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Experiment 002 / Essential electronics

Ohm’s law

Relate voltage, current, resistance and power. Two known values make the circuit easier to understand.

Workbench / Ideal model ● In your browser
01 / Adjust the inputs

Sliders with a wide range use a logarithmic scale. When zero is allowed, it has its own position. Type a number to set an exact value.

Try an example
02 / Connect the quantities Ohmic resistor
+ − 5 V R I →
Voltage 5 V
Current 5 mA
Resistance 1 kΩ
Power 25 mW
I = V / R = 5 / 1000 = 0.005 A
P = V × I = 5 × 0.005 = 0.025 W

Nonnegative quantities. Resistance must be positive and finite.

Use it in your classroom

Share this exercise with its current values. The link contains only numbers and model options; do not add names or email addresses.

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Five volts across one kilohm

Connect an ideal 1 kΩ resistor across 5 V. Current is 5 / 1000 = 0.005 A, or 5 mA. The supply delivers 5 × 0.005 = 0.025 W, which becomes heat in the resistor.

Keep 5 V and double the resistance: current and power halve. Double voltage at fixed resistance: current doubles and power quadruples.

Two relationships, six possible pairs

V = I × R   ·   P = V × I
I = V / R   ·   R = V / I
P = I² × R = V² / R
V = √(P × R)   ·   I = √(P / R)

Choose the two quantities you know. The workbench shows the relevant formulas and substitutes values in volts, amperes, ohms and watts. Changing a field’s unit preserves its physical quantity.

Zero does not always determine an answer

The tool assumes a positive, finite ohmic resistor and nonnegative quantities. With R = 100 Ω and V = 0 V there is a unique solution: I = 0 A and P = 0 W. Knowing only V = 0 and I = 0 leaves R undetermined: any positive resistance fits.

With V = 0 and a positive current, this model has no solution. Ideal short circuits, infinite open-circuit resistance, active components and negative resistances are excluded. AC and impedance are not modeled.

Common mistakes

  • Applying V = IR to a diode as if R were constant.
  • Confusing power with current: a 1 W resistor does not always consume 1 W.
  • Mixing mA and A in a substitution without converting first.
  • Treating 0 / 0 as zero instead of recognizing an undetermined result.

Try it yourself

A 100 Ω resistor dissipates 0.25 W. What are V and I?

I = √(0.25 / 100) = 0.05 A. V = 0.05 × 100 = 5 V. Reproduce this using the power example.

Can you find R from V = 0 V and P = 0 W?

Not uniquely. Every positive, unpowered resistance satisfies those two values. You need another independent quantity, such as R.

Reference: Ohm’s law, OpenStax. This page’s explanations and exercises are original.