An engineer is asked to design a 10mH inductor. They design an inductor with the following parameters 500 turns of 30 AWG wire, the lengh of the coil i 1.9 cm and the diameter o the coil i 1 cm, and the core is ferrite with a permeability of 20 micro H/m. What i the actual impedance o the coil design?Also, which statement would most correctly describe how you would fix the design to the inductance of the inductor is 10mH? Why?A) Double the diameter of the coil to 2cmB) Halve the number o turns to 250C) Decrease the number o turns to 350D) Increase the number o turns to 650E) Halve the diameter of the coil to .5 cm
** C) Decrease the number of turns to 350. What is he actual INDUCTANCE, not impedence, which would only occur at a specific frequency, ok. Using the formula or calculator at source, gives inductance for values as 20.667mH. Now adjust design to yield 10 mH . Because the inductance is proportional to the square of the number of turns. So 500? 250000 Make this half 125,000 Get √125,000 353.55, Which is close enough to 350, Answer Check on calculator to confirm. Please request clarification, if required.
