Question

How fast can the \(150{\rm{ }}A\) current through a \(0.250{\rm{ }}H\) inductor be shut off if the induced emf cannot exceed \(75.0{\rm{ }}V\)?

Short answer

The time in which current through a inductor can be shut off is \(500\;ms\).

Step-by-step solution

Concept Introduction

A passive electrical component that dampens current variations is an inductor. Other names for inductors are coils and chokes. In electrical nomenclature, the letter \(L\) stands in for an inductor

Information Given

• The current value:\(150{\rm{ }}A\)
• The inductance value:\(0.250{\rm{ }}H\)
• The emf value: \(75.0{\rm{ }}V\)

Calculating the Force

Whenthecurrentinaninductorchange,weknowthattheinducedelectromotiveforceisgivenby

\(\varepsilon = \frac{{L\Delta I}}{{\Delta t}}\)

Asaresult,whentheotherfactorsareknown,wecansolveforthetime.

\(L = \frac{{\varepsilon \Delta t}}{{\Delta I}}\)

Numerically, we will have

\(\begin{array}{c}\Delta t = \frac{{0.25 \times 150}}{{75}}\\ = 0.5\;s\end{array}\)

Therefore, the required solution is \(500\;ms\).