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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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