Joule heating

Q=I2RtQ=I^2Rt

Result unit: J (joule)

What it describes

The heat dissipated by a current-carrying conductor (Joule–Lenz law).

Symbols

QQ
heat
[J]
II
current
[A]
RR
resistance
[Ω]
tt
time
[s]

Where it comes from

Current flowing through a resistor dissipates heat proportional to the square of the current — that is why overloaded wires heat up. It is the operating principle of heaters and fuses.

When to use it

Heaters, resistive heating, energy losses in wires.

Example

A current of 2 A flows through a 10 Ω resistor for 30 s. Calculate the heat dissipated.
SubstitutionI=2I=2 A, R=10R=10 Ω, t=30t=30 s.
ResultQ=22⋅10⋅30=1200Q=2^2\cdot10\cdot30=1200 J.

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