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Chapter 30: Q90P (page 902)

How long would it take, following the removal of the battery, for the potential difference across the resistor in an RL circuit (with L = 2.00H, R = 3.00) to decay to 10.0% of its initial value?

Short Answer

Expert verified

The time required for the current to decay to 10% of its initial value is t = 1.54s

Step by step solution

01

Given

L=2.00HR=3.00i=0.1i0

02

Understanding the concept

When the source of constant emf is removed, the current decay from a valueaccording to the equation of decay of current given by 30.45. So we can use that equation to find the time.

Formula:

i=i0e-t/τLτL=LR

03

Calculate the time required for the current to decay to 10% of its initial value.

We have, the equation of decay of current as,

i=i0e-tτLii0=e-tτLi0i=etτL

Taking logofboththe sides

lni0i=lne-tτLlni0i=tτL

Where τL=LRandi=0.1i0

lnlni00.1i0=tRLt=LRlnln(10)t=2.003.00lnln(10)t=1.54s

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Most popular questions from this chapter

In Figure (a), a uniform magnetic field increases in magnitude with time t as given by Figure (b), where the vertical axis scale is set by Bs=9.0mTand the horizontal scale is set by ts=3.0sA circular conducting loop of area 8.0×10-4m2lies in the field, in the plane of the page. The amount of charge q passing point A on the loop is given in Figure (c) as a function of t, with the vertical axis scale set by qs=3.0s and the horizontal axis scale again set by localid="1661854094654" ts=3.0s. What is the loop’s resistance?

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