66. II FIGURE P30.66 shows the current through a10mH inductor. Draw a graph showing the potential differenceΔVL across the inductor for these6ms .

Short Answer

Expert verified

The potential difference across the inductor is

ΔVL=10Vfor0ms<t<2msΔVL=0V           for2ms<t<3msΔVL=20V         for3ms<t<5msΔVL=0V                    fort>5ms

Step by step solution

01

Step 1. Introduction

It is given that the value of inductor,L=10mH=10×103H

We have potential difference across the inductor asΔVL=LdIdt

02

Step 2. Explanation

From t=0msto t=2ms, current Ichanges from 0 A to 2A.

dIdt=(2A0A)(2ms0ms)=2A(2ms)103s/ms=1000A/sΔVL=10×103H(1000A/s)=10V

Step 3.

From t=2msto role="math" t=3ms,role="math" Iremains constant

Therefore voltage becomes zero.

Step 3.

Now from t=3msto 5ms, the current decreases from 2Ato 2A

dIdt=(2A2A)(5ms3ms)=4A(2ms)103s/ms=2000A/s

So, the inductor voltage is

ΔVL=110×103H(2000A/s)=20V

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