Question

In a purely inductive AC circuit as shown in the figure, Delta V_max = 100 V. The maximum current is 8.00 A at 48.0 Hz. Calculate the inductance L. At what angular frequency omega is the maximum current 2.90 A?

Answer 1

a) 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 R  in the first equation for Imax  and solve for L:

Imax = Vmax / (2*pi*f) *L  

so, L = Vmax / Imax (2*pi*f)= 100/8(2*3.14*48)=0.041 H= 41 mH approx.

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

ω= 100/0.041*2.90= 841.04 rad/s