# IND_CAP_notes - EE201.3 INDUCTANCE Resistance opposes...

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EE201.3 Denard Lynch Page 1 of 12 Revised: Oct. 1, 2008 INDUCTANCE Resistance opposes current; Capacitance opposes changes in voltage ; Inductance opposes changes in current . Recall Faraday’s Law: E M = ! d " dt for the voltage induce in a loop of wire. (Which resulted in: E m = l vel X B ) If we have N turns, it’s like multiplying the flux linkages by N, so: dt N d dt d N E M ) ( ! = ! = , and since BA = ! dt BA d N E M ) ( = ; and A is constant, so ! ) ( , define ; ) ( , constant , , ; ) ( , ; ) ( ; 2 2 dt I d L E l A N L dt I d l A N E l N dt l NI d NA E l NI H Hl NI dt H d NA E H B dt dB NA E M M M M M = = = = = = = = = μ Where the letter ‘L’ is used to represent “self inductance”, usually referred to as just “inductance”, and the unit is the Henry (H). (E.g. A rate of change of current of 1 ampere per second in a 1 Henry inductor will produce a voltage of 1 Volt across the inductor.) Now looking at the expression for Inductance, ) valid! (always (2) that so ; that recall and ; (1) 2 2 ! = = ! = N L A l l A N L For an air-core coil, the reluctance “outside” the coil is small compared to the reluctance “inside” the coil (R i >> R o ). Thus, for an air-core inductor where the length is >> than the diameter (i.e. l/d > 10), the inductance can be estimated using: 4%. typically is error the , 10 where ; 2 0 < ! = d l l A N L (Note: for l/d < 10, you can apply Nagoaka’s correction factor) V V ! R O R I

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EE201.3 Denard Lynch Page 2 of 12 Revised: Oct. 1, 2008 Now go back for a moment to: L = N 2 ! ; and " = NI = #! , (similar to V = IR) and substituting ! = NI # we are left with L = N # I (also, always valid!). Example 1: An air-core coil… find the inductance, L. Check: l/d = 12.5 – OK. Example 2: A Laminated sheet steel core (S.F. = .93) with dimensions as shown. Calculate the inductance, L. (Note: Assume we are operating in the linear region of the B-H curve). (Use a linear approximation from the B-H curves. Jackson’s range from .00135 - .0025 depending where you graph. Boylestad is always around .001 for B up to about .8) L = μ AN 2 l L = .001 ( ) .01 2 ( ) .93 ( ) 100 2 ( ) .10 ( ) L = 9.3 mH 12mm 15cm 120t 1cm X 1cm 10cm S.F. = .93 100t L = o AN 2 l L = o ( ) .006 2 ( ) ! ( ) 120 2 ( ) .15 ( ) L = 13.64 H
EE201.3 Denard Lynch Page 3 of 12 Revised: Oct. 1, 2008 Example 3: A laminated Sheet Steel ‘C’ core inductor (S.F. = .96) with an air gap cut to prevent saturation. Again, use estimate for permeability assuming it is linear. .001 fr Boylestad i) Assume total reluctance is in air gap: Consider fringing, so A = (1cm+1mm) 2 ii) Calculate the total reluctance first, including the metal 1cm X 1cm 1mm 6cm 6cm 100t L = μ o AN 2 l L = o ( ) .011 2 ( ) 100 2 ( ) .001 ( ) L = 1.52 mH ! = l core o A + l gap A ! = .239 ( ) .001 ( ) .01 2 .96 ( ) ( ) + .001 ( ) o ( ) .011 2 ( ) ! = 2.489 X 10 6 + 6.577 X 10 6 ! = 9.066 X 10 6 rels L = N 2 ! = 100 2 ( ) 9.066 X 10 6 = 1.103 mH

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EE201.3 Denard Lynch Page 4 of 12 Revised: Oct. 1, 2008 Back to the expression for voltage across an inductor: dt di L V = . For nice “straight line” (constant slope) changes in current, the calculation of voltage across an inductor is straight forward. Example 4: Determine the voltage across a 50mH inductor for the current shown in the following graph. Graph the voltage on a comparable time-scale.
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IND_CAP_notes - EE201.3 INDUCTANCE Resistance opposes...

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