Final answer:
To determine how quickly a current through an inductor can be shut off, we use Faraday's law of induction. For a 150 A current through a 0.250 H inductor, the current can be shut off in 0.500 seconds without the induced emf exceeding 75.0 V.
Step-by-step explanation:
Calculating the Maximum Rate to Shut Off a Current Through an Inductor
To answer how fast the 150 A current through a 0.250 H inductor can be shut off without exceeding a 75.0 V induced emf, we will use Faraday's law of induction. This law informs us that the induced electromotive force (emf) in an inductor is equal to the negative change in magnetic flux over time. Since an inductor's magnetic flux is directly proportional to the current flowing through it, we can write Faraday's law as:
E = -L (dI/dt)
where E is the induced emf, L is the inductance, dI is the change in current, and dt is the change in time. Rearranging for dt, the time to shut off the current without exceeding the maximum emf, we have:
dt = -L (dI/E)
Plugging in the given values:
dt = -0.250 H (-150 A / 75.0 V) = 0.500 seconds
Therefore, the current can be shut off in 0.500 seconds without exceeding the 75.0 V induced emf limit.