Ohm's law

In a fixed loop made of one ideal source and one positive resistor, the current is set by both the voltage and the resistance: I = V / R. A larger voltage V gives a larger current I, and a larger resistance R gives a smaller current. The power P = V I = V² / R is the rate at which the resistor turns electrical energy into heat. This experiment only uses R > 0, so the current always stays finite.

Questions

In a loop with only a source and a resistor, how are the voltage, the resistance and the current related?

If the resistance stays the same and the voltage is doubled, what happens to the current?

If the voltage stays the same and the resistance is doubled, what happens to the current and to the power?

Predict

Before moving anything, predict: if the voltage V rises from 6 V to 12 V while the resistance stays 3 Ω, how many times larger does the current I become? And the power P? Write your prediction down and compare it with the experiment.

Try

Keep R = 3 Ω and set the voltage V to 3 V, 6 V, 9 V and 12 V in turn, recording the current I and the power P each time. Then fix V = 6 V and set the resistance R to 1 Ω, 2 Ω, 3 Ω and 6 Ω, recording again. Notice that the axes of the plot on the right stay at 0–12 V and 0–6 A: only the position of the operating point changes.

Observe

With R fixed, I is proportional to V: doubling V doubles I, but the power P = V I becomes four times larger. With V fixed, I is inversely proportional to R: doubling R halves I, and P = V² / R halves too. A smaller R gives a steeper line, so the current at a given voltage is larger. When the current passes 6 A the operating point leaves the fixed frame and a short hint appears.

Explain

Ohm's law

For a linear resistor the current is proportional to the voltage and inversely proportional to the resistance, I = V / R. This is the only relation in the experiment; every other readout follows from it.

Where the power comes from

Power is the product of the voltage and the current: P = V I. Substituting I = V / R gives P = V² / R. So doubling the voltage doubles the current and quadruples the power, while doubling the resistance halves both the current and the power.

Why R must stay positive

If R = 0 then I = V / R would grow without bound. That is an ideal short circuit, not a finite state of this model, so the experiment keeps R ≥ 1 Ω and the current finite. The frame is fixed at 0–6 A: beyond it the picture shows a hint instead of changing its scale.

Check

Take V = 6 V and R = 3 Ω and confirm I = 2 A and P = 12 W. Take V = 12 V and R = 6 Ω and confirm I = 2 A and P = 24 W: doubling both the voltage and the resistance leaves the current unchanged and doubles the power. Take V = 6 V and R = 1 Ω and confirm I = 6 A and P = 36 W. Check each step with both P = V I and P = V² / R.

Explore further

Two resistors in series add up to a single equivalent resistance, and the same Ohm's law still applies. You can also compare with Thin lens and images to see how different experiments share the idea of linking three quantities with one formula.