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

A 3.56H inductor is placed in series with a 12.8Ω resistor. An emf of 3.24V is then suddenly applied across the RL combination. At 0.278s after the emf is applied what is the rate at which energy is being delivered by the battery?

Open in App
Solution

τL=LR=3.5612.8=0.278s
i0=VR=3.2412.8=0.253A
After one time constant (t=0.278s=τC)
i=(11e)i0=0.63i0=0.16A
Power supplied by battery =Ei
P=(3.24×(0.16)=0.518W

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Series and Parallel Combination of Capacitors
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon