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Community > Groups > Component Learning > Building my own load using wire wound resistors
Building my own load using wire wound resistors
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Ben

Does anybody have any tips or suggestions on how to create a noninductive load resistor (or low-inductance load resistor) using cheap 10W wirewound resistors in cement block packages?
I am thinking I'll put N of them in parallel (reducing inductance by a factor of N) and then arrange the current paths to be in opposite directions. This will create a mutual inductance between adjacent resistors, and since the currents flow in opposite directions, the effective inductance will be reduced to (L - M) where L is the self inductance and M is the mutual inductance.
I just purchased qty=28 of 10 watt, 56 ohm resistors. I'll put 14 of them in parallel (giving a 4.0 ohm resistor), and connect that in series with another set of 14 ohms in parallel. End-to-end resistance is 8.0 ohms, and inductance is (hopefully) a lot less than (1/14th) the inductance of a single resistor.
Figure 1 below shows a schematic of my connection. Figure 2 illustrates the mutual inductance between a pair of adjacent resistors. Of course it's a distributed system but for compactness I've drawn it as two "lumps".
What do you think of the idea? Is there a far, far better way to get low end-to-end inductance when using wirewound resistors?
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