# What is Joule’s Law? The Joule effect.

## Joule effect

It is known as Joule effect to the phenomenon by which if in a wire circulates an electrical current, part of the energy is transformed into heat. This law (Joule’s Law) is named after the British physicist James Prescott Joule.

## Joule’s Law

Joule’s law shows the relationship that exists among

- The current that flows through a wire.
- The heat generated by the current that flows through a wire.
- The resistance of the wire and …
- The time the current passes through the wire.

Joule’s law formula: Q = I^{2} . R . t, where:

- Q is the amount of heat, in Joules (J)
- I is the electric current that flows through the wire, in amperes (A)
- R is the value of the electrical resistance of the wire, in ohms (R)
- t is the amount of time that current passes through the wire, in seconds (s).

Joule’s law can be established as the amount of heat (Q) generated in a wire with resistance (R), when a current (I) passes through it for a period of time (t).

This heat is directly proportional to:

- The square of the current.
- The resistance of wire.
- The time the current flows through the wire.

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## What are the heating effects caused by electric current?

When a current flows in a wire, thermal energy is generated in it. The heating effects caused by electric current depend on three factors:

- The resistance of the wire. Higher resistance produces more heat.
- The time that current flows through the wire. The greater the time, the greater the heat generated.
- The greater the current, the more heat generation.

## Some applications of Joule’s law

- Water heater.
- The incandescent bulb (also generates light).
- Fuse (the fuse melts, burns when the current exceeds a set limit).
- Electric iron.
- Electric stove.
- Thermistors: Thermistors are resistors whose resistance changes when the temperature changes.
- etc.

**Example:** How much heat does a 2 amp current generate in a wire with a 50 ohm resistor for 2 seconds?

We have: I = 2 amps, R = 5 ohms, t = 2 seconds.

Using the formula: J = I^{2} . R . t , we get: J = (2)^{2} x (5) x (2) = (4) (5) (2) = 40 Joules. Then: 40 Joules of heat are produced.

Notes:

- 1 joule = 0.24 calories of thermal energy
- 1 Kilowatt-hour = 1 KWh = 1 000 watt x 3 600 seconds = 3.6 x 10
^{6}joules (1 KWh = 3 600 000 joules)