Question

A 1.85 H inductor carries a steady current of 0.450 A. When the switch in the circuit is thrown open, the current is effectively zero in 10.0 ms. What is the average induced emf in the inductor during this time?

Answer 1

Answer:i

Induced emf will be equal to 83.25 volt

Step-by-step explanation:

We have given value of inductance L = 1.85 H

Initially the current flow is equal to 0.450 A

When switch is open current become 0 in 10 ms

So change in current di = 0.450-0=0.450 A

Time taken to change the current dt = 10 ms = 0.01 sec

Induced emf is equal to
`e=L(di)/(dt)`

So
`e=1.85* (0.45)/(0.01)=83.25volt`

Induced emf will be equal to 83.25 volt