Question

2. A 33-mH inductor is directly connected to a varying current supply. a. During a “ramp up” period of 40. ms, the current through the inductor increases linearly from 0.50 to 1.00 A. What is the average emf induced across the inductor during this time period? A. 0.19 V D. 0.41 V B. 0.26 V E. 0.52 V C. 0.33 V b. The self-induced emf (across the inductor) will be... A. in the same direction as the current B. opposite the direction of the current c. What is the change in energy stored in the inductor during this time period? A. 0.18 mJ D . 12 m) B. 0.64 mJE. 48 mJ C. 2.4 m) d. How would your above answers change if the current were to decrease linearly from 1.00 A to 0.50 A instead of increase? Strength of average emf induced: A. stronger B. unchanged C. weaker Direction of emf induced: A. same as original case B. reversed from original case

Answer 1

A)

emf = L*di /dt = 33 mH * ( 1-0.5) /( 40 ms) = 33/40 *0.5 =0.4125 V D

B)

opposite

C)

ΔE = 1/2* 33* ( 1^2 - 0.5^2) =12.375 = 12 m J

d)

i ) unchanged

ii) revere from original case

****************************************************************************
Goodluck for exam Comment in case any doubt, will reply for sure..