The maximum current is 3.00 A at 40.0 Hz. Calculate the inductance L.

+1 vote

In a purely inductive AC circuit as shown in the figure, ΔVmax = 100 V.

image
(a) The maximum current is 3.00 A at 40.0 Hz. Calculate the inductance L.
 H

(b) At what angular frequency ω is the maximum current 3.50 A?
 rad/s
asked Mar 19, 2012 in Physics by yoshi ~Expert~ (919 points)
    

2 Answers

+3 votes

Imax = Vmax / R   (*where R = inductive resistance XL)

XL = ω*L     (*where XL = inductive reactance , ω = angular frequency, L = inductance)

ω = 2*pi*f       (*where f = frequency)

plugging ω  back into equation for inductive reactance we get:

XL = (2*pi*f) *L  

Then plug this back in for in the first equation for Imax  and solve for L:

Imax = Vmax / (2*pi*f) *L  

so

L = Vmax / Imax (2*pi*f)

 

Part B:

you can find  ω using the L from above and the previous equation:

L = Vmax / Imax (2*pi*f)

ω =  (2*pi*f)

so

L = Vmax / Imax*ω

solving for ω

ω = Vmax / L*Imax   (*where Imax is the new Imax given in Part B)

answered Mar 24, 2012 by pokemonmaster ~Expert~ (3,856 points)
edited Mar 25, 2012 by pokemonmaster
+2 votes
A.) For this circuit we have:

XL=V/I

XL=100V/3.00A

XL=33.33[omega]

XL=2[pie] f L

L=XL/2[pie] f

L=33.33[omega]/2[pie](40.0)Hz

L=132.6mH

B.) This time we have

XL=V/I

XL=100V/35A

XL=2.85 [omega]

XL=2[pie] f L

f=XL/2[pie]L

f=2.85/2[pie](132.6E^-3)

f=3.42Hz
answered Mar 20, 2012 by oboebabe21 ~Rookie~ (105 points)
I believe Part B is looking for angular frequency ω rather than f.

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