Question

How fast can the 140 a current through a 0.200 h inductor be shut off if the induced emf cannot exceed 80.0 v?

Answer 1

Recall that to compute for the emf of a circuit given current and inductance, we must recall that


`emf = - M (\Delta I )/(\Delta t)`

where I is the current (A), M is the mutual inductance (h), and t is the time (ms). Since the current must not exceed 80.0 V, we have


`80.0 \geq 0.200((140)/(t))`

`t \geq (28.0)/(80)`

`t \geq 0.35`

From this, we see that it must take at least 0.35 ms so it doesn't exceed 80 V.
Answer: 0.35 ms