You have both 45-gauge and 50-gauge wires that are 00028


Air-Coil Inductor Design

In a particular radio frequency (RF) application, you determine there is a need for a small inductor of 150 μH and rather than trying to order one and wait for it to arrive, you decide to wind it yourself. The applicable equation is

L = (r2N2 ) / (9r + 10l)

where L is the inductance in μH, r is the radius of the coil in inches, l is the length of the coil in inches, and N is the number of turns. A maximum length of 1 inch and a maximum coil diameter of 0.25 inches are required in order to fit in the space available. You have both 45-gauge and 50-gauge wires that are 0.0028 and 0.001 inches in diameter, respectively, but the thinner wire is difficult to wind without breaking. Design your coil.

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Physics: You have both 45-gauge and 50-gauge wires that are 00028
Reference No:- TGS02144652

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