By Team Aakash Byju's | 19th January 2023
When current in a coil changes from 5 A to 2 A in 0.1 s, an average of 50 V is produced. The self-inductance of the coil is
A.
B.
C.
D.
6 H
0.67 H
1.67 H
3 H
Detailed Explanation
Note that, the current in the coil is changing from 5 A to 2 A. This means the current is decreasing. But the use of the inductor is to oppose the change in current in the loop.
So, the inductor converts its magnetic energy into electrical energy by creating a potential difference.
So, the drop in voltage across the inductor is given by
Where , V - drop in voltage = -50 V L - Self-inductance of the inductor di - change in current = 2 A - 5 A = -3 A dt - change in time = 0.1 s
V = L
di
dt
The negative sign for V indicates that there is an increase in voltage instead of dropping across the inductor. Thus,
V
-50 = L
-3
0.1
=> L =
5
3
~
1.67 H.