Question

A 0.660-mH inductor and a 120-Ω resistor are placed in series with a 24-V battery. What is the maximum energy stored in the inductor?

Answer 1

Given is:-

H OT X 099 0 = 1

R= 12022

€= 241

Now,

the maximum current in the ckt is given by

Imar = R = 12022

which gives us

Imar = 0.2A

Now,

the maximum energy stored in the inductor is given by

Umar = -LIMAR

by plugging all the values we get

Umar = }(0.660 x 10-3H)(0.24)

which gives us

Umar = 13.2 x 10-6

or

Umar = 13.2uJ