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Main > Application > Sample problems

Ohm's 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

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 Ohm's 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 Ohm's law* the current is determined as:
I = V / R [A],
where the conductor resistance is R = (1/g) * (L/A)

Joule heat Q = R*I2

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:



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 heat [W]

Ohms' law


DC conduction


Time harmonics


Transient electric (time-average)


*Wikipedia: Ohm's law.

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