Question

At t = 0, the current through a 65.0 mH inductor is 52.0 mA and is increasing at the rate of 100 mA/s. What is the initial energy stored in the inductor?
________8.788e-5____________ J

How long does it take for the energy to increase by a factor of 10 from the initial value?
_________________ s

Help for the second part please, if you could show an equation that would be great! Thanks!

Answer 1

Formula for energy stored in inductor is, E=112 Here. L is inductance and Iis current Substitute the given values in above equation E-L(65x103 H(52x10-3 A -8.78x10- J Calculate the energy increased by factor 10, E-(8.78 × 10-5 J)10=87.8x 10-5 J New current in inductor is, 0.164 A 2 65 x10-3 HH Thus, the difference in current is given as, d =164 mA-52 mA =112 mA Given that the current increases by 100 mA per a second and the time within that the current by 112 mA as well as the energy increased is given as, (100 mA)t=112 mA 112 mA 100 mA =1.125

Answer 2

E=½LI²

Where E=energy
L=Inductance
I=current

So finding the energy isn't hard.

To find the time for part b, first find the current needed for 10 times the enrgy (also using E=½LI², but this time with 10 times the energy).

Then, if we assume that the rate of increase is constant (seems to be true from the problem statement) finding the time is a matter of dividing the current difference by the rate of change (the 100 mA/s).