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Main >> Applications >> Sample problems >> Joule-Lenz law

Joule-Lenz law

This verification example compares the current and the Joule heat generated in conductor calculated by the Ohm's law and Joule's law, and calculated in QuickField using three formulations: DC conduction, AC conduction and Transient electric.

All dimensions are in millimeters

Ohm law verification with FEM

V = 0.1 V - voltage applied;
f = 50 Hz - frequency in time harmonics and transient magnetics problems;
g = 1e6 S/m - conductivity of conductor material.
L = 0.4 m - conductor length.
A = 0.02*0.005 m² - conductor cross-section area.

Calculate the current and Joule heat inside the conductor and compare with the value given by the Joule-Lenz law.

To maintain the same value of Joule heat across all formulations the voltage value is adjusted:
In the time harmonics problem peak amplitude value of voltage is set √2·V.
In the transient electric voltage is set via formula V(t) = √2·V · sin(2·180·50·t).

According to the Joule-Lenz law* the power of heating generated by an electrical current I:
W = R*I², where the conductor resistance is R = (1/g) * (L/A).
The electric current value could be calculated I = V / R [A],

*Wikipedia: Joule heating.

Conductor resistance R = (1/1e6) * (0.4/0.02*0.005) = 0.004 Ohm
Current: I = 0.1 / 0.004 = 25 A
Joule heat Q = 0.004 * 25 * 25 = 2.5 W.

DC conduction:

Joule-Lenz law direct current

Time harmonics (peak current value is presented, RMS value is √2 times smaller, 35.355/1.4142 = 25 A):

Joule-Lenz law alternating current

Transient electric:



Joule heat

0.01 s

0 A

0 W

0.0125 s

25 A

2.5 W

0.015 s

35.355 A

5 W

Joule-Lenz law transient magnetic analysis


Heat power [W]

Joule-Lenz law


DC conduction


Time harmonics


Transient electric (time-average)


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    QuickField 6.4