2.3 Ohm’s Law

Use voltage, current, and resistance to predict circuit behaviour.

The relationship

Ohm’s law connects voltage, current, and resistance. [1, pp. 76-84]

I = V / R

Rearranged:

V = I × R

R = V / I

Where:

  • V is voltage in volts
  • I is current in amperes
  • R is resistance in ohms

What the relationship means

For a resistive load:

  • increasing voltage while resistance stays constant increases current
  • increasing resistance while voltage stays constant decreases current

We will introduce resistor components in detail when we actually start using them. For now, resistance is the idea of opposition to current.

A practical computer example

Suppose a USB device is operating at:

  • 5.00 V
  • 0.900 A

Its equivalent resistance at that operating point is:

R = V / I

R = 5.00 V / 0.900 A

R = 5.56 Ω

Both measured values have 3 significant figures, so the final result is reported with 3 significant figures.

NoteA real USB device is not just a resistor

Digital electronics can change how much current they draw as they operate. The calculated resistance is an equivalent resistance at that operating point, not necessarily a fixed physical resistor inside the device.

Open and short circuits are limits of the same model

Ohm’s law helps explain the two failure cases from the previous lesson:

  • open circuit: resistance becomes extremely large, so current approaches zero
  • short circuit: resistance becomes very small, so current can become very large

The source and wiring always have some real resistance, so practical short-circuit current is finite.

Significant figures still apply

Use the number-system rules from Unit 1:

  • keep full calculator precision during the calculation
  • for multiplication or division, round the final answer to the fewest significant figures in the inputs
  • round once, at the end

Practice

  1. A 12.0 V load has an effective resistance of 48.0 Ω. Find the current.
  2. A device draws 0.250 A from a 12.0 V supply. Find its equivalent resistance.
  3. A circuit draws 2.00 A through a 5.00 Ω load. Find the voltage.
  4. Explain why a short circuit can create much more current than a normal load.
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