I was trying to simulate two serpentine resistors on COMSOL multiphysics with exactly the same length, width and thickness. I drew them on AutoCAD. The only difference between the two resistors is the number of bends. For resistor(1), number of bends is only 2, while for resistor(2), number of bends is 18.

After running simulation on AC/DC module and Electric Currents (ec) sub module, resistance values came as the following:

Resistor(1) = 13.221 ohm (2 number of bends)

Resistor(2)= 12.654 ohm (18 number of bends)

I expected that the number of bends will not affect the resistance value, but it actually does in an inverse manner (though not much, but it does)

Is anyone aware why would the number of bends affect the resistance value?

Simulation parameters:

Resistor(1):

Length:337mm; Width: 1mm; Thickness: 0.2mm; Number of bends:2

Resistor(2):

Length:337mm; Width: 1mm; Thickness: 0.2mm;Number of bends:18

Meshing:

Resolution: Extra fine.

Minimum Element Size: 0.18mm.

Maximum Element Size: 4.2mm.

Maximum Element Growth Rate: 1.35

Curvature Factor:0.3

Resolution of Narrow Regions: 0.8

Effect of meshing resolution on resistance values:

## Best Answer

You are measuring the length through the center, but the current will not follow that path, rather it will crowd into the corners and, on average, take a somewhat shorter path through each bend, so your total resistance will be less for the serpentine pattern with more bends. Below is a field solver simulation showing a current density at the inside corner that is 5x higher than the average in the straightaways.

One rule of thumb is that a corner square is equivalent to 0.56 squares rather than one square. Your first example has 4 corners, the second 36.

If I back-calculate the effect from the ratios in your simulation I get 0.55 as the effect of one square, which is pretty close.