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 

$emf = - M \frac{\Delta I }{\Delta t}$

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

$80.0 \geq 0.200(\frac{140}{t})$
$t \geq \frac{28.0}{80}$
$t \geq 0.35$

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