an ideal resistor, the voltage difference across its terminals equals its resistance multiplied by the current through it. This is a statement about one element, not automatically about the entire circuit. Mark the resistor’s two ends, choose a direction for current, and identify the voltage between those ends. A 1 kΩ part connected directly across a regulated 5 V source has 5 V across it; a 1 kΩ part elsewhere in a network might not.

Voltage is measured in volts (V), current in amperes (A), and resistance in ohms (Ω). The kilo prefix means 1,000, while milli means one thousandth. Keeping those units visible prevents a common thousandfold mistake.

V = I × R; I = V / R; R = V / I

Worked example: 5 V across 1 kΩ

Assume an ideal 5 V DC source, one 1,000 Ω resistor, negligible lead resistance, and a closed loop. Current is 5 V / 1,000 Ω = 0.005 A = 5 mA. The resistor dissipates V × I = 5 V × 0.005 A = 0.025 W = 25 mW. With these assumptions, the entire supply voltage appears across that single resistor. The same arithmetic can describe a branch in a larger circuit only after you establish its actual branch voltage.

A nominal quarter-watt resistor has a rating above 25 mW, but a rating is not a promise of safe operation under every ambient temperature or mounting condition. Check the part’s data sheet and leave margin.

I = 5 V / 1,000 Ω = 5 mA; P = 5 V × 5 mA = 25 mW

Run your own numbers with the Ohm’s law calculator.

Find the voltage before dividing

Suppose two 1 kΩ resistors are in series across 5 V. If nothing else is connected and both behave ideally, the total is 2 kΩ, so loop current is 2.5 mA and each resistor drops 2.5 V. Dividing 5 V by either individual 1 kΩ resistor would incorrectly claim 5 mA in the same series loop. Shared current in series and shared voltage in parallel are useful checks before applying a formula to any component.

On a breadboard, holes that look close may not share a node. A loose lead or an unbridged power rail can make the assumed 5 V across a resistor disappear entirely.

Choosing a resistor from a target current

If a simple resistive branch should draw 2 mA at 5 V, calculate R = 5 V / 0.002 A = 2,500 Ω. A nearby standard value changes the current, so recalculate with the actual part and its tolerance. This calculation assumes the resistor alone sees 5 V. If an LED or another element sits in series, subtract its operating voltage first and acknowledge that the LED drop is an approximation, not a fixed universal number.

R = Vresistor / Itarget

Where the simple equation stops helping

An LED, diode, lamp filament, transistor, or heating element need not have constant resistance. In particular, treating an LED as a fixed resistor and writing I = Vsupply / Rled is not a safe way to select its operating current. Sources can have current limits or internal resistance; component temperature and tolerance can shift measured values. Ohm’s law still applies to an explicitly modeled ohmic resistor, but it does not by itself solve a nonlinear network or make a real supply ideal.

Debug an unexpected current

If a 1 kΩ resistor does not appear to carry about 5 mA in the one-resistor example, disconnect power and check the resistor marking and continuity of the intended loop. With a properly rated meter in voltage mode, measure across the resistor under power. A reading near 0 V points to a missing supply, a bypass, or both ends on one node; a reading near 5 V with little apparent current suggests an open path or a measurement error. Do not put a meter in current mode across the source to check this.

  1. Verify the source setting and component value with power off.
  2. Trace the path from source positive, through the resistor, and back to source negative.
  3. Measure the resistor’s voltage in parallel, then compare I = V / R with a correctly wired series current measurement if needed.

Try a bounded example

Sketch a low-voltage 5 V source and one 1 kΩ resistor, label the two resistor nodes, and predict voltage, current, and dissipation before building anything. NPNLab offers browser-based electronics learning and a circuit Lab; use a simple supported resistor circuit to compare a prediction with available observations, not as certification that real parts will match an ideal calculation. Change the resistance to 2 kΩ and predict which quantities halve and which do not. Keep the physical experiment on an appropriate isolated low-voltage supply.

Build a resistor loop

Keep this recipe open while building in the Lab. The link opens the workspace; it does not import a prebuilt circuit.

Ideal predictions, not measurements: compare these assumptions with the supported model, then verify any real hardware independently.

  • R1 voltage: 5 V. Ideal 5 V source; no other loads.
  • R1 current: 5 mA. Ideal resistor: 5 V / 1 kΩ.
  • R1 power: 25 mW. Ideal DC estimate: 5 V × 5 mA.
Parts and connections
PartSetting or value
DC supply (DC1)5 V; set a suitable current limit above 5 mA
Resistor (R1, pins 1 and 2)1 kΩ, rated at least 0.25 W
NodeTerminals joined
A (+5 V)DC1 + (5V pin); R1 lead 1
B (return, 0 V)R1 lead 2; DC1 − (GND pin)
  1. Set the DC supply to 5 V and R1 to 1 kΩ; keep the source off while connecting parts.
  2. Join the terminals at A, then join the terminals at B. Keep A and B separate except through R1.
  3. Run the supported DC model and compare the resistor voltage and current with your prediction.

Check your build

  • Trace + → R1 → −; do not wire + directly to −.
  • Check voltage across R1 with a voltage probe in parallel, never a current-mode meter across the supply.
  • Compare the model observation with the ideal prediction; a real part also has tolerance and ratings.

Try one change: Change R1 to 2 kΩ. With the same ideal 5 V source, predict 2.5 mA and 12.5 mW before running again.

Sources and further reading

Diagrams are simplified teaching illustrations, not measured results or construction plans. See how these guides are prepared.