wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two coils have self-inductances L1 = 8mH and L2 = 4mH. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the current, the induced voltage and the energy stored in the first coil are i1, V1 and W1. Corresponding values for the second coil are i2, V2 and W2 respectively. Then:
(IIT-JEE 1994)

A
i1i2=4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
V1V2=14
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
W1W2=12
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
i1i2=14
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C W1W2=12
and di1dt = di2dt --- (1)
Also given, V1i1 = V2i2
So since V1=L1di1dt and V2=L2di2dt
We know, V1V2=L1L2 = i2i1=2

This doesn't match 3 of the options so the fourth one has to be right. Let's solve it anyway.
W1W2=0.5L1i120.5L2i22=12

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Self and Mutual Inductance
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon