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Main >> Applications >> Sample problems >> JouleLenz 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.
Geometry:
All dimensions are in millimeters
Given:
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 crosssection area.
Task:
Calculate the current and Joule heat inside the conductor and compare with the value given by the JouleLenz law.
Solution:
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 JouleLenz 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],
Results:
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:
Time harmonics (peak current value is presented, RMS value is √2 times smaller, 35.355/1.4142 = 25 A):
Transient electric:
Time 
Current 
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 

Heat power [W] 
JouleLenz law 
2.5 
DC conduction 
2.5 
Time harmonics 
2.5 
Transient electric (timeaverage) 
2.5 