Question

When the \(20.0{\rm{ }}A\) current through an inductor is turned off in \(1.50{\rm{ }}ms\), an \(800{\rm{ }}V\) emf is induced, opposing the change. What is the value of the self-inductance?

Short answer

The value of the self-inductance \(60{\rm{ }}mH\).

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:\(20.00{\rm{ }}A\)
• The time value:\(1.50{\rm{ }}ms\)
• The emf value: \(800{\rm{ }}V\)

Calculating the Value of Self-Inductance

The electromotive force is provided by when current changes through a solenoid.

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

As a result, we may solve for a solenoid's inductance as follows:

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

Inourcase,numerically,wewillhave

\(\begin{array}{c}L = \frac{{800 \times 0.0015}}{{20}}\\ = 0.06{\rm{ }}H\end{array}\)

Therefore, the required solution is \(60{\rm{ }}mH\).