Final answer:
The current through a 0.250 H inductor can be safely shut off in 0.50 seconds without exceeding the maximum allowed induced emf of 75.0 V, which does not correspond to any of the provided options (a) through (d).
Step-by-step explanation:
The speed at which a 150 A current through a 0.250 H inductor can be shut off without exceeding a 75.0 V induced emf can be determined using Faraday's law of induction. Faraday's law states that the magnitude of the induced emf in a coil is equal to the rate of change of magnetic flux through the coil. The formula for the induced emf is № = -L * (ΔI/Δt) where № is the induced emf, L is the inductance, ΔI is the change in current, and Δt is the change in time. Thus, the maximum rate of change of the current that does not exceed the emf limit can be found by rearranging the equation to solve for Δt: Δt = -L * (ΔI/№).
By substituting the given values into the formula, we get: Δt = (0.250 H * 150 A)/ 75 V = 0.50 seconds. Therefore, we can see that the current through the inductor can be turned off in 0.50 seconds without the induced emf exceeding 75 volts, meaning none of the provided options are correct nor close enough to the calculated value.